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10336546 No.10336546 [Reply] [Original] [archived.moe]

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

>> 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 consider

a + b = 2k + 2l = 2(k+l)

Since k+l is again an int. It must be that a+b is even.

>> No.10336581

>>10336558
Bump

>> 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

>> No.10336877

>>10336546
https://www.amazon.com/Proofs-BOOK-Martin-Aigner/dp/3642008550

>> No.10336883
File: 61 KB, 674x646, 1547782747920.jpg [View same] [iqdb] [saucenao] [google] [report]
10336883

>>10336860
(-0.1, 0.6), (0.4, 1.1) and random open infinite sets contained in [1, 2].

>> No.10336898

>>10336883
Ah, a subset of the family. My bad.

>> No.10336911
File: 285 KB, 1122x1281, 13.jpg [View same] [iqdb] [saucenao] [google] [report]
10336911

If [math] G [/math] is a group such that [math] a \in G \implies a^2 = e[/math], [math] e [/math] is the identity, then [math] G [/math] is abelian.

Let [math] a,b \in G [/math] then [math]ab \in G[/math] and [math] (ab)^2 = e [/math] and [math] (ab)^2 = abab = e[/math]
then [math] a(abab)b = aeb = ab[/math],
[math] aababb = a^2 ba b^2 = ab[/math],
[math] ebae = ba = ab[/math]
[math]a, b[/math] were both arbitrary so all elements commute.

>> 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 as
x= 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 get
y = 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.

>> 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!

>> No.10336926

>>10336911
abba=aea=e
But abab=e, then ab=ba

>> No.10336936

>>10336926
Trivially destructed. Thanks for proving again that Ruki posters have low iq.

>> No.10336942

>>10336926
>>10336936
using the inverse is just fucking boring

>> No.10336943

>>10336942
>using the inverse is just fucking boring
what did he mean by this?

>> No.10336953

>>10336936
>Ruki
Huh?

>> No.10337021

>>10336936
>not wanting the proof to work for Monoids too
what did he mean by this

>> No.10337042

>>10337021
Are you retarded? Your claim assumes that every element has an inverse (itself). Jeez

>> No.10337265

>>10336546
bump lets keep this one going guys good shit so far i took acreenshots because i'm a antichad brainsplat

>> No.10337283

>>10336911
Trash proof

>> No.10337477
File: 60 KB, 955x1023, Euler's_formula.png [View same] [iqdb] [saucenao] [google] [report]
10337477

>>10336546
[math]
\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*}
[/math]

>> No.10337487
File: 44 KB, 383x499, duckscrete math.jpg [View same] [iqdb] [saucenao] [google] [report]
10337487

proposition: 2^(1/n) is irrational for n > 2
proof: 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.

>> No.10337757

>>10337477
Figuring this out for myself was so fucking satisfying.

>> No.10337773

>>10337487
see I read this at first, thought to myself, "hey that's pretty cool"

then I was like, wait, no, that's troll math

but then I read through it again and yeah that seems to work

gg anon

>> No.10337790

>>10337773
well, i'm not sure whether FLT requires that result, but i doubt it does

>> No.10338893
File: 73 KB, 612x792, Logez02.jpg [View same] [iqdb] [saucenao] [google] [report]
10338893

>>10337757
It was reading that original proof for [math]e^{i\theta} = \cos{\theta} + isin{\theta}[/math] and then deducing from there that [math]\ln{a + bi} = \frac{1}{2} \ln{\left(a^2 + b^2 \right)} + i\arctan{\frac{b}{a}}[/math] 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 [math]e^{i\theta} = \cos{\theta} + i\sin{\theta}[/math]. I call it slightly bullshit because it pretty much already required me to know the end result. But, hey, it's valid. I think.
[math]
\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*}
[/math]
If I fucked up the [math]\LaTeX[/math], it's because I'm a dumb mobileposter at the moment.

>> No.10338984

>>10337790
Lmao. 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.

>> No.10339008

>>10337477
cool, now prove [eqn]\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}[/eqn]
plz

>> No.10339236

What are the steps to putting together a proof?

>> No.10339282

>>10339008
check wikipedia page Gaussian integral, brainlet

>> No.10339294

>>10339236
1. Think
2. Write down your thoughts

>> No.10339318
File: 37 KB, 600x485, 5zeyQks.jpg [View same] [iqdb] [saucenao] [google] [report]
10339318

>>10336546
Yes

>> No.10340538

>>10339008
Just let I be the integral and solve for I^2 in polar coordinates

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