text: This is summarized in the following chart: The above table can be used to conclude that boron (a Group III element) will typically have an oxidation state of +3, and nitrogen (a group V element) an oxidation state of -3. Ions containing cerium in the +4 oxidation state are oxidizing agents, capable of oxidizing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). As stated in rule number four above, the sum of the oxidation states for all atoms in a molecule or polyatomic ion is equal to the charge of the molecule or ion. In List of oxidation states of the elements Per element: {{List of oxidation states of the … Wikipedia Oxidation states of p block elements Some p block elements have lot of oxidation numbers such as chlorine (-1, 0, +1, +3, +5, +7) and sulfur (-2, 0, +4, +6). There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia. The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion. Oxidation results in an increase in the oxidation state. This is the most common function of oxidation states. The more electronegative element in a substance is given a negative oxidation state. What is the oxidation state of chromium in Cr2+? It looks at the increasing tendency of the elements to form compounds in which their oxidation states are +2, particularly with reference to tin and lead. In an ionic compound, the potassium's oxidation state is always +1. Negative Oxidation States. Start with oxygen, which is more electronegative than any element except fluorine. For example, the charge on the nitrogen atom in ammonium ion NH4+ is 1+, but the formal oxidation state is -3—the same as it is for nitrogen in ammonia. ); therefore, the ion is more properly named the sulfate(VI) ion. The algebraic sum of the oxidation numbers of elements in a compound is zero. This applies regardless of the structure of the element: Xe, Cl 2, S 8, and large structures of carbon or silicon each have an oxidation state of zero. Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference. Oxidation state 0 occurs for all elements – it is simply the element in its elemental form. In ions, the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion. This page explores the oxidation states (oxidation numbers) shown by the Group 4 elements - carbon (C), silicon (Si), germanium (Ge), tin (Sn) and lead (Pb). The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1. Therefore, the oxidation state of the cerium must decrease by 4 to compensate. For example, in a sulfite ion (SO32-), the total charge of the ion is 2-, and each oxygen is assumed to be in its usual oxidation state of -2. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0). As the table shows, the presence of the other oxidation states varies, but follows some patterns. The common oxidation states of all of the metals in the periodic table are all positive. Wiktionary In the process of transitioning to manganese(II) ions, the oxidation state of manganese decreases by 5. And so their oxidation state is typically negative 2-- once again, just a rule of thumb-- or that their charge is reduced by two electrons. Because there are three oxygen atoms in sulfite, oxygen contributes $3\times-2=-6$ to the total charge. The sulfite ion is SO32-. However, its transition is more complicated than previously-discussed examples: it is both oxidized and reduced. The oxidation state of the sulfur is +4. Wiktionary Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. This type of reaction, in which a single substance is both oxidized and reduced, is called a disproportionation reaction. Each time the vanadium is oxidized (and loses another electron), its oxidation state increases by 1. In this case, it is probable that the oxygen will end up in water, which must be balanced with hydrogen. Rank these species by their ability to act as an oxidizing agent. The change in oxidation state of an element during a reaction determines whether it has been oxidized or reduced without the use of electron-half-equations. The NaCl chlorine atom is reduced to a -1 oxidation state; the NaClO chlorine atom is oxidized to a state of +1. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Oxidation states are typically represented by integers, which can be positive, negative, or zero. CC BY-SA 3.0. http://en.wiktionary.org/wiki/oxidation The reaction between chlorine and cold dilute sodium hydroxide solution is given below: $\ce{2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O} \nonumber$. It denotes the electrons gained by a substance when it is reduced, and the electrons lost when a substance is oxidized. Predict the oxidation states of common elements by their group number. Here are two ways of approaching this problem: You will have come across names like iron(II) sulfate and iron(III) chloride. Have questions or comments? [ "article:topic", "vanadium", "oxidation numbers", "authorname:clarkj", "showtoc:no", "oxidation states" ], Former Head of Chemistry and Head of Science, 2+ ion will be formed from vanadium metal by, . If we were to go right over here to the Group 5 elements, typical oxidation state is negative 3. Cl2 + 2Li rightarrow 2LiCl Oxidation half-reaction: Cl2 + 2e- rightarrow 2Cl- Reduction half-reaction: 2LI-2e- rightarrow 2Li + Use the symbol e- for an electron. For some elements, the oxidation number is just another way of stating what its most stable cation or anion will be. The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion. Reduction results in a decrease in the oxidation... Unpaired Electrons of d-orbitals. (adsbygoogle = window.adsbygoogle || []).push({}); Oxidation state indicates the degree of oxidation for an atom in a chemical compound; it is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Clearly, each atom in H 2, Cl2, P4, Na, Al, O2, O3, S8, and Mg, has an oxidation number zero. The remaining atoms and the charges must be balanced using some intuitive guessing. This periodic table contains the oxidation numbers of the elements. What is the oxidation state of chromium in the dichromate ion, Cr2O72-? If the oxidation state of one substance in a reaction decreases by 2, it has gained 2 electrons. This applies regardless of the structure of the element: Xe, Cl. Legal. Elements in the middle are determined by conservation of charge. http://apchemcyhs.wikispaces.com/ The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. This helps determine the oxidation state of any one element in a given molecule or ion, assuming that we know the common oxidation states of all of the other elements. The concept of oxidation states expresses the combining power exercised by elements in their compounds and is thus related to their valences. If electrons are added to an elemental species, its oxidation number becomes negative. Separate this redox reaction into its component half-reactions. Here the hydrogen exists as a hydride ion, H-. Oxidation states can be useful in working out the stoichiometry for titration reactions when there is insufficient information to work out the complete ionic equation. … Oxidation state (or oxidation number) refers to the number of electrons added to or removed from an element when it forms a chemical compound. The oxidation state of the molybdenum increases by 4. Let n equal the oxidation state of chromium: What is the oxidation state of chromium in Cr(H2O)63+? The 2+ ion will be formed from vanadium metal by oxidizing the metal and removing two electrons: $\ce{V \rightarrow V^{2+} + 2e^{-}} \label{1}$. Assigning oxidation numbers to organic compounds There are seven elements that can achieve +8 (Ru, Xe, Os, Ir, Pu, Cm and Hs). The compounds in which cobalt exhibits the +2 oxidation state (Co 2+, the ion being stable in water) are called cobaltous, while those in which cobalt exhibits the +3 oxidation state (Co 3+) are called cobaltic. For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion). The exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals (such as LiH), and an oxidation state of −1 in peroxides (such as H. The algebraic sum of oxidation states for all atoms in a neutral molecule must be zero. This ion is more properly named the sulfate(IV) ion. This table also contains the element number, element symbol, element name and atomic weights of each element. The oxidation state of the sulfur is +6 (work it out! The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. Oxidation states simplify the process of determining what is being oxidized and what is being reduced in redox reactions. When oxygen is in its elemental state (O 2), its oxidation number is 0, as is the case for all elemental atoms. The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. Alternatively, the sum of the oxidation states in a neutral compound is zero. CC BY-SA 3.0. http://en.wikipedia.org/wiki/Oxidation_state, http://simple.wikipedia.org/wiki/File:Plutonium_in_solution.jpg, https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/. In other cases, it is not so obvious, so we have rules. This is not a redox reaction. Therefore, there must be 4 cerium ions involved for each molybdenum ion; this fulfills the stoichiometric requirements of the reaction. Assume carbon atom has x oxidation state , and we know that Nitrogen has regular -3 oxidation state and oxygen has -2 oxidation state. The reaction between sodium hydroxide and hydrochloric acid is: None of the elements are oxidized or reduced. It is probable that the elemental chlorine has changed oxidation state because it has formed two ionic compounds. Cerium is reduced to the +3 oxidation state (Ce3+) in the process. The oxidation state of a pure element is always zero. The less electronegative element is assigned a positive oxidation state. Oxidation state is a number assigned to an element in a compound. Element Symbol. These elements are termed as chalcogens/ ore-forming elements. For example, Cl, When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2. It is possible to remove a fifth electron to form another the $$\ce{VO_2^{+}}$$ ion with the vanadium in a +5 oxidation state. The algebraic sum of the oxidation states in an ion is equal to the charge on the ion. reductionthe gain of electrons, which causes a decrease in oxidation state, oxidationthe loss of electrons, which causes an increase in oxidation state. The oxidation state is +3. Oxidation states are typically represented by integers which may be positive, zero, or negative. The oxidation state of the manganese in the manganate(VII) ion is +7, as indicated by the name (but it should be fairly straightforward and useful practice to figure it out from the chemical formula). General Rules Regarding Oxidation States The oxidation state of a free element (uncombined element) is zero. Any oxidation state decrease in one substance must be accompanied by an equal oxidation state increase in another. This would be essentially the same as an unattached chromium ion, Cr3+. Use oxidation states to work out the equation for the reaction. This can also be extended to negative ions. Do not confuse the formal charge on an atom with its formal oxidation state, as these may be different (and often are different, in polyatomic ions). Question b) According to oxidation rules, we would first assign the oxidation state of -1 to each Cl as it is a group 7A element. $\ce{VO^{2+} + H_2O \rightarrow VO_2^{+} + 2H^{+} + e^{-}}$. Now, we could keep going. The less electronegative one is given a positive oxidation state. An atom’s increase in oxidation state through a chemical reaction is called oxidation, and it involves a loss of electrons; an decrease in an atom’s oxidation state is called reduction, and it involves the gain of electrons. In the process, the manganate(VII) ions are reduced to manganese(II) ions. The oxidation number or the oxidation states shows the number of electrons that an atom uses or receives when it forms a compound. In its compounds cobalt nearly always exhibits a +2 or +3 oxidation state, although states of +4, +1, 0, and −1 are known. Notice that the oxidation state is not always the same as the charge on the ion (true for the products in Equations \ref{1} and \ref{2}), but not for the ion in Equation \ref{3}). Therefore, sulfur must have an oxidation state of +4 for the overall charge on sulfite to be 2-: $(+4-6=-2).$. The positive oxidation state is the total number of electrons removed from the elemental state. This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero. Every reactive iron(II) ion increases its oxidation state by 1. The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state. The oxidation state of hydrogen has decreased—hydrogen has been reduced. So these are typically reduced. Here is a more common example involving iron(II) ions and manganate(VII) ions: A solution of potassium manganate(VII), KMnO4, acidified with dilute sulfuric acid oxidizes iron(II) ions to iron(III) ions. Bold numbers represent the more common oxidation states. This is an ion and so the sum of the oxidation states is equal to the charge on the ion. This is the reaction between magnesium and hydrogen chloride: $\ce{Mg + 2HCl -> MgCl2 +H2} \nonumber$. Oxidation States of Transition Metals Introduction. Values in italics represent theoretical or unconfirmed oxidation numbers. Iron(II) sulfate is FeSO4. Solution for Explain how the oxidation state of group 17 elements affect the acidic strength of their compound The ion could be reduced back to elemental vanadium, with an oxidation state of zero. The oxidation number of a Group 1 element in a compound is +1. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. CC BY-SA 3.0. http://simple.wikipedia.org/wiki/File:Plutonium_in_solution.jpg Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The sum of the oxidation states in the attached neutral molecule must be zero. The vanadium in the $$\ce{V^{2+}}$$ ion has an oxidation state of +2. Element. The sulfate ion is SO42-. This particular resource used the following sources: http://en.wiktionary.org/wiki/reduction Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. ); therefore, the ion is more properly named the. In general, hydrogen has an oxidation state of +1, while oxygen has an oxidation state of -2. The oxidation state of an atom is equal to the total number of electrons which have been removed from an element (producing a positive oxidation state) or added to an element (producing a negative oxidation state) to reach its present state. The charge on the polyatomic ion is −1. The fully balanced equation is displayed below: $MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} \nonumber$. Using oxidation states to identify what has been oxidized and what has been reduced, Using oxidation states to determine reaction stoichiometry, information contact us at info@libretexts.org, status page at https://status.libretexts.org, oxidation and reduction in terms of electron transfer, The oxidation state of an uncombined element is zero. FeSO4 is properly named iron(II) sulfate(VI), and FeSO3 is iron(II) sulfate(IV). In the case between ammonium and ammonia, the formal charge on the N atom changes, but its oxidation state does not. We can work from the above rule to find the unknown oxidation state. In CNO- ion here, we can calculate the oxidation state of Carbon atom. This is because halides and oxides are the most electronegative elements and can only stabilize the higher oxidation states The elements that form halides are T i , V , C r , M n and F e The type of halides formed are T i X 4 , V X 5 , C r X 6 , M n X 4 , F e X 3 , C o X 3 etc This number describes oxidation-reduction reactions, and balancing redox chemical reactions. CC BY-SA 3.0. http://en.wikipedia.org/wiki/Oxidation_state Checking all the oxidation states verifies this: Chlorine is the only element to have changed oxidation state. If the process is reversed, or electrons are added, the oxidation state decreases. What is the oxidation state of copper in CuSO4? Keep in mind that oxidation states can change, and this prediction method should only be used as a general guideline; for example, transition metals do not adhere to any fixed rules and tend to exhibit a wide range of oxidation states. Element. These rules provide a simpler method. All the elements of Group 17 form compound in odd oxidation states (-1, +1, +3, +5, +7) but down the group importance of the higher oxidation states generally decreases. But in its peroxides like Na2O2  and H2O2, it has -1 as its oxidation state What is the oxidation state of chromium in CrCl3? Positive Oxidation States. The vanadium in the $$\ce{V^{3+} }$$ ion has an oxidation state of +3. Sodium metal, for example, has an oxidation state of 0 in the elemental state. When oxygen is part of a peroxide, its oxidation number is -1. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. Each time an oxidation state changes by one unit, one electron has been transferred. The alkali metals (group I) always … Oxygen has an oxidation state of -2. However, for the purposes of this introduction, it would be useful to review and be familiar with the following concepts: To illustrate this concept, consider the element vanadium, which forms a number of different ions (e.g., $$\ce{V^{2+}}$$ and $$\ce{V^{3+}}$$). There are 3 Cl ions, so the sum of them would be -3. The highest known oxidation state is +8 in the tetroxides of ruthenium, xenon, osmium, iridium, hassium, and some complexes involving plutonium; the lowest known oxidation state is −4 for some elements in the carbon group. Libretexts content is licensed by CC BY-NC-SA 3.0: Xe, Cl name. Is usually -2 except in compounds with fluorine, oxygen contributes [ ]... Atoms in an ion is equal to the +3 oxidation state is equal the. Hydrides include compounds like sodium hydride, NaH metal, for example, has an oxidation number of simple. As a hydride ion, Cr2O72- us at info @ libretexts.org or out... 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Has formed two ionic compounds sulfate ( IV ) VII ) ion, the oxygen has oxidation! Not so obvious, so the sum of the oxidation state of manganese decreases 2. Ignore them when oxidation states of elements do the sum of the elements ) ions are reduced to (... Stable cation or anion will be −1 function of oxidation states varies, but follows patterns... Us at info @ libretexts.org or check out our status page at:! The N atom changes, but follows some patterns number of -1 ( no fluorine or oxygen atoms are )! And +3 respectively at the chlorate, ClO− 3 hydrogen chloride: \ \ce. Simply the element number, element name and atomic weights of each element in a substance is given negative! Our status page at https: //status.libretexts.org cases like these, some chemical intuition is useful \ \ce. Is in a compound oxidation states of elements standard oxidation states by 4 to compensate present! Then discover the positive limits of the oxidation state in peroxides: peroxides include peroxide! Hydrogen ions in its free or uncombined state holds up an oxidation state the. Being reduced in redox reactions element symbol, element symbol, element name and atomic weights of element! Species, its oxidation state of hydrogen ions peroxide, H2O2 one is given positive...


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# Freedom and constraints of a metal atom encapsulated in fullerene cages

Fullerenes are a beautiful example of hollow structures that give rise to endohedral host--guest complexes with a wide range of potential applications. Only very recently was the first solid metallofullerene characterized \cite{shino}, but key long-standing questions concerning metallofullerenes remain unanswered. Are there general rules for metal-cage chemistry? At room temperature, does the metal atom move? If so, how and on which time scale?'' We present {\it ab-initio} molecular dynamics simulations of La- and Y-fullerenes, which reveal how the fullerene structure determines the chemisorption sites, and to what extent chemical constraints limit the freedom of the metal atom. In C$_{60}$, high symmetry and low barriers allow it to diffuse and make a round trip in $\sim$1 psec. C$_{82}$ offers only few and unforeseen attractive nests to the metal atom. Trapped there, it moves with characteristic times of $\sim$0.1 psec. Our results shed light on the strikingly higher stability of the C$_{82}$ system, and suggest its ability to be used as an electronic switch.

By: Wanda Andreoni and Alessandro Curioni

Published in: Physical Review Letters, volume 77, (no 5), pages 834-7 in 1996

Please obtain a copy of this paper from your local library. IBM cannot distribute this paper externally.

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# Rate

A rate is a type of ratio where something of one unit is compared with something else of another unit. Rates are applied in many real-world scenarios like unit conversions, speed/velocity, and per-item price.

Unit cancellation is a common strategy used in rate and conversion problems where if there are two instances of a unit written -- one on the numerator and the other on the denominator, then the instances are crossed out. For instance $\frac{60 \text{ kilometers}}{1 \text{ hour}} \cdot 2 \text{ hours} = 120 \text{ kilometers}$.

## Problems

• Practice Problems on Alcumus
• Unit Conversions
• Speed and Other Rates


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# Milnor immersion of circle, disks, and a ball

Milnor surprisingly found an immersion of a circle in the plane which bounds two "incompatible" immersed disks (see [1] for example, picture included). I think I can glue these two disks together to define an immersion of a sphere $$f:S^2\to\mathbb{R}^3$$ where Milnor's immersed circle (the equator) sits in the $$xy$$-plane and contracts to a point (in the $$\pm z$$ directions) by running along either immersed disk.

Does $$f(S^2)$$ bound an immersed ball, i.e. does $$f$$ extend to an immersion of a 3-ball?

This question was sparked by glancing at Gromov's book [1] and a related paper of Eliashberg-Mishachev which argues that the projection $$S^2\to\mathbb{R}^2$$ (composing $$f$$ above with projection onto $$xy$$-plane) is not homotopic through folded maps to the "standard" folded map $$(x,y,z)\mapsto (x,y)$$ on the unit 2-sphere (the fold being the equator).

If one assumes a couple of simple conditions on $$f$$, namely that the restriction of $$z$$ on $$f(S^2)$$ is a Morse function with two critical points and that projections of two halves of $$f(S^2)$$ ($$z>0$$, $$z<0$$) onto the plane $$z=0$$ realize two original different immersions then $$f$$ can not be extended to an immersion of the ball.

Indeed, assume by contradiction that such an immersion $$F:B^3\to \mathbb R^3$$ exists. Let $$m<0$$ be the minimum of $$z$$ on $$f(S^2)$$ and $$M>0$$ be the max. In this case the preimages of planes $$z=c$$ under $$F$$ will be $$2$$-disks in $$B^3$$ for all $$c\in (m,M)$$. And by continuity, (changing $$c$$ from $$M$$ to 0, or from $$m$$ to $$0$$) we will get that the map $$F$$ applied to the disk $$F^{-1}(z=0)$$ realizes both immersions, which is absurd.

One can relax the two conditions by requiring that there is a regular function $$z'$$ on $$\mathbb R^3$$ such that $$z'=0$$ is the plane $$z=0$$ and $$z'$$ restricts to $$f(S^2)$$ as a Morse function with two critical points. If I understand correctly this condition is implicit in your construction of $$f$$. (One should express what it means that the two halves of $$S^2$$ realize two different immersions of a disk.)

PS. To clarify the above reasoning I'll give a proof in the most restrictive case. This case already contains the main idea. So the setting will be the following

Setting. Let $$S^2$$ be the unit $$2$$-sphere, $$S^1\subset S^2$$ be the equator, $$O_+$$ be the north pole and $$O_-$$ the south. Let $$S^2_+$$ be the open upper half-sphere and $$S^2_-$$ the lower one. Let us assume that the immersion $$f: S^2\mathbb\to \mathbb R^3$$ has the following properties.

1) The restriction of $$f$$ to the equator $$S^1\subset S^2$$ realizes the Milnor circle.

2) Function $$z$$ restricts to $$f(S^2)$$ as a Morse function with two critical points so that $$\max_z f(S^2)=1$$, $$\min_z f(S^2)=-1$$ and moreover the poles are sent to critical points $$z(f(O_+))=1$$, $$z(f(O_-))=-1$$.

3) Let $$\pi: (x,y,z)\to (x,y)$$ be the projection. Then the maps $$\pi\circ f: S^2_+\to \mathbb R^2$$ and $$\pi\circ f: S^2_-\to \mathbb R^2$$ realize the two Milnor immersions.

So, suppose all these conditions hold. Then each set $$F^{-1}(z=c)$$ is a disk $$D_c\subset B^3$$ for $$c\in (-1,1)$$. Consider the family of immersed disks $$\pi(F(D_c))\subset \mathbb R^2$$ for $$c$$ varying from $$1$$ to $$0$$. I claim that this family of immersed disks is the same as the following second family of immersed disks:

Second family. Let $$D_c^+\subset S_+^2$$ be the disk composed of points with $$z\ge c$$. Then we can consider the family of immersed disks $$\pi(F(D_c^+))$$.

So the claim is that the two families $$\pi(F(D_c))$$ and $$\pi(F(D_c^+))$$ coincide.

But then, the immersion $$\pi(F(D_0))$$ is the same as $$\pi(F(D_0^+))=\pi(F(S_+^2))$$.

Just in the same way we prove that $$\pi(F(D_0))$$ is the same as $$\pi(F(S_-^2))$$. This contradicts 3).


text:
# Can't argue with success? Looking for “bad math” that “gets away with it”

I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6s in

$$\frac{64}{16}.$$

Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs). Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for are cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

• @rschwieb: I don't understand your comment at all. The OP is asking for examples where, by wrong means, you end up with a correct result, and not where by "seemingly" valid means, you end up with a wrong result. It does not matter how you interpret "seemingly"; any "proof" that ends up with a wrong conclusion should not be posted as an answer here. That's the OP's choice. – TMM Dec 17 '12 at 14:30
• This is reminding me of an anecdote about a physicist from the early 20th century with a reputation for making arithmetic errors who as a joke intentionally made a huge order of magnitude error (10^10???) in a published paper; and then published a correction the next month noting that the error didn't affect the results of the computation. Unfortunately I'm failing to Google it so I can't see if what he did would be relevant to this question or not. – Dan Neely Dec 17 '12 at 16:52
• The College Mathematics Journal used to have a section entitled "Fallacies, Flaws, and Flimflam." It regularly featured exactly these kinds of things. – Mike Spivey Dec 17 '12 at 18:54
• Why is this on topic? – Anko Dec 17 '12 at 21:11
• There is a whole book about this "Mathematical Fallacies, Flaws, and Flimflam". This book is exactly what you are looking for. It has scores of such examples in a whole bunch of categories like algebra, calculus, multi-variable calculus, and so on. Love this book! – Fixed Point Dec 17 '12 at 21:17

Question is : Find $\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{\ln(x)}\right)$

My answer is : Divide and rule!!

$\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{ln(x)}\right) = \lim_{x \to 1}\left(\frac{x}{x-1}\right)-\lim_{x \to 1}\left(\frac{1}{ln(x)}\right)$

Now using L'Hospital rule!

$\lim_{x \to 1}\left(\frac{x}{x-1}-\frac{1}{ln(x)}\right) = \lim_{x \to 1}\left(\frac{x}{x-1}\right)-\lim_{x \to 1}\left(\frac{1}{ln(x)}\right)=\frac{1}{1}-0=1$

When I was in kinder-garden I was sick one day so I missed class and the next day the teacher gave us a quiz on basic multiplication. But I didn't know what multiplication was so when I got the test paper I thought that she wrote the + signs really badly. I didn't want to make her feel bad by pointing it out (not that my writing was any good either).

So I took the test. She graded it with me and explained that multiplication was something different so I said that I must have missed every question. Then she went through and checked every problem. Except it turned out that I got one problem right. 2x2 which happens to equal 2+2. If 0x0 was on there I would have gotten 2 problems right :)

• … yawn. But, – you did multiplication in kindergarten? – k.stm Jun 30 '16 at 15:56

Consider: $$\int_{-\infty}^{+\infty}\dfrac{e^{2x}-e^x}{x(1+e^{2x})(1+e^x)}dx =ln2$$ This integral has been worked out here: Finding value of integration with infinite limit But it is rather difficult because the linear $x$ term in the denominator. So I am deciding to omit $x$ and instead do $$\int_{-\infty}^{+\infty}\dfrac{e^{2x}-e^x}{(1+e^{2x})(1+e^x)}dx ?$$ This one is LOTS easier and the answer is still $ln2$. Works for me...

• This is a general principle since it works also for $$\int_0^{\infty}\frac x{x^3+1}dx=\int_0^{\infty}\frac 1{x^3+1}dx=\frac{2\pi\sqrt{3}}9$$ – zwim Sep 27 '17 at 1:18
• $Lim_{x\rightarrow\infty}x.e^{-x}=\infty ×0=0$
• $Lim_{x\rightarrow\infty}log(x+1)-log(x)=\infty -\infty=0$

Well, using $\frac{dy}{dx}$ as a ratio is also one such example. It carries no meaning at all to use $\frac{dy}{dx}$ as ratio but it works each time to get you to the right result instead of appreciating the limit process behind it. I think it is as much abuse of notation as is eliminating 6's from a fraction, it is just that in this case the notation is designed such that each time you abuse it, you will get the right answer but you won't know actually why if you just rely on the notation. For example: $\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$ due to chain rule and not due to treating $\frac{dy}{dx}$ as a fraction just like $\frac{x^2-y^2}{x-y}=x+y$ by algebra and not by

$$\frac{x^2-y^2}{x-y}$$

"cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then conclude that "two negatives make a positive", so the final answer is be $x+y$.

Now if you see a student prove $\frac{5^2-3^2}{5-3}=8$ by the wrong method, will he be correct as this method of proof will hold for all $x,y \in R$ until $x \neq y$.

Also in three dimensions you have stuff like $\frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=-1.$

Also, $dy$ isn't even defined , let alone be manipulated like a real number.

• Depending on your perspective, often such use is not simply (shockingly) wrong, just lacking in proper justification and rigor. If it works each time, there is something to it. – Jonas Meyer Nov 30 '13 at 10:13
• There are multiple definitions of $\mathrm dy\ldots$ – Mark S. Dec 10 '13 at 0:11
• -1, what about non-standard analysis? – JMCF125 Apr 8 '14 at 10:20

I notice that most of the answers are cute and humorous, so I may incur the wrath of the forum by posting an answer that is neither cute nor humorous.

I would have to say that most informal proofs are not valid, but there are formal proofs that come to the same conclusion, so the conclusion is correct, but informal proofs tend to be incomplete and lack rigor.

Cantor's diagonal argument is a good example. How long will it take to get a seventh grader to understand something like this, http://us.metamath.org/mpegif/ruc.html ? I don't even know if that is right.

Yet, most seventh graders will understand vsauce's 2 minute explanation of the theory. But it is incomplete. https://www.youtube.com/watch?v=s86-Z-CbaHA&feature=youtu.be&t=4m35s

Vsauce is correct when he says (paraphrased) "If there is a 1 to 1 correspondence between the two, then we can match one whole number to each real number, then the sets are the same size".

The hidden assumption is that we must use the whole numbers in order of magnitude. Which is not true.

So he is wrong when he says "We have used up every single whole number, the entire infinity of them, and yet we can still come up with more real numbers." The truth is that he did not need to "use up every single whole number".

To avoid this, all he had to do is match a subset of the whole numbers with his purported list of all real numbers. Like this

$4 \to r_1$

$44 \to r_2$

$444 \to r_3$

$4444 \to r_4$

and so on.

Then vsauce can generate another infinite list of real numbers that were not on his first list and again we can match them to a subset of the whole numbers like this

$14 \to rr_1$

$144 \to rr_2$

$1444 \to rr_3$

$14444 \to rr_4$

and so on.

And we can repeat this a countable infinite number of times, while vsauce can repeat his operation an uncountable infinite number of times.

But that is what vsauce needs to prove, and he can only do it with the formal proof, and that will take a very long and involved video.

The problem : solve the equation $$e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)}$$

### Incorrect method

As everybody knows $$e^a+e^b=e^{ab}$$

So let's apply it to solve the equation :

$\begin{array}{lll} e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)} & \iff & e^{(x-2)(x+8)}=e^{(4-x)(3x+2)} \\ & \iff & (x-2)(x+8)=(4-x)(3x+2) \\ & \iff & x^2+6x-16=10x-3x^2+8 \\ & \iff & 4x^2-4x-24=0 \\ & \iff & x^2-x-6=0 \\ & \iff & (x+2)(x-3)=0 \\ \end{array}$

Thus $x=-2$ or $x=3$.

### Correct method:

Let's set $X=e^x$ and $a=e^2$ and substitute

$\frac Xa+a^4X=\frac {a^2}X+aX^3\iff X^2+a^5X^2=a^3+a^2X^4\iff(a^2X^2-1)(X^2-a^3)=0$

Leading also to $x=-2$ and $x=3$.

The problem : simplify the expression $\sin(x+y)\sin(x-y)$

### Incorrect method

As everybody knows $$f(a+b)f(a-b)=f(a)^2-f(b)^2$$

Applying the formula gives immediately

$\sin(x+y)\sin(x-y)=\sin^2(x)-\sin^2(y)$

### Correct method

$\begin{array}{lll} \sin(x+y)\sin(x−y) &=& \big[\sin(x)\cos(y)+\cos(x)\sin(y)\big]\ \big[\sin(x)\cos(y)−\cos(x)\sin(y)\big]\\ &=&\sin^2(x)\cos^2(y)−\cos^2(x)\sin^2(y)\\ &=&\sin^2(x)(1−\sin^2(y))−(1−\sin^2(x))\sin^2(y)\\ \require{cancel}&=&\sin^2(x)−\cancel{\sin^2(x)\sin^2(y)}−\sin^2(y)+\cancel{\sin^2(x)\sin^2(y)}\\ &=&\sin^2(x)−\sin^2(y)\\ \end{array}$

## protected by Eric NaslundDec 18 '12 at 17:55

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


text:
# TriMatch

#### Example One: New Student Outreach

The data in example one (data(students)) represent newly enrolled students in a distance education program. By the nature of the program all students are considered part-time. Moreover, enrollment in the institution does not necessarily mean students are making active progress towards degree completion. To address this issue the institution began an outreach program whereby academic advisors would regularly contact new students within the first six months of enrollment until either six months have passed, or the student enrolled in some credit earning activity (generally a course or an examination for credit). Since randomization to receive the outreach was not possible, a comparison group was defined as students who enrolled six months prior to the start of the outreach. The treatment group is identified as students who enrolled six months after the start of the outreach and who received at least one academic advisor contact.

Covariates for estimating propensity scores were retrieved from the student information system. The dependent variable of interest is the number of credits attempted within the first seven months of enrollment.

During the implementation phase it was identified that the outreach being conducted was substantially different between the two academic advisors responsible for the treatment. As such, it became necessary to treat each academic advisor as a separate treatment. The analysis of propensity score models for more than two groups (i.e. two treatments and one control) has relied on conducting three separate analyses. We outline here an approach to conducting propensity score analysis with three groups.

data(students)
names(students)

We will create a treat variable that identifies our three groups.

treat <- students$TreatBy table(treat, useNA = "ifany") treat Control Treatment1 Treatment2 200 83 91 describeBy(students$CreditsAttempted, group = list(students$TreatBy), mat = TRUE, skew = FALSE) item group1 var n mean sd median trimmed mad min max range se 11 1 Control 1 200 7.225 7.903 5 6.069 7.413 0 26 26 0.5588 12 2 Treatment1 1 83 4.361 6.156 0 3.254 0.000 0 27 27 0.6757 13 3 Treatment2 1 91 5.923 7.148 3 4.767 4.448 0 25 25 0.7493  The following boxplot shows unadjusted results. ggplot(students, aes(x = TreatBy, y = CreditsAttempted, colour = TreatBy)) + geom_boxplot() + coord_flip() + geom_jitter()  #### Estimate Propensity Scores The trips function will estimate three propensity score models. form <- ~Military + Income + Employment + NativeEnglish + EdLevelMother + EdLevelFather + HasAssocAtEnrollment + Ethnicity + Gender + Age tpsa <- trips(students, students$TreatBy, formu = form)

Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

Warning: glm.fit: fitted probabilities numerically 0 or 1 occurred

id      treat model1 model2 model3       ps1       ps2 ps3 strata1 strata2 strata3
1  1    Control  FALSE  FALSE     NA 1.462e-01 1.687e-01  NA       1       1    <NA>
2  2 Treatment1   TRUE     NA   TRUE 8.304e-01        NA   1       5    <NA>       5
3  3    Control  FALSE  FALSE     NA 3.934e-01 3.693e-01  NA       4       4    <NA>
4  4    Control  FALSE  FALSE     NA 2.965e-08 1.779e-08  NA       1       1    <NA>
5  5    Control  FALSE  FALSE     NA 2.965e-08 1.779e-08  NA       1       1    <NA>
6  6    Control  FALSE  FALSE     NA 2.220e-16 1.650e-08  NA       1       1    <NA>

(p <- plot(tpsa))

#### Matched Triplets

tmatch <- trimatch(tpsa, exact = students[, c("Military", "DegreeLevel")])

Triangle plot of the results. We can see how the propensity scores translate from one model to another.

head(tmatch)

Treatment2 Treatment1 Control      D.m3     D.m1    D.m2  Dtotal
1        321        105      91 0.0004357 0.010718 0.01161 0.02276
2        342        326     198 0.0150613 0.002718 0.04715 0.06493
3         30        310     203 0.0408941 0.004873 0.02937 0.07514
4         95        310     248 0.0454191 0.020088 0.01894 0.08444
5        104        105     129 0.0007764 0.070095 0.01655 0.08742
6        352        105     109 0.0091104 0.013030 0.06693 0.08907

plot(tmatch, rows = c(2), line.alpha = 1, draw.segments = TRUE)

We can plot the distances. We can specify other calipers to see how may matched triplets we eliminate if we specify a small caliper to the trimatch function.

plot.distances(tmatch, caliper = c(0.15, 0.2, 0.25))

Standard deviations of propensity scores: 0.17 + 0.15 + 0.23 = 0.55
Percentage of matches exceding a distance of 0.15 (caliper = 0.15): 11%
Percentage of matches exceding a distance of 0.2 (caliper = 0.2): 0.56%
Percentage of matches exceding a distance of 0.25 (caliper = 0.25): 0%

The numbers on the left edge are the row numbers from tmatch. We can then use the plot.triangle.matches function with specifying the rows parameters to any or all of these values to investigate that matched triplet. The following figures shows that the large distances in due to the fact that only one data point has a very large propensity score in both model 1 and 2.

tmatch[tmatch$Dtotal > 0.6, ] Treatment2 Treatment1 Control D.m3 D.m1 D.m2 Dtotal 179 295 50 184 0.1842 0.2494 0.1959 0.6294 tmatch[466, ] Treatment2 Treatment1 Control D.m3 D.m1 D.m2 Dtotal NA <NA> <NA> <NA> NA NA NA NA plot(tmatch, rows = c(466), line.alpha = 1, draw.segments = TRUE) Error: 'data' must be of a vector type  #### Checking balance. plot.balance(tmatch, students$Age, label = "Age")

Using propensity scores from model 3 for evaluating balance.

Friedman rank sum test

data:  Covariate and Treatment and ID
Friedman chi-squared = 10.39, df = 2, p-value = 0.00555

Repeated measures ANOVA

Effect DFn DFd    F        p p<.05     ges
2 Treatment   2 356 7.01 0.001033     * 0.02293

plot.balance(tmatch, students$Age, label = "Age", nstrata = 8) Using propensity scores from model 3 for evaluating balance. Friedman rank sum test data: Covariate and Treatment and ID Friedman chi-squared = 10.39, df = 2, p-value = 0.00555 Repeated measures ANOVA Effect DFn DFd F p p<.05 ges 2 Treatment 2 356 7.01 0.001033 * 0.02293  plot.balance(tmatch, students$Military, label = "Military", nstrata = 10)

Using propensity scores from model 3 for evaluating balance.

Friedman rank sum test

data:  Covariate and Treatment and ID
Friedman chi-squared = NaN, df = 2, p-value = NA

plot.balance(tmatch, students$Gender, label = "Gender") Using propensity scores from model 3 for evaluating balance. Friedman rank sum test data: Covariate and Treatment and ID Friedman chi-squared = 13.57, df = 2, p-value = 0.00113  plot.balance(tmatch, students$Ethnicity, label = "Ethnicity")

Using propensity scores from model 3 for evaluating balance.

Friedman rank sum test

data:  Covariate and Treatment and ID
Friedman chi-squared = 0.215, df = 2, p-value = 0.8981

covs <- students[, c("Military", "NativeEnglish", "Gender", "Age", "Income", "Employment",
"EdLevelMother", "EdLevelFather")]
covs$Gender <- as.integer(covs$Gender)
covs$Income <- as.integer(covs$Income)
covs$Employment <- as.integer(covs$Employment)
covs$EdLevelMother <- as.integer(covs$EdLevelMother)
covs$EdLevelFather <- as.integer(covs$EdLevelFather)
plot.multibalance(tpsa, covs, grid = TRUE)

#### Examine unmatched students.

# Look at the subjects that could not be matched
unmatched <- attr(tmatch, "unmatched")
nrow(unmatched)/nrow(tpsa) * 100

[1] 53.74

# Percentage of each group not matched
table(unmatched$treat)/table(tpsa$treat) * 100

Control Treatment1 Treatment2
66.50      42.17      36.26

unmatched[unmatched$treat != "Control", ] id treat model1 model2 model3 ps1 ps2 ps3 strata1 strata2 strata3 2 2 Treatment1 TRUE NA TRUE 0.8304 NA 1.000e+00 5 <NA> 5 23 23 Treatment1 TRUE NA TRUE 1.0000 NA 1.000e+00 5 <NA> 5 38 38 Treatment1 TRUE NA TRUE 0.5186 NA 5.457e-01 5 <NA> 3 65 65 Treatment1 TRUE NA TRUE 0.7518 NA 1.000e+00 5 <NA> 5 70 70 Treatment1 TRUE NA TRUE 0.3445 NA 6.562e-01 4 <NA> 4 80 80 Treatment1 TRUE NA TRUE 0.5788 NA 8.082e-01 5 <NA> 5 83 83 Treatment1 TRUE NA TRUE 0.7625 NA 1.000e+00 5 <NA> 5 87 87 Treatment2 NA TRUE FALSE NA 0.4371 2.259e-01 <NA> 5 1 119 119 Treatment2 NA TRUE FALSE NA 0.5751 1.641e-01 <NA> 5 1 123 123 Treatment2 NA TRUE FALSE NA 0.6137 1.316e-01 <NA> 5 1 125 125 Treatment2 NA TRUE FALSE NA 0.5315 1.413e-08 <NA> 5 1 138 138 Treatment1 TRUE NA TRUE 0.4051 NA 6.757e-01 4 <NA> 5 139 139 Treatment2 NA TRUE FALSE NA 0.3737 3.892e-08 <NA> 4 1 143 143 Treatment2 NA TRUE FALSE NA 0.3551 6.714e-01 <NA> 4 4 144 144 Treatment1 TRUE NA TRUE 0.5993 NA 7.979e-01 5 <NA> 5 149 149 Treatment2 NA TRUE FALSE NA 0.5527 2.139e-01 <NA> 5 1 213 213 Treatment2 NA TRUE FALSE NA 0.1445 5.528e-01 <NA> 1 4 214 214 Treatment1 TRUE NA TRUE 0.5998 NA 8.225e-01 5 <NA> 5 216 216 Treatment2 NA TRUE FALSE NA 0.2992 3.410e-01 <NA> 3 2 240 240 Treatment2 NA TRUE FALSE NA 0.3969 2.137e-01 <NA> 4 1 267 267 Treatment2 NA TRUE FALSE NA 0.6377 2.153e-01 <NA> 5 1 277 277 Treatment2 NA TRUE FALSE NA 0.4930 1.852e-01 <NA> 5 1 282 282 Treatment2 NA TRUE FALSE NA 0.4925 3.043e-01 <NA> 5 2 284 284 Treatment2 NA TRUE FALSE NA 0.4910 2.230e-08 <NA> 5 1 285 285 Treatment2 NA TRUE FALSE NA 0.6094 1.195e-01 <NA> 5 1 286 286 Treatment2 NA TRUE FALSE NA 0.3559 2.083e-01 <NA> 4 1 289 289 Treatment1 TRUE NA TRUE 0.2875 NA 3.651e-01 3 <NA> 2 290 290 Treatment1 TRUE NA TRUE 0.2381 NA 5.260e-01 2 <NA> 3 296 296 Treatment1 TRUE NA TRUE 0.1898 NA 1.883e-01 2 <NA> 1 299 299 Treatment1 TRUE NA TRUE 0.3980 NA 4.123e-01 4 <NA> 2 302 302 Treatment2 NA TRUE FALSE NA 0.3743 6.308e-08 <NA> 4 1 308 308 Treatment1 TRUE NA TRUE 0.3783 NA 7.631e-01 4 <NA> 5 309 309 Treatment2 NA TRUE FALSE NA 0.4612 3.000e-01 <NA> 5 2 311 311 Treatment1 TRUE NA TRUE 0.4735 NA 5.857e-01 5 <NA> 4 312 312 Treatment1 TRUE NA TRUE 0.5615 NA 6.758e-01 5 <NA> 5 313 313 Treatment1 TRUE NA TRUE 0.2626 NA 5.589e-01 3 <NA> 4 314 314 Treatment1 TRUE NA TRUE 0.5754 NA 7.471e-01 5 <NA> 5 315 315 Treatment1 TRUE NA TRUE 0.1773 NA 3.214e-01 2 <NA> 2 317 317 Treatment1 TRUE NA TRUE 0.4762 NA 8.001e-01 5 <NA> 5 318 318 Treatment2 NA TRUE FALSE NA 0.4372 5.909e-01 <NA> 5 4 319 319 Treatment2 NA TRUE FALSE NA 0.5894 1.260e-01 <NA> 5 1 323 323 Treatment2 NA TRUE FALSE NA 0.3476 2.220e-16 <NA> 4 1 324 324 Treatment2 NA TRUE FALSE NA 0.3517 6.188e-08 <NA> 4 1 325 325 Treatment1 TRUE NA TRUE 0.5885 NA 7.360e-01 5 <NA> 5 327 327 Treatment1 TRUE NA TRUE 0.5673 NA 5.496e-01 5 <NA> 3 332 332 Treatment2 NA TRUE FALSE NA 0.6212 3.449e-01 <NA> 5 2 335 335 Treatment1 TRUE NA TRUE 1.0000 NA 1.000e+00 5 <NA> 5 336 336 Treatment1 TRUE NA TRUE 0.6168 NA 5.726e-01 5 <NA> 4 338 338 Treatment2 NA TRUE FALSE NA 0.2720 4.562e-01 <NA> 2 3 339 339 Treatment2 NA TRUE FALSE NA 0.6239 1.097e-01 <NA> 5 1 345 345 Treatment1 TRUE NA TRUE 0.3443 NA 7.315e-01 4 <NA> 5 350 350 Treatment1 TRUE NA TRUE 0.3303 NA 7.407e-01 4 <NA> 5 353 353 Treatment1 TRUE NA TRUE 0.5003 NA 8.108e-01 5 <NA> 5 356 356 Treatment1 TRUE NA TRUE 0.3477 NA 5.409e-01 4 <NA> 3 360 360 Treatment1 TRUE NA TRUE 0.2902 NA 3.978e-01 3 <NA> 2 361 361 Treatment1 TRUE NA TRUE 0.3101 NA 6.108e-01 3 <NA> 4 362 362 Treatment1 TRUE NA TRUE 0.4252 NA 5.813e-01 5 <NA> 4 363 363 Treatment2 NA TRUE FALSE NA 0.5417 3.601e-01 <NA> 5 2 364 364 Treatment2 NA TRUE FALSE NA 0.2766 5.596e-01 <NA> 2 4 365 365 Treatment1 TRUE NA TRUE 0.8931 NA 1.000e+00 5 <NA> 5 366 366 Treatment2 NA TRUE FALSE NA 0.2139 6.202e-01 <NA> 2 4 367 367 Treatment1 TRUE NA TRUE 0.3604 NA 1.000e+00 4 <NA> 5 368 368 Treatment2 NA TRUE FALSE NA 0.4665 2.543e-01 <NA> 5 2 369 369 Treatment1 TRUE NA TRUE 0.4517 NA 1.000e+00 5 <NA> 5 370 370 Treatment2 NA TRUE FALSE NA 0.3318 6.000e-01 <NA> 3 4 372 372 Treatment2 NA TRUE FALSE NA 0.5336 1.678e-01 <NA> 5 1 373 373 Treatment2 NA TRUE FALSE NA 0.5506 1.345e-01 <NA> 5 1 374 374 Treatment2 NA TRUE FALSE NA 0.6451 2.147e-01 <NA> 5 1  We can create a triangle plot of only the unmatched students by subsetting tpsa with those students in the unmatched data frame. plot(tpsa[tpsa$id %in% unmatched$id, ])  #### Loess Plot plot.loess3(tmatch, students$CreditsAttempted, plot.points = geom_jitter, ylab = "Credits Attempted")

geom_smooth: method="auto" and size of largest group is <1000, so using loess. Use
'method = x' to change the smoothing method.

plot.loess3(tmatch, students$CreditsAttempted, plot.points = geom_jitter, ylab = "Credits Attempted", points.alpha = 0.5, plot.connections = TRUE) geom_smooth: method="auto" and size of largest group is <1000, so using loess. Use 'method = x' to change the smoothing method.  #### Parrellel Plot plot.parallel(tmatch, students$CreditsAttempted)

#### Friedman Rank Sum Test

tmatch.out <- merge(tmatch, students$CreditsAttempted) outcomes <- grep(".out$", names(tmatch.out), perl = TRUE)
tmatch.out$id <- 1:nrow(tmatch.out) out <- melt(tmatch.out[, c(outcomes, which(names(tmatch.out) == "id"))], id.vars = "id") names(out) <- c("ID", "Treatment", "Outcome") head(out) ID Treatment Outcome 1 1 Treatment2.out 17 2 2 Treatment2.out 0 3 3 Treatment2.out 0 4 4 Treatment2.out 13 5 5 Treatment2.out 0 6 6 Treatment2.out 6 set.seed(2112) friedman.test(Outcome ~ Treatment | ID, out) Friedman rank sum test data: Outcome and Treatment and ID Friedman chi-squared = 2.257, df = 2, p-value = 0.3235  #### Repeated Measures ANOVA rmanova <- ezANOVA(data = out, dv = Outcome, wid = ID, within = Treatment) Warning: Converting "ID" to factor for ANOVA. print(rmanova)$ANOVA
Effect DFn DFd     F      p p<.05      ges
2 Treatment   2 356 1.993 0.1378       0.007381

$Mauchly's Test for Sphericity Effect W p p<.05 2 Treatment 0.9674 0.05342$Sphericity Corrections
Effect    GGe  p[GG] p[GG]<.05    HFe  p[HF] p[HF]<.05
2 Treatment 0.9685 0.1394           0.9789 0.1389

#### Pairwise Wilcoxon Rank Sum Tests

pairwise.wilcox.test(x = out$Outcome, g = out$Treatment, paired = TRUE, p.adjust.method = "bonferroni")

Pairwise comparisons using Wilcoxon signed rank test

data:  out$Outcome and out$Treatment

Treatment2.out Treatment1.out
Treatment1.out 0.094          -
Control.out    1.000          0.533

#### Posthoc t-tests

(t1 <- t.test(x = tmatch.out$Treatment1.out, y = tmatch.out$Control.out, paired = TRUE))

Paired t-test

data:  tmatch.out$Treatment1.out and tmatch.out$Control.out
t = -1.536, df = 178, p-value = 0.1262
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.7950  0.3481
sample estimates:
mean of the differences
-1.223

(t2 <- t.test(x = tmatch.out$Treatment2.out, y = tmatch.out$Control.out, paired = TRUE))

Paired t-test

data:  tmatch.out$Treatment2.out and tmatch.out$Control.out
t = 0.2968, df = 178, p-value = 0.7669
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.420  1.923
sample estimates:
mean of the differences
0.2514

(t3 <- t.test(x = tmatch.out$Treatment2.out, y = tmatch.out$Treatment1.out, paired = TRUE))

Paired t-test

data:  tmatch.out$Treatment2.out and tmatch.out$Treatment1.out
t = 2.04, df = 178, p-value = 0.04283
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
0.04813 2.90159
sample estimates:
mean of the differences
1.475

#### Boxplot of differences

plot.boxdiff(tmatch, students\$CreditsAttempted)

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text:
## Groups and group homomorphisms

• A group is a set $G$ with a function  $\begin{array}{ccc}G×G& \to & G\\ \left({g}_{1},{g}_{2}\right)& ⟼& {g}_{1}{g}_{2}\end{array}$
such that
(a)   If ${g}_{1},{g}_{2},{g}_{3}\in R$ then $\left({g}_{1}{g}_{2}\right){g}_{3}={g}_{1}\left({g}_{2}{g}_{3}\right)$,
(b)   There exists an identity, $1\in G$, such that if $g\in G$ then $1g=g1=g$,
(c)   If $g\in G$ then there exists an inverse to $g$, ${g}^{-1}\in G$, such that ${g}^{-1}g=g{g}^{-1}=1$.

HW: Show that the identity $1\in G$ is unique.
HW: Show that if $g\in G$ then the inverse ${g}^{-1}\in G$ is unique.
HW: Why isn't $\left\{1,2,3,4,5\right\}$ a group?

Important examples of groups are:

(a)   The integers $ℤ$ with the operation of addition,
(b)   The integers mod $n$, $ℤ/nℤ$, with operation addition,
(c)   The symmetric group ${S}_{n}$,
(c)   The general linear group of invertible $n×n$ matrices ${\mathrm{GL}}_{n}\left(ℂ\right)$.

Group homomorphisms are for comparing groups.

Let $G$ and $H$ be groups with identities ${1}_{G}$ and ${1}_{H}$, respectively.

• A group homomorphism from $G$ to $H$ is a function $f:G\to H$ such that
(a)   If ${g}_{1},{g}_{2}\in G$ then $f\left({g}_{1}{g}_{2}\right)=f\left({g}_{1}\right)f\left({g}_{2}\right)$,
(b)   $f\left({1}_{G}\right)={1}_{H}$.
• A group isomorphism is a bijective group homomorphism.
• Two groups $G$ and $H$ are isomorphic, $G\simeq H$, if there exists a group isomorphism $f:G\to H$ between them.

Two groups are isomorphic if the elements of the rings and their operations match up exactly. Think of groups that are isomorphic as being "the same". When classifying groups we put two groups in the same class only if they are isomorphic. This is what is meant by classifying groups "up to isomorphism".

Let $f:G\to H$ be a group homomorphism. Let ${1}_{G}$ and ${1}_{H}$ be the identities in $G$ and $H$, respectively. Then

(a)   $f\left({1}_{G}\right)={1}_{H}$.
(b)   If $g\in G$ then $f\left({g}^{-1}\right)={f\left(g\right)}^{-1}$.

 Proof. (a)   Multiply both sides of the following equation by $f{\left({1}_{G}\right)}^{-1}$. $f( 1G ) =f( 1G· 1G) =f( 1G ) f(1G).$ (b)   Since $f\left(g\right)f\left({g}^{-1}\right)=f\left(g{g}^{-1}\right)=f\left({1}_{G}\right)={1}_{H}$ and $f\left({g}^{-1}\right)f\left(g\right)=f\left({g}^{-1}g\right)=f\left({1}_{G}\right)={1}_{H},$ $f(g) -1 =f( g-1).$ $\square$

• A subgroup $H$ of a group $G$ is a subset $H\subseteq G$ such that
(a)   If ${h}_{1},{h}_{2}\in H$ then ${h}_{1}{h}_{2}\in H$,
(b)   $1\in H$,
(c)   If $h\in H$ then ${h}^{-1}\in H$.
• The trivial group $\left\{1\right\}$ is the set containing only $1$ with the operation given by $1\cdot 1=1$.

#### Cosets

Let $G$ be a group and let $H$ be a subgroup of $G$. We will use the subgroup $H$ to divide up the group $G$.

• A left coset of $H$ in $G$ is a set $gH=\left\{gh\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}h\in H\right\}$, where $g\in G$.
• $G/H$ (pronounced "$G$ mod $H$") is the set of left cosets of $H$ in $G$.
• A right coset of $H$ in $G$ is a set $Hg=\left\{hg\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}h\in H\right\}$, where $g\in G$.
• $H\G$ is the set of right cosets of $H$ in $G$.

Unless specified otherwise we will always work with left cosets and just call them cosets.

HW: Let $G$ be a group and let $H$ be a subgroup of $G$. Let $x$ and $g$ be two elements of $G$. Show that $x\in gH$ if and only if $gH=xH$.

Let $G$ be a group and let $H$ be a subgroup of $G$. Then the cosets of $H$ in $G$ partition $G$.

 Proof. To show: (a)   If $g\in G$ then there exists $g\prime \in G$ such that $g\in g\text{'}H$. $\phantom{\text{To show:}}$ (b)   If ${g}_{1}H\cap {g}_{2}H\ne \varnothing$ then ${g}_{1}H={g}_{2}H$. (a)   Let $g\in G$. $\phantom{\text{(a)}}$   Then $g=g·1\in gH$, since $1\in H$. $\phantom{\text{(a)}}$   So $g\in gH$. (b)   Assume ${g}_{1}H\cap {g}_{2}H\ne \varnothing$. $\phantom{\text{(b)}}$   To show: (ba)   ${g}_{1}H\subseteq {g}_{2}H$. $\phantom{\text{(b)}}$   $\phantom{\text{To show:}}$ (bb)   ${g}_{2}H\subseteq {g}_{1}H$. $\phantom{\text{(b)}}$   Let $k\in {g}_{1}H\cap {g}_{2}H$. $\phantom{\text{(b)}}$   Suppose $k={g}_{1}{h}_{1}={g}_{2}{h}_{2}$, where ${h}_{1},{h}_{2}\in H$. $\phantom{\text{(b)}}$   Then $g1 = g1 h1 h1-1 =k h1-1 = g2 h2 h1-1 ,and g2 = g2h2 h2-1 = k h2 -1 = g1h1 h2 -1.$ (ba)   Let $g\in {g}_{1}H$. $\phantom{\text{(ba)}}$   Then there exists $h\in H$ such that $g={g}_{1}h$. $\phantom{\text{(ba)}}$   Then $g = g1h = g2h2 h1-1 h ∈g2H ,since h2 h1-1 h∈H.$ $\phantom{\text{(ba)}}$   So ${g}_{1}H\subseteq {g}_{2}H$. $\phantom{\text{(b)}}$   (bb)   Let $g\in {g}_{2}H$. $\phantom{\text{(bb)}}$   Then there exists $h\in H$ such that $g={g}_{2}h$. $\phantom{\text{(bb)}}$   Then $g=g2h =g1h1 h2-1h ∈g1H ,since h1 h2-1 h ∈H.$ $\phantom{\text{(bb)}}$   So ${g}_{2}H\subseteq {g}_{1}H$. So ${g}_{1}H={g}_{2}H$. So the cosets of $H$ in $G$ partition $G$. □

Let $G$ be a group and let $H$ be a subgroup of $G$. If ${g}_{1},{g}_{2}\in G$ then $Card(g1H) =Card(g2H) .$

 Proof. To show: There is a bijection from ${g}_{1}H$ to ${g}_{2}H$. Define a map $\phi$ by $φ: g1H → g2H x ↦ g2 g1-1 x.$ To show: (a)   $\phi$ is well defined. $\phantom{\text{To show:}}$ (b)   $\phi$ is a bijection. (a)   To show: (aa)   If $x\in {g}_{1}H$ then $\phi \left(x\right)\in {g}_{2}H$. (ab)   If $x=y$ then $\phi \left(x\right)=\phi \left(y\right)$. (aa)   Assume $x\in {g}_{1}H$. Then $x={g}_{1}h$ for some $h\in H$. So $\phi \left(x\right)={g}_{2}{{g}_{1}}^{-1}{g}_{1}h={g}_{2}h\in {g}_{2}H$. (ab)   This is clear from the definition of $\phi$. So $\phi$ is well defined. (b)   By virtue of Theorem 2.2.3 Part I, ???????????????????? we want to construct an inverse map for $\phi$. Define $ψ: g2H → g1H y ↦ g1 g2-1 y.$ HW: Show (exactly as in (a) above) that $\psi$ is well defined. Then $ψ(φ(x) = g1 g2-1 φ(x) = g1 g2-1 g2 g1-1 x =x, and$ $φ(ψ)y) =g2 g1-1 φ(y) =g2 g1-1 g1 g2-1 y =y. and$ So $\psi$ is an inverse function to $\phi$. So $\phi$ is a bijection. $\square$

Let $H$ be a subgroup of a group $G$. Then $Card(G) =Card(G/H) Card(H).$

 Proof. By Proposition 2.2, all cosets in $G/H$ are the same size as $H$. Since the cosets of $H$ partition $G$, the cosets are disjoint subsets of $G$, and $G$ is a union of those subsets. So $G$ is the union of $\mathrm{Card}\left(G/H\right)$ disjoint subsets all of which have size $\mathrm{Card}\left(H\right)$. $\square$

The above results show that the cosets of a subgroup $H$ divide the group $G$ into equal size pieces, one of these pieces being the subgroup $H$ itself.

• A set of coset representatives of $H$ in $G$ is a set of distinct elements $\left\{{g}_{i}\right\}$ of $G$ such that
1. each coset of $H$ is of the form ${g}_{i}H$ for some ${g}_{i}$ and
2. ${g}_{i}H\ne {g}_{j}H$ unless ${g}_{i}={g}_{j}$.
• The index of a subgroup $H$ in a group $G$, $|G:H|$, is the number of cosets of $H$ in $G$, $|G:H| =Card(G/H).$

HW:Show that $|G:\left\{1\right\}|=\mathrm{Card}\left(G\right)$.

#### Quotient groups $↔$ Normal subgroups

Let $H$ be a subgroup of a group $G$. We can try to make the set of cosets of $H$ into a group by defining an operation on the cosets. The only problem is that this doesn't work for the cosets of just any subgroup, the subgroup has to have special properties.

• A normal subgroup $N$ is a subset of a group $G$ such that if $n\in N$ and $g\in G$ then $gn{g}^{-1}\in N$.

HW: Show that a subgroup $N$ of a group $G$ is normal if and only if $N$ satisfies: if $g\in G$ then $gN=Ng$.

Let $N$ be a subgroup of a group $G$. Then $N$ is a normal subgroup of $G$ if and only if $G/N$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$ is a group.

 Proof. Proof $⇒$ Assume $N$ is a normal subgroup of $G$. To show: $\left(aN\right)\left(bN\right)=\left(abN\right)$ is a well defined operation on $\left(G/N\right)$. $N$ is the identity element of $G/N$. ${g}^{-1}N$ is the inverse of $gN$. We want the operation on $G/N$ given by $G/N×G/N → G/N (aN,bN) ↦ abN$ to be well defined. To show: If $\left({a}_{1}N,{b}_{1}N\right),\left({a}_{2}N,{b}_{2}N\right)\in G/N×G/N$ and $\left({a}_{1}N,{b}_{1}N\right)=\left({a}_{2}N,{b}_{2}N\right)$ then ${a}_{1}{b}_{1}N={a}_{2}{b}_{2}N$. Assume $\left({a}_{1}N,{a}_{2}N\right),\left({a}_{2}N,{b}_{2}N\right)\in G/N×G/N$ and $\left({a}_{1}N,{b}_{1}N\right)=\left({a}_{2}N,{b}_{2}N\right)$. Then ${a}_{1}N={a}_{2}N$ and ${b}_{1}N={b}_{2}N$. To show: aa. ${a}_{1}{b}_{1}N\subseteq {a}_{2}{b}_{2}N.$ $\phantom{\text{To show:}}$ ab. ${a}_{2}{b}_{2}N\subseteq {a}_{1}{b}_{1}N$. Since ${a}_{1}N={a}_{2}N$, we know ${a}_{1}={a}_{1}·1\in {a}_{2}N$. So there exists ${n}_{1}\in N$ such that ${a}_{1}={a}_{2}{n}_{1}$. Similarly, there exists ${n}_{2}\in N$ such that ${b}_{1}={b}_{2}{n}_{2}$. Let $k\in {a}_{1}{b}_{1}N$. Then there exists $n\in N$ such that $k={a}_{1}{b}_{1}n$. So $k = a1b1n = a2n1 b2 n2n = a2 b2 b2-1 n1b2 n2 n.$ Since $N$ is normal, ${{b}_{2}}^{-1}{n}_{1}{b}_{2}\in N,$ and therefore $\left({{b}_{2}}^{-1}{n}_{1}{b}_{2}\right){n}_{2}n\in N$. So $k={a}_{2}{b}_{2}\left({{b}_{2}}^{-1}{n}_{1}{b}_{2}\right){n}_{2}n\in {a}_{2}{b}_{2}N.$ So ${a}_{1}{b}_{1}N\subseteq {a}_{2}{b}_{2}N$. To show: ${a}_{2}{b}_{2}N\subseteq {a}_{1}{b}_{1}N$. Since ${a}_{1}N={a}_{2}N$, there exists ${n}_{1}\in N$ such that ${a}_{1}{n}_{1}={a}_{2}$. Since ${b}_{1}N={b}_{2}N$, there exists ${n}_{2}\in N$ such that ${b}_{1}{n}_{2}={b}_{2}$. Let $k\in {a}_{2}{b}_{2}N$. Then there exists $n\in N$ such that $k={a}_{2}{b}_{2}n$. So $k = a2b2n = a1n1 b1n2n = a1b1 b1-1 n1b1 n2n.$ Since $N$ is normal, ${{b}_{1}}^{-1}{n}_{1}{b}_{1}\in N$, and therefore $\left({{b}_{1}}^{-1}{n}_{1}{b}_{1}\right){n}_{2}n\in N$. So $k={a}_{1}{b}_{1}\left({{b}_{1}}^{-1}{n}_{1}{b}_{1}\right){n}_{2}n\in {a}_{1}{b}_{1}N$. So ${a}_{2}{b}_{2}N\subseteq {a}_{1}{b}_{1}N$. So $\left({a}_{1}{b}_{1}\right)N=\left({a}_{2}{b}_{2}\right)N$. So the operation is well defined. The coset $N=1N$ is the identity since if $g\in G$ then $(N) (gN) = (1G)N =gN = (g1)N = (gN)(N),$ If $gN$ is a coset then its inverse is ${g}^{-1}N$ since $(gN) (g-1N) = (gg-1)N = N = g-1gN = (g-1N) (gN).$ So $G/N$ is a group. $⇐$ Assume $G/N$ is a group with operation $\left(aN\right)\left(bN\right)=abN$. To show: If $g\in G$ and $n\in N$ then $gn{g}^{-1}\in N$. First we show: If $n\in N$ then $nN=N$. Assume $n\in N$. To show: a. $nN\in N$. $\phantom{\text{To show:}}$ b. $N\in nN.$ Assume $x\in nN$. Then there exists $m\in N$ such that $x=nm$. Since $N$ is a subgroup, $nm\in N$. So $x\in N$. So $nN\subseteq N$. Assume $m\in N$. Then, since $N$ is a subgroup, $m=n{n}^{-1}m\in nN$. So $N\subseteq nN$. Now let $g\in G$ and $n\in N$. Then, by definition of the operation, $gng-1N = (gN) (nN) (g-1N) = (gN) (N) (g-1N) = g1g-1N = N.$ So $gn{g}^{-1}\in N$. So $N$ is a normal subgroup of $G$. $\square$

• The quotient group $G/N$ is the set of cosets of the normal subgroup $N$ of the group $G$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$.

Wow!! We actually made this weird set of cosets into a group!!

Let $N$ be a subgroup of a group $G$. Then $N$ is a normal subgroup of $G$ if and only if $G/N$ with operation given by $\left(aN\right)\left(bN\right)=\left(abN\right)$ is well defined.
HW: Show that if $G=N$ then $G/N\simeq \left\{1\right\}$.

#### Kernel and image of a homomorphism

• The kernel of a group homomorphism $f:G\to H$ is the set $kerf={g∈G | f(g)=1H},$ where ${1}_{H}$ is the identity in $H$.
• The image of a group homomorphism $f:G\to H$ is the set $imf={f(g)∈H | g∈G }.$

Let $f:G\to H$ be a group homomorphism. Then

(a)   $\mathrm{ker}f$ is a normal subgroup of $G$.
(b)   $\mathrm{im}f$ is a subgroup of $H$.

 Proof. To show: $\mathrm{ker}f$ is a normal subgroup of $G$. $\mathrm{im}f$ is a subgroup of $G$. To show: aa. $\mathrm{ker}f$ is a subgroup. $\phantom{\text{To show:}}$ ab. $\mathrm{ker}f$ is normal. To show: aaa. If ${k}_{1},{k}_{2}\in \mathrm{ker}f$ then ${k}_{1}{k}_{2}\in \mathrm{ker}f$. $\phantom{\text{To show:}}$ aab. ${1}_{G}\in \mathrm{ker}f$. $\phantom{\text{To show:}}$ aac. If $k\in \mathrm{ker}f$ then ${k}^{-1}\in \mathrm{ker}f$. Assume ${k}_{1},{k}_{2}\in \mathrm{ker}f$. Then $f\left({k}_{1}\right)={1}_{H}$ and $f\left({k}_{2}\right)={1}_{H}$. So $f\left({k}_{1}{k}_{2}\right)=f\left({k}_{1}\right)f\left({k}_{2}\right)={1}_{H}$. So ${k}_{1}{k}_{2}\in \mathrm{ker}f$. Since $f\left({1}_{G}\right)={1}_{H},{1}_{G}\in \mathrm{ker}f$. Assume $k\in \mathrm{ker}f$. So $f\left(k\right)={1}_{H}$. Then $f(k-1) =f (k)-1 =1H -1 =1H .$ So ${k}^{-1}\in \mathrm{ker}f$. So $\mathrm{ker}f$ is a subgroup. To show: If $g\in G$ and $k\in \mathrm{ker}f$ then $gk{g}^{-1}\in \mathrm{ker}f$. Assume $g\in G$ and $k\in \mathrm{ker}f$. Then $f( gkg-1) = f(g)f(k) f(g-1) = f(g) f(g-1) = f(g) f(g) -1 = 1.$ So $gk{g}^{-1}\in \mathrm{ker}f$. So $\mathrm{ker}f$ is a normal subgroup of $G$. To show: $\mathrm{im}f$ is a subgroup of $H$. To show: ba. If ${h}_{1},{h}_{2}\in \mathrm{im}f$ then ${h}_{1}{h}_{2}\in \mathrm{im}f$. $\phantom{\text{To show:}}$ bb. ${1}_{H}\in \mathrm{im}f$. $\phantom{\text{To show:}}$ bc. If $h\in \mathrm{im}f$ then ${h}^{-1}\in \mathrm{im}f.$ Assume ${h}_{1},{h}_{2}\in \mathrm{im}f$. Then there exist ${g}_{1},{g}_{2}\in G$ such that ${h}_{1}=f\left({g}_{1}\right)$ and ${h}_{2}=f\left({g}_{2}\right)$. Then $h1h2 =f(g1) f(g2) = f(g1g2) ,$ since $f$ is a homomorphism. So ${h}_{1}{h}_{2}\in \mathrm{im}f$. By Proposition 2.1 (a), $f\left({1}_{G}\right)={1}_{H}$. So ${1}_{H}\in \mathrm{im}f$. Assume $h\in \mathrm{im}f$. Then there exists $g\in G$ such that $h=f\left(g\right)$. Then, by Proposition 2.1 (b), $h-1 =f(g) -1 =f( g-1).$ So ${h}^{-1}\in \mathrm{im}f$. So $\mathrm{im}f$ is a subgroup of $H$. $\square$

Let $f:G\to H$ be a group homomorphism. Let ${1}_{G}$ be the identity element of $G$. Then

(a)   $\mathrm{ker}f=\left\{{1}_{G}\right\}$ if and only if $f$ is injective.
(b)   $\mathrm{im}f=H$ if and only if $f$ is surjective.

 Proof. Let ${1}_{G}$ and ${1}_{H}$ be the identities for $G$ and $H$, respectively. $⇒$ Assume $\mathrm{ker}f=\left({1}_{G}\right)$. To show: If $f\left({g}_{1}\right)=f\left({g}_{2}\right)$ then ${g}_{1}={g}_{2}$. Assume $f\left({g}_{1}\right)=f\left({g}_{2}\right)$. Then, by Proposition 2.1 (b), and the fact that $f$ is a homomorphism, $1H =f(g1) f(g2) -1 =f( g1 g2 -1 ) .$ So ${g}_{1}{{g}_{2}}^{-1}\in \mathrm{ker}f$. Since $\mathrm{ker}f=\left({1}_{G}\right)$, then ${g}_{1}{{g}_{2}}^{-1}={1}_{G}$. So ${g}_{1}={g}_{2}$. So $f$ is injective. $⇐:$ Assume $f$ is injective. To show: aa. $\left({1}_{G}\right)\subseteq \mathrm{ker}f$. $\phantom{\text{To show:}}$ ab. $\mathrm{ker}f\subseteq \left({1}_{G}\right)$. Since $f\left({1}_{G}\right)={1}_{H},{1}_{G}\in \mathrm{ker}f$. So $\left({1}_{G}\right)\subseteq \mathrm{ker}f$. (Proof) Let $k\in \mathrm{ker}f$. Then $f\left(k\right)={1}_{H}$. So $f\left(k\right)=f\left({1}_{G}\right)$. Thus, since $f$ is injective, $k={1}_{G}$. So $\mathrm{ker}f\subseteq \left({1}_{G}\right)$. So $\mathrm{ker}f=\left({1}_{G}\right)$. $⇒$ Assume $\mathrm{im}f=H$. To show: If $h\in H$ then there exists $g\in G$ such that $f\left(g\right)=h$. Assume $h\in H$. Then $h\in \mathrm{im}f$. So there exists $g\in G$ such that $f\left(g\right)=h$. So $f$ is surjective. $⇐$ Assume $f$ is surjective. To show: ba. $\mathrm{im}f\subseteq H$. $\phantom{\text{To show:}}$ bb. $H\subseteq \mathrm{im}f$. Let $x\in \mathrm{im}f$. Then there exists $g\in G$ such that $x=f\left(g\right)$. By the definition of $f,f\left(g\right)\in H$. So $x\in H$. So $\mathrm{im}f\subseteq H$. (Proof) Assume $x\in H$. Since $f$ is surjective there exists a $g$ such that $f\left(g\right)=x$. So $x\in \mathrm{im}f$. So $H\subseteq \mathrm{im}f$. So $\mathrm{im}f=H$. $\square$

Notice that the proof of Proposition (gpinjsur)(b) does not use the fact that $f:G\to H$ is a homomorphism, only the fact that $f:G\to H$ is a function.

HW: Show that if $S$ and $T$ are sets and $f:S\to T$ is a functions then $\mathrm{im}f=T$ if and only if $f$ is surjective.

(a)   Let $f:G\to H$ be a group homomorphism and let $K=\mathrm{ker}f$. Define $f^: G/kerf ⟶ H gK ⟼ f(g).$ Then $\stackrel{^}{f}$ is a well defined injective group homomorphism.
(b)   Let $f:G\to H$ be a group homomorphism and define $f′: G ⟶ imf g ⟼ f(g).$ Then $f\prime$ is a well defined surjective group homomorphism.
(c)   If $f:G\to H$ is a group homomorphism then $G/kerf≃imf,$ where the isomorphism is a group isomorphism.

 Proof. To show: aa. $\stackrel{^}{f}$ is well defined. $\phantom{\text{To show:}}$ ab. $\stackrel{^}{f}$ is injective. $\phantom{\text{To show:}}$ ac. $\stackrel{^}{f}$ is a homomorphism. To show: aaa. If $g\in G$ then $\stackrel{^}{f}\left(gK\right)\in H$. $\phantom{\text{To show:}}$ aab. If ${g}_{1}K={g}_{2}K$ then $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{2}K\right)$. Assume $g\in G$. Then $\stackrel{^}{f}\left(gK\right)=f\left(g\right)$ and $f\left(g\right)\in H$ by the definition of $\stackrel{^}{f}$ and $f$. Assume ${g}_{1}K={g}_{2}K$. Then there exists $k\in K$ such that ${g}_{1}={g}_{2}k$. To show: $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{2}K\right)$. To show: $f\left({g}_{1}\right)=f\left({g}_{2}\right)$. Since $k\in \mathrm{ker}f$, then $f\left(k\right)=1$ and so $f(g1) =f(g2k) =f(g2) f(k) =f(g2).$ So $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{2}K\right)$. So $\stackrel{^}{f}$ is well defined. To show: If $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{1}K\right)$ then ${g}_{1}K={g}_{2}K$. Assume $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{2}K\right)$. Then $f\left({g}_{1}\right)=f\left({g}_{2}\right)$. So $f\left({g}_{1}\right)f{\left({g}_{2}\right)}^{-1}=1$. So $f\left({g}_{1}{{g}_{2}}^{-1}\right)=1$. So ${g}_{1}{{g}_{2}}^{-1}\in \mathrm{ker}f$. So ${g}_{1}{{g}_{2}}^{-1}=k$ for some $k\in \mathrm{ker}f$. So ${g}_{1}={g}_{2}k$ for some $k\in \mathrm{ker}f$. To show: aba. ${g}_{1}K\subseteq {g}_{2}K$. $\phantom{\text{To show:}}$ abb. ${g}_{2}K\subseteq {g}_{1}K$. Let $g\in {g}_{1}K$. Then there exists ${k}_{1}\in K$ such that $g={g}_{1}{k}_{1}$. So $g={g}_{2}k{k}_{1}\in {g}_{2}K,$ since $k{k}_{1}\in K$. So ${g}_{1}K\subseteq {g}_{2}K$. Let $g\in {g}_{2}K$. Then $g={g}_{2}{k}_{2}$ for some ${k}_{2}=k$. So $g={g}_{1}{k}^{-1}{k}_{2}\in {g}_{1}K$, since ${k}^{-1}{k}_{2}\in K$. So ${g}_{2}K\subseteq {g}_{1}K$. So ${g}_{1}K={g}_{2}K$. So $\stackrel{^}{f}$ is injective. To show: $\stackrel{^}{f}\left({g}_{1}K\right)\stackrel{^}{f}\left({g}_{2}K\right)=\stackrel{^}{f}\left(\left({g}_{1}K\right)\left({g}_{2}K\right)\right)$. Since $f$ is a homomorphism, $f^(g1K) f^(g2K) = f(g1) f(g2) = f(g1g2) = f^ (g1g2K) = f^( (g1K) (g2K)).$ So $\stackrel{^}{f}$ is a homomorphism. To show: ba. $f\prime$ is well defined. $\phantom{\text{To show:}\phantom{}}$ bb. $f\prime$ is surjective. $\phantom{\text{To show:}\phantom{}}$ bc. $f\prime$ is a homomorphism. This is proved in Ex 2.2.3, Part I ?????????????? This is proved in Ex 2.2.3, Part I ?????????????? Since $f$ is a homomorphism, $f′(g) f′(h) = f(g)f(h) = f(gh) = f′(gh).$ So $f\text{'}$ is a homomorphism. Let $K=\mathrm{ker}f.$ By (a), the function $f^: G/K → H gK ↦ f(g)$ is a well defined injective homomorphism. By (b), the function $f^′: G/K → imf^ gK ↦ f^(gK) =f(g)$ is a well defined surjective homomorphism. To show: ca. $\mathrm{im}\stackrel{^}{f}=\mathrm{im}f$. $\phantom{\text{To show:}}$ cb. $\stackrel{^}{f}\text{'}$ is injective. To show: caa. $\mathrm{im}\stackrel{^}{f}\subseteq \mathrm{im}f$. $\phantom{\text{To show:}}$ cab. $\mathrm{im}f\subseteq \stackrel{^}{f}$. Let $h\in \mathrm{im}\stackrel{^}{f}$. Then there exists $gK\in G/K$ such that $\stackrel{^}{f}\left(gK\right)=h$. Let $g\prime \in gK$. Then there exists $k\in K$ such that $g\prime =gk$. Then, since $f$ is a homomorphism and $f\left(k\right)=1$, $f(g′) = f(gk) = f(g)f(k) = f(g) = f^(gK) = h.$ So $h\in \mathrm{im}f$. So $\mathrm{im}\stackrel{^}{f}\subseteq \mathrm{im}f$. Let $h\in \mathrm{im}f$. Then there exists $g\in G$ such that $f\left(g\right)=h$. So $\stackrel{^}{f}\left(gK\right)=f\left(g\right)=h$. So $h\in \mathrm{im}\stackrel{^}{f}$. So $\mathrm{im}f\subseteq \mathrm{im}\stackrel{^}{f}$. To show: If $\stackrel{^}{f}\text{'}\left({g}_{1}K\right)=\stackrel{^}{f}\text{'}\left({g}_{2}K\right)$ then ${g}_{1}K={g}_{2}K$. Assume $\stackrel{^}{f}\prime \left({g}_{1}K\right)=\stackrel{^}{f}\prime \left({g}_{2}K\right)$. Then $\stackrel{^}{f}\left({g}_{1}K\right)=\stackrel{^}{f}\left({g}_{2}K\right)$. Then, since $\stackrel{^}{f}$ is injective, ${g}_{1}K={g}_{2}K$. So $\stackrel{^}{f}\prime$ is injective. Thus $f^′: G/K → imf^ gK ↦ f(g)$ is a well defined bijective homomorphism. $\square$

#### Direct products

Suppose $H$ and $K$ are groups. The idea is to make $H×K$ into a group.

• The direct product $H×K$ of two groups $H$ and $K$ is the set $H×K$ with the operation given by $(h1,k1) (h2,k2) = (h1h2, k1k2) ,$ for ${h}_{1},{h}_{2}\in H$, ${k}_{1},{k}_{2}\in K$.
• More generally, given groups ${G}_{1},{G}_{2},\dots ,{G}_{n}$, the direct product ${G}_{1}×{G}_{2}×\cdots ×{G}_{n}$ is the set ${G}_{1}×{G}_{2}×\cdots ×{G}_{n}$ with the operations given by $(h1,…, hi,…, hn) + (k1,…, ki,…, kn) = (h1k1 ,…, hiki,…, hnkn)$ where ${h}_{i},{k}_{i}\in {G}_{i}$, and ${h}_{i}{k}_{i}$ is given by the operation in the group ${G}_{i}$. The operation in the direct product is just the operations from the original groups acting componentwise.

HW: Show that these are good definitions, i.e. that, as defined above, $H×K$ and ${G}_{1}×{G}_{2}×\cdots ×{G}_{n}$ are groups with identities given by $\left({1}_{H},{1}_{K}\right)$ and $\left({1}_{{G}_{1}},\dots ,{1}_{{G}_{n}}\right)$, respectively (${1}_{{G}_{i}}$ denotes the identity in the group ${G}_{i}$).

#### Further Definitions

• An abelian group is a group $G$ such that if ${g}_{1},{g}_{2}\in G$ then ${g}_{1}{g}_{2}={g}_{2}{g}_{1}$.
• The center $Z\left(G\right)$ of a group $G$ is the set $Z(G) ={c∈G | if g∈G then cg=gc }.$

HW: Give an example of a non-abelian group.

HW: Prove that every subgroup of an abelian group is normal.

HW: Prove that $Z\left(G\right)$ is a normal subgroup of $G$.

HW: Prove that $Z\left(G\right)=G$ if and only if $G$ is abelian.

• The order $|G|$ of a group $G$ is the number of elements in $G$.
• Let $G$ be a group and let $g\in G$. The order $o\left(g\right)$ of $g$ is the smallest positive integer $n$ such that ${g}^{n}=1$. If no such integer exists then $o\left(g\right)=\infty$.

• Let $G$ be a group and let $S$ be a subset of $G$. The subgroup generated by $S$ is the subgroup $⟨S⟩$ of $G$ such that
(a)   $S\subseteq ⟨S⟩,$
(b)   If $H$ is a subgroup of $G$ and $S\subseteq H$ then $⟨S⟩\subseteq H.$
$⟨S⟩$ is the smallest subgroup of $G$ containing $S$. Think of $⟨S⟩$ as gotten by adding to $S$ exactly those elements of $G$ that are needed to make a group.

HW: Let $G$ be a group and let $S$ be a subset of $G$. Show that the subgroup generated by $S$ exists and is unique.

## Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

## References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.


text:
### Reverse use of the simple interest formula

There are four pronumerals in the formula $$I = PRT$$. As with all such formulae, given the values of any three, the value of the fourth can be found.

#### Example 5

Michael invested $80 000 in a building society that pays 6.5% p.a. simple interest. Over the years the investment has paid him$57 200 in interest. How many years has he had the investment?

#### Solution

I = 57 200, P = 80 000 and R = 6.5% = 0.065

\begin{align}I &= PRT\\\\ 57\ 200 &= 80\ 000 × 0.065 × T\\\\ 57\ 200 &= 5200 × T\\\\ T &= \dfrac{57\ 200}{5200}\\\\ &= 11\ \text{years}\end{align}


text:
Open letter to Mr. Phisher

I received a very amusing e-mail today:

Hello,
It is with profound sense of sadness i wrote this email to you. I don’t know how you will find this but you just have to forgive me for not telling you before leaving. I traveled down to United Kingdom Yesterday for a short vacation but unfortunately,i was mugged at a gun point on my way to the hotel i lodged all my money and all other vital documents including my credit card and my cell phone have been stolen by muggers.
I’ve been to the embassy and the Police here but they’re not helping issues at all,Things are difficult here and i don’t know what to do at the moment that why i email to ask if you can lend me £1,500.00 so i can settle the hotel bill and get a returning ticket back home. Please do me this great help and i promise to refund the money as soon as i get back home.
I look forward to your positive response.
Thanks

This coming supposedly coming from the Gmail of a fellow engineer that happens to live in my town and is at 2 degrees of separation from me in the LinkedIn network.

Sorry, Mr. Phisher, I don’t know the guy enough to send the funds, my bad for not being more proactive in extending my network, he was just a step away! (if he really is the one who set the account with his name). It will help your cause if you read the newspapers, for the last few days there have been no flights to or from London, you know, the volcano thing.

I heard stories of people getting similar e-mails from people they know, some of the stories may be more verisimilar even. The would be phishers make good use of the information that can be gathered from social networks to craft these targeted e-mails.

Google problems may have bigger problems than people creating accounts to send phishing e-mails.

Hackers expose slew of Hotmail acount passwords

Again,

Hackers expose slew of Hotmail acount passwords

Social Engineering and phishing really work.

See what I wrote on Passwords

I recently discussed the problems associated with weak passwords here. Since then, there have been a few cases of hackers publishing stolen passwords form popular sites such as phpbb or the passwords that the conficker worm uses to spread across shares. Some researches report that people often use the same password on many websites making themselves vulnerable to serious attack if the password for a low value website is the same as the one used in a high value target

Password selection tips abound and as long as your password has enough entropy, users data is somewhat out of reach of most hackers.

Despite the advice of security gurus, the manifest limitations of the average human brain for generating and remembering more than a few passwords is a physical barrier to a widespread adoption secure practices. Password managers may help to keep your passwords organized. They have functions to generate strong passwords and can connect directly with browsers or e-mail programs.

Another way around is the OpenID network that allow users to have one identity for multiple on-line services. The OpenID protocol is inclusive enough that can work as an Authenticator using biometrics or smart-tokens.  Open ID is still in the adoption phase, not all online services accept it.

Authentication – Part IV Password Protocols

To be useful, passwords need to be transmitted or negotiated between the server and the client.

Transmission of the password in the clear is subjected to eavesdropping and therefore very insecure. The password storage on the servers side must also be protected from the possibility that the file falls on the wrong hands, compromising the security of the system.

There are several constraints to the design of password protections protocols, one of the most important being the limited amount of entropy that user memorized passwords necessarily have. Computation time is another big constraint. Even small delays in the response time can make the difference between a system the user is happy to interact with, and a system in which security features will be disabled for the sake of interactivity.

The key features of a password protection protocol are described below:

1. The transmission and storage of passwords should be non plain-text equivalent: This means, the protocol should be such that even if an attacker obtains the database containing the password or eavesdrop the exchange between client and server, this will not compromise the security of the exchange.
2. The protocol must be resistant to replay attack: That is, if an eavesdropper successfully record a login session, the information can not be used to compromise a future (or past) exchange between the Client and the server.
3. The protocol must be resistant to the Denning-Sacco attack: In this attack, by capturing the session key (not the raw password) the eavesdropper has enough information to successfully mount a brute force attack or at least to successfully impersonate the user.
4. The protocol must be resistant to active attacks:In these situations the protocol leaks enough information that allows the attacker to impersonate the server to the client, make a guess of the correct password and then, by faking a failure, obtain confirmation from the client when the guessed password is correct.
5. Protocols that work on the base of zero-knowledge proof of password possession are preferable: Zero-knowledge means that the server does not need to know the password to prove that the client knows the password. Passwords are never stored on the server therefore they cannot be stolen.

Some protocols encrypt the exchange of information to avoid the plain-text equivalence. Others used a form of asymmetric key exchange (a la Diffie-Helmann) that are generated based on the password but do not leak any information about it.

The following is a list of the some of the commonly used password schemes, classified by its strength:

Weak (not to be used)

• Encoded passwords (HTTP Basic Authentication)
• Classic challenge-response protocols (HTTP Digest Authentication, Windows NTLM Authentication, APOP, CRAM, CHAP, etc.)
• One-Time Password schemes based on a human memorable (low entropy) secret (S/Key, OPIE)
• Kerberos V4

Pseudo-Strong (they have known vulnerabilities in some implementations)

• SSH Public Key Authentication or “Secure Shell”,

• Kerberos V5 – To fix the password security weaknesses in Kerberos V4, version 5 added preauthentication, (the client needs to prove knowledge of the password before the server starts authentication).

Strong

• Secure Remote Password (SRP) – Developed in 1997 by Wu, is a strong password authentication protocol now widespread among Open Source and commercial products. SRP does not expose passwords to either passive or active network intruders, and it stores passwords as a “non-plaintext-equivalent” one-way hash on the server. SRP is available as part of standard Telnet and FTP implementations, and is being rapidly incorporated into Internet protocols that require strong password authentication.
• Encrypted Key Exchange(EKE) – Developed by Bellovin & Merritt in 1992 is one of the earliest examples of secure password protocols.
• Strong Password Exponential Key Exchange (SPEKE) developed by David Jablon . It is licensed by Entrust for their TruePass product.
• Diffie-Helmann Encrypted Key Exchange (DH-EKE)
• AMP
• SNAPI
• AuthA
• OKE
• Variations of all of the above.

Passwords have been the main tool for verifying identity and granting access to computer resources.In general, as FIPS 112 defines it, a password is a sequence of characters that can be used for several authentication purposes.

There are two security problems with a password; 1) somebody other than the legitimate user can guess it; or 2) it can be intercepted (sniffed) during transmission by a third party. Over the years, many different kinds of password generation methods and password protection protocols were designed to address  hese two weaknesses.

The security (strength) of a password is determined by its Shannon entropy , which is a measure of the difficulty in guessing the password. This entropy is measured in Shannon bits. For example, a random 10-letter English text would have an estimated entropy of around 15 Shannon bits, meaning that on average we might have to try $\frac{1}{2}(2^{15}) = 2^{14} =16384$ possibilities to guess it. In practice, the number of attempts needed would be considerably less because of side information available and redundancy (patterns and lack of randomness).  Since most human users cannot remember long random strings, a major weakness of passwords is that the entropy is usually too small for security. Even if the user can construct long passwords using an algorithm or a fix rule, the same rule may be know or guess by hackers.

It is well-known to hackers that users commonly select passwords that include variations of the user name, make of the car they drive, name of some family member, etc.  Social engineering is one of the most powerful tools being used by hackers.

Because of limitations in the underlying infrastructure, some authentication systems (notably banks) limit the number of characters and the alphabet from which the characters can be chosen. These kind of limitations are susceptible to dictionary-type attacks. In a dictionary  or brute force attack, the hacker will attempt to gain access using words from a list or dictionary. If the actual password is in the list, it can be obtained (in average)  when about half of the total number of possibilities have been tried. Even with systems that limit the number of trials for a user this is a potential security risk, because the hacker has good chances at gaining access by randomly trying names and passwords until a valid combination is found.

It is commonly accepted that with current tools, up to 30% of the passwords in a system can be recovered within hours. Moreover it is predicted that even random (perfect) passwords of 8 characters will be routinely cracked with technology available to most users by the year 2016.

There are many documents that give rules and policies for good password selection such as NIST Special Publication 800-63 and SANS security Project.

CrypTool  (ver 1.4)  has a very good tool to check the strength of a password against several criteria such as the amount of entropy and the resistance to dictionary attacks.

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