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# /sci/ - Science & Math

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Can someone show me what a proof is? Like simple, intermediate, and difficult examples? Are you supposed to prove theorems, lemmas, corollaries or what? I know the different types of proofs but I need a source for good examples. Anything in Logic, Abstract Algebra or Real Analysis
t. Brainless

 >> Anonymous Mon Jan 28 17:08:53 2019 No.10336558 Claim: An even number plus and even number is even. Proof: Let a, b be even numbers. Then a = 2k for k is an int and b = 2l for l an int. Then considera + b = 2k + 2l = 2(k+l)Since k+l is again an int. It must be that a+b is even.
 >> Anonymous Mon Jan 28 17:18:08 2019 No.10336581 >>10336558Bump
 >> Anonymous Mon Jan 28 19:09:52 2019 No.10336860 Easy example:If B is an infinite family of open sets whose union contains [0,1], then there is a finite subset of B which also has this property. Proof:1. Let x be the lowest upper bound of all real numbers in the interval [0,1] such that there exists a finite subfamily of B that covers [0,x].2. There is an open interval covering 0, so x is larger than 0.3. Assume x!=1. We will derive a contradiction. Pick an interval in B which contains x. Then it contains a number a bit smaller than x, call it a. Then there is a finite subfamily covering [0,a], so a finite family covering [0,x]. Then the open interval containing x also extends a bit further than x, so x is not the upper bound.QED
 >> Anonymous Mon Jan 28 19:15:58 2019 No.10336877
 >> Anonymous Mon Jan 28 19:17:51 2019 No.10336883 File: 61 KB, 674x646, 1547782747920.jpg [View same] [iqdb] [saucenao] [google] [report] >>10336860(-0.1, 0.6), (0.4, 1.1) and random open infinite sets contained in [1, 2].
 >> Anonymous Mon Jan 28 19:21:17 2019 No.10336898 >>10336883Ah, a subset of the family. My bad.
 >> Anonymous Mon Jan 28 19:25:40 2019 No.10336911 File: 285 KB, 1122x1281, 13.jpg [View same] [iqdb] [saucenao] [google] [report] If $G$ is a group such that $a \in G \implies a^2 = e$, $e$ is the identity, then $G$ is abelian.Let $a,b \in G$ then $ab \in G$ and $(ab)^2 = e$ and $(ab)^2 = abab = e$then $a(abab)b = aeb = ab$,$aababb = a^2 ba b^2 = ab$,$ebae = ba = ab$$a, b$ were both arbitrary so all elements commute.
 >> Anonymous Mon Jan 28 19:25:45 2019 No.10336912 Intermediate example:in ZF, the assertion that every vector space has a basis implies the axiom of multiple choice. This is the claim that for every family of sets (X_i), we can find a family of finite subsets (X'_i) such that X'_i is a subset of X_i for each i.Proof:1. Take a collection of nonempty sets X_i ( indexed by some set J e.g.)2. Let k(X) be the field of rational functions with indeterminates all elements of X_i.3. Let K be a subfield of k(X) consisting of all rational functions whose monomials in the numerator and denominator all have the same degree for each indeterminate. 4. Let V be the vector space over K spanned by all the indeterminates x in X_i (for all i in J).5. By assumption, this vector space has a basis.6. Fix one set in the collection, X_i. Pick an element x in X_i. Express it in the basis asx= sum a_i b_i, where a_i is in K and b_i is a basis element. Note that the sum is finite.7. For any other y in X_i, note that y/x is in the field K, so you can multiply the previous expression to gety = sum y (a_i /x) b_i. This is the unique way to express y since b_i are in the basis.8. This works for any element in X_i, meaning that (a_i/x) is dependent only on the set X_i and not on x. Same with b_i.9. Define F_i a subset of X_i to be those variables which occur in the aforementioned a_j/x for some j.10. For each i in J, this gives a unique finite subset of X_i.
 >> Anonymous Mon Jan 28 19:31:21 2019 No.10336924 Hard example:there are, like, a lot of numbers...Proof:1. Assume there aren't a lot of numbers, call the set {x_i}2. Pick the largest number y in {x_i}3. Now form the set {x_i + y}. Note that the sets are disjoint.4. Now we got twice as many numbers!
 >> Anonymous Mon Jan 28 19:31:33 2019 No.10336926 >>10336911abba=aea=eBut abab=e, then ab=ba
 >> Anonymous Mon Jan 28 19:34:42 2019 No.10336936 >>10336926Trivially destructed. Thanks for proving again that Ruki posters have low iq.
 >> Anonymous Mon Jan 28 19:37:15 2019 No.10336942 >>10336926>>10336936using the inverse is just fucking boring
 >> Anonymous Mon Jan 28 19:38:09 2019 No.10336943 >>10336942>using the inverse is just fucking boringwhat did he mean by this?
 >> Anonymous Mon Jan 28 19:41:08 2019 No.10336953 >>10336936>RukiHuh?
 >> Anonymous Mon Jan 28 20:04:44 2019 No.10337021 >>10336936>not wanting the proof to work for Monoids toowhat did he mean by this
 >> Anonymous Mon Jan 28 20:12:33 2019 No.10337042 >>10337021Are you retarded? Your claim assumes that every element has an inverse (itself). Jeez
 >> Anonymous Mon Jan 28 21:49:25 2019 No.10337265 >>10336546bump lets keep this one going guys good shit so far i took acreenshots because i'm a antichad brainsplat
 >> Anonymous Mon Jan 28 21:56:56 2019 No.10337283 >>10336911Trash proof
 >> Anonymous Mon Jan 28 23:49:58 2019 No.10337477 File: 60 KB, 955x1023, Euler's_formula.png [View same] [iqdb] [saucenao] [google] [report] >>10336546\begin{alignat*}{1}e^x &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \\&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\\therefore e^{ix} &= \sum_{k=0}^{\infty} \frac{\left(ix\right)^k}{k!} \\&= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \\\cos{x} &= \sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k}}{\left(2k\right)!} \\&= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \\\sin{x} &= \sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k+1}}{\left(2k+1\right)!} \\&= x - \frac{x^3}{3!} + \dots \\\therefore i\sin{x} &= i\sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k+1}}{\left(2k+1\right)!} \\&= ix - \frac{ix^3}{3!} + \dots \\\cos{x} + i\sin{x} &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \\&= e^{ix} \\e^{ix} &= \cos{x} + i\sin{x} \quad \square \\&\text{It follows that any nonzero complex number } z \text{ can be represented in standard form as } a + bi \text{ or in polar form as } re^{i\theta} \text{.} \\z = a + bi &= re^{i\theta} \\&= r\left(\cos{\theta} + i\sin{\theta}\right) \\a &= r\cos{\theta} \\b &= r\sin{\theta} \\\frac{b}{a} &= \frac{\sin{\theta}}{\cos{\theta}} \\&= \tan{\theta} \\\arctan{\left(\frac{b}{a}\right)} &= \theta \\a^2 + b^2 &= r^2\cos^2{\theta} + r^2\sin^2{\theta} \\&= r^2\left(\cos^2{\theta} + \sin^2{\theta}\right) \\&= r^2 \\\sqrt{a^2 + b^2} &= r \\z = a + bi &= \sqrt{a^2 + b^2}e^{i\arctan{\left(\frac{b}{a}\right)}} \quad \square \\&\text{It follows that any nonzero complex number } z \text{, being the product of a real constant } r \text{ and some power of } e \text{, has some natural logarithm } \ln{z} \text{.} \\ \ln{z} = \ln{\left(a + bi\right)} &= \ln{\left(\sqrt{a^2 + b^2}e^{i\arctan{\left(\frac{b}{a}\right)}}\right)} \\&= \ln{\sqrt{a^2 + b^2}} + i\arctan{\left(\frac{b}{a}\right)} \quad \square \\\end{alignat*}
 >> Anonymous Mon Jan 28 23:57:31 2019 No.10337487 File: 44 KB, 383x499, duckscrete math.jpg [View same] [iqdb] [saucenao] [google] [report] proposition: 2^(1/n) is irrational for n > 2proof: suppose for the sake of contradiction that there exist integers p, q such that 2^(1/n) = p/q.then 2 = p^n / q^n.but then p^n = 2q^n = q^n + q^n.by Fermat's Last Theorem, no such integers exist for n > 2.we have reached a contradiction, therefore 2^(1/n) is not rational for n > 2.
 >> Anonymous Tue Jan 29 01:55:52 2019 No.10337757 >>10337477Figuring this out for myself was so fucking satisfying.
 >> Anonymous Tue Jan 29 02:03:37 2019 No.10337773 >>10337487see I read this at first, thought to myself, "hey that's pretty cool"then I was like, wait, no, that's troll mathbut then I read through it again and yeah that seems to workgg anon
 >> Anonymous Tue Jan 29 02:20:53 2019 No.10337790 >>10337773well, i'm not sure whether FLT requires that result, but i doubt it does
 >> Anonymous Tue Jan 29 12:46:35 2019 No.10338893 File: 73 KB, 612x792, Logez02.jpg [View same] [iqdb] [saucenao] [google] [report] >>10337757It was reading that original proof for $e^{i\theta} = \cos{\theta} + isin{\theta}$ and then deducing from there that $\ln{a + bi} = \frac{1}{2} \ln{\left(a^2 + b^2 \right)} + i\arctan{\frac{b}{a}}$ that originally got me really interested in math.All because I wanted to show up to class the next day and smugly tell my teacher and the class that you CAN take the logarithm of a negative number after all! Well, you can, but the few who cared to stick around for my proof couldn't understand it.I did later come up with my own slightly bullshit proof for $e^{i\theta} = \cos{\theta} + i\sin{\theta}$. I call it slightly bullshit because it pretty much already required me to know the end result. But, hey, it's valid. I think.\begin{align*}e^{i\theta} &= a + bi \\\frac{d}{d\theta} \left(e^{i\theta} \right) &= \frac{d}{d\theta} \left(a + bi \right) \\&= ie^{i\theta} \\&= i\left(a + bi \right) \\\frac{d}{d\theta} \left(a \right) + i\frac{d}{d\theta} \left(b \right) &= -b + ai \\\frac{d}{d\theta} \left(a \right) &= -b \\\frac{d}{d\theta} \left(b \right) &= a \\a &= \cos{\theta} \\b &= \sin{\theta} \\a + bi &= r \left(\cos{\theta} + i\sin{\theta} \right) \\e^{i\theta} = \cos{\theta} + i\sin{\theta} \qquad \square\end{align*}If I fucked up the $\LaTeX$, it's because I'm a dumb mobileposter at the moment.
 >> Anonymous Tue Jan 29 13:20:33 2019 No.10338984 >>10337790Lmao. Look and learn:Assume an integer x is not a perfect n'th power. Then say x^(1/n) is rational p/q. x q^n = p^n. Do prime decomposition and notice that on the rhs every prime has a power =0 mod n, while on the left that's not true, contradiction. Hence x^(1/n) is irrational.
 >> Anonymous Tue Jan 29 13:24:42 2019 No.10339008 >>10337477cool, now prove [eqn]\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}[/eqn]plz
 >> Anonymous Tue Jan 29 14:20:04 2019 No.10339236 What are the steps to putting together a proof?
 >> Anonymous Tue Jan 29 14:27:47 2019 No.10339282 >>10339008check wikipedia page Gaussian integral, brainlet
 >> Anonymous Tue Jan 29 14:31:06 2019 No.10339294 >>103392361. Think2. Write down your thoughts
 >> Anonymous Tue Jan 29 14:38:44 2019 No.10339318 File: 37 KB, 600x485, 5zeyQks.jpg [View same] [iqdb] [saucenao] [google] [report] >>10336546Yes
 >> Anonymous Tue Jan 29 22:01:51 2019 No.10340538 >>10339008Just let I be the integral and solve for I^2 in polar coordinates
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