Proof thread

[Anonymous No.6749084]

Fellow mathematics students...

What was the first proof you proved that gave you that feeling of being a mathematician?

For me, it would have been that all metrizable topological spaces are Hausdorff spaces.

Also, any specific proofs you delight in? Share!

[Anonymous No.6749093]

Inclusion exclusion. I was 16.

[Anonymous No.6749101]

>>6749093
For how many sets?

[Anonymous No.6749115]

I've never proved anything on my own. I'm a solutions manual cheater. I check the answer and then memorize the proof.

[Anonymous No.6749117]

>>6749115
Your exams must be easy if they are just proofs you have already seen. :(

[Anonymous No.6749124]

>>6749101
General case proof.

[Anonymous No.6749130]

I have done proofs in every subject I had on mathematics, but majority was form probability and stochastic processes.

One inequility stuck in my mind from calculus I. It's really not so trivial:

sin(tan(x)) >= tan(sin(x)) for x in [0,pi/4]

[Anonymous No.6749132]

>>6749124
Union of infinite sets, O.K.

[Anonymous No.6749135]

The first not-so-trivial proof I've found on my own was to show that:
If (X,d) is a metric space, then:
X sequentially compact --> X compact
I was quite proud of it because the proof is completely different from what our assistants suggested.

[Anonymous No.6749141]

>>6749135
My professor had a neat way of teaching. He always did proofs in another way that was arguably more difficult. That way we could try to come up with the simple proofs for it.

[Anonymous No.6749153]

>>6749141
This sounds like a good method for more advanced students who already know a lot of simple proofs. Where I'm studying, there's a lot of weekly exercices, so you wouldn't have time to find simpler proofs. Sounds like fun though :)

[Anonymous No.6749156]

>>6749153
Do the exercises pertain to proofs?

[Anonymous No.6749183]

>>6749156
There are only few numerical exercices, so almost all exercices are somehow about proving something. While doing them, it's mostly exasperating. But once you've found all the proofs, it's a really good feeling ;)

[Anonymous No.6749804]

>>6749084
For my intro to proofs homework we had to show that sqrt(2) is irrational. That was fun.

[Anonymous No.6749845]

>>6749084
That <span class="math">\mathcal{B}(\matbb{R}^n) = \sigma(\mathcal{E})[/spoiler]. In other words the Borel sets on <span class="math">\mathbb{R}^n[/spoiler] are equal to the smallest sigma algebra containing the set <span class="math">\mathcal{E} = \{(a_1,b_1)\times \cdots \times (a_n,b_n) : -\infty < a_j < b_j < \infty \}[/spoiler]

Proving that the open sets of <span class="math">\mathcal{R}^n[/spoiler] are contained in <span class="math">\sigma(\mathcal{E})[/spoiler] is the fun part.

[Anonymous No.6749848]

>>6749084
>>6749084
That <span class="math">\mathcal{B}(\mathbb{R}^n) = \sigma(\mathcal{E})[/spoiler]. In other words the Borel sets on <span class="math">\mathbb{R}^n[/spoiler] are equal to the smallest sigma algebra containing the set <span class="math">\mathcal{E} = \{(a_1,b_1)\times \cdots \times (a_n,b_n) : -\infty < a_j < b_j < \infty \}[/spoiler]

Proving that the open sets of <span class="math">\mathcal{R}^n[/spoiler] are contained in <span class="math">\sigma(\mathcal{E})[/spoiler] is the fun part.

[Anonymous No.6749856]

>>6749848
>>6749084
That <span class="math">\mathcal{B}(\matbb{R}^n) = \sigma(\mathcal{E})[/spoiler]. In other words the Borel sets on <span class="math">\mathbb{R}^n[/spoiler] are equal to the smallest sigma algebra containing the set <span class="math">\mathcal{E} = \{(a_1,b_1)\times \cdots \times (a_n,b_n) : -\infty < a_j < b_j < \infty \}[/spoiler]

Proving that the open sets of <span class="math">\mathbb{R}^n[/spoiler] are contained in <span class="math">\sigma(\mathcal{E})[/spoiler] is the fun part.

[Anonymous No.6749864]

>>6749084

That <span class="math">\mathcal{B}(\mathbb{R}^n) = \sigma(\mathcal{E})[/spoiler]. In other words the Borel sets on <span class="math">\mathbb{R}^n[/spoiler] are equal to the smallest sigma algebra containing the set <span class="math">\mathcal{E} = \{(a_1,b_1)\times \cdots \times (a_n,b_n) : -\infty < a_j < b_j < \infty \}[/spoiler]

Proving that the open sets of <span class="math">\mathbb{R}^n[/spoiler] are contained in <span class="math">\sigma(\mathcal{E})[/spoiler] is the fun part.

Also since I've been at this for a while I've contracted LaTeX autism. It sucks, but at least I have about 200 useful macros.

[Anonymous No.6749867]

fellow mathematics students:

when does a pentagon have 4 sides?

when it is intersected by a plane

[Anonymous No.6749872]

irrationality of all square roots of primes when i was 15

[Anonymous No.6749896]

f is a function that counts the prime factors of a natural number (example, f(12) = 3 because 12 = 3 * 2 * 2). For x < 2^n, f(x) < n. I remember I was really excited.

[Anonymous No.6749919]

the yoneda lemma

:^)

[Anonymous No.6749941]

I still find myself writing it - two dimensional waves and the Laplace equation. Incredibly long but amazingly satisfying

[Anonymous No.6750002]

When I proved FLT independently.

[Anonymous No.6750113]

• undergrad
• get an idea about number theory
• no sleep for three days, prove theorem
• go to library, search for related, find one
• ask hot librarian for reprint from 125 years ago
• my theorem proved by Georg Cantor
• depression sets in

[Anonymous No.6750137]

>that feel when you have many elegant, encompassing notions and ideas
>that feel when they most are lost in the autism of rigor and the reiteration of concepts that you got the first time

[Anonymous No.6750458]

>>6750113
Man I would be sad.

[Anonymous No.6750644]

>>6750113
If you did the research in the right order it would have saved you from losing that sleep. With our level of development almost anything really easy to prove is already proven or has a lot of evidence in support of it.

[Anonymous No.6751019]

not quite a feeling of being a mathematician, but the first one that stands out (which of course now seems trivial) is a proof that there are infinitely many pairs of natural numbers, a and n, such that the sum of the natural numbers up to a equals the sum of the natural numbers between a and n. Wasn't exactly elagant, but it had bugged 16-year-old me for weeks and there was a definite feeling of relief when it came together

[Anonymous No.6751023]

>everything easy to prove is proven
Kek

[Anonymous No.6751028]

>>6749084

The area of a circle


I tried, guys

[Anonymous No.6751040]

Cayley's theorem in group theroy.

[Anonymous No.6751416]

>>6751019

err i tried it and your theorem seems false, sorry

[Anonymous No.6751672]

>>6751416
>>6751019

my mistake it seems true

[Anonymous No.6752025]

Every mathematical proof is beautiful in its own way. There are those which are light and delightly, like a single raindrop upon the window pane; some are meant to be follow thread by thread, until the complete pattern is woven, the mystic fabric in your hands, each stream hidden by the others but made more vibrant than itself; still others are like the sounds of a waterfall, majestic, roaring, only to be beheld in awe and never to be approximated by human mind.

[Anonymous No.6752093]

{(1,2),(2,3),...,(n-1,n)} generates S_n, without making reference to disjoint cycle or product of transposition representations.

[Anonymous No.6752131]

My favorite proofs were the ones I did in high school geometry. By the end of the semester, I was a god at them. They were simple and easy to understand. 12+ line proofs impressed my runescape gfs. I learned a lot of fundamentals in that class, though my teacher was kind of a dick. Also I dropped out of school the next semester and didn't touch math for ~4 years.

[Anonymous No.6752515]

>>6751672
>>6751416
that's funny because I started out trying to show there was finitely many - since I could only find two and it seemed like something that should be fairly common - but in doing so proved the opposite

[Anonymous No.6752615]

>>6749115
indian?

[Anonymous No.6752637]

>>6752515

1 and 1
6 and 8
35 and 49
204 and 288
1189 and 1681
6930 and 9800
40391 and 57121

and probably many more

[Anonymous No.6752656]

>>6750113
Well at least it wasn't some no-name dude.

[Anonymous No.6752661]

>>6749124
You did measure theory at age 16?

[Anonymous No.6752665]

Not a proof per-se but in grade 6 or 7 we were learning about geometric shapes and I finished ahead of my class so my teacher told me to find a pattern in the number of edges/vertices/etc of a shape.

I came up with a general formula to find the number of vertices/edges given a shape and I remember I felt so proud and smart. I think I believed I was the first in the world to do it (being a naive middle school student).

Of course later I found out people have probably done much better than me at half my age, but I think pure mathematics would have a lifelong passion for me if I didn't go into neuroscience.

Sometimes I childishly wish I could live multiple lifetimes to experience and learn everything I'm missing out on.

[Anonymous No.6753875]

>>6752637
>>6752515
>>6751672
>>6751416
>>6751019
>>6751019


i found proof of that, anyone want it?

[Anonymous No.6753895]

Is there an elegant proof of this?:

Prove that given any sequence of 79 consecutive natural numbers, there is at least one number such that the sum of its digits is divisible by 13.

[Anonymous No.6753917]

>>6753895
Why do you think it's likely to be true?

[Anonymous No.6753918]

>>6753895
Wouldn't any thirteen consecutive numbers have that?

[Anonymous No.6753934]

>>6753918
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4

[Anonymous No.6753961]

>>6753917
Because it is...it was the second problem in a math olympiad I assisted on thursday. I proved it, but it is an ugly and long proof. The "Bonus track" was exhibit a sequence of
78 consecutive numbers such that none of them has its digits divisible by 13.

An example would be the sequence from 9999999961 to 10000000038. It's not an easy problem. (It's the only problem that was not exhibited in the oral exposition, meaning that nobody got it right).

>>6753918
No, see >>6753934

[Anonymous No.6753962]

>>6753961
*has the sum of its digits divisible by 13

[Anonymous No.6754023]

>>6753875
>>6752637
>>6752515
>>6751672
>>6751416
>>6751019
>>6751019

i'll give da proof anyway:
assume a and n being as you say: <span class="math"> \sum_{k=1}^a k = \sum_{k=a}^n k [/spoiler].
That is to say <span class="math"> \sum_{k=1}^a k = \sum_{k=1}^n k - \sum_{k=1}^{a-1} k [/spoiler].
It means <span class="math"> \frac{a(a+1)}{2} = \frac{n(n+1)}{2}-\frac{(a-1)a}{2} [/spoiler],
which also means that <span class="math"> \frac{n(n+1)}{2} = a^2 [/spoiler].

We are hence looking for an unbounded set of n such that <span class="math"> \frac{n(n+1)}{2} [/spoiler] is a squared natural number.
Let's build a sequence <span class="math"> n_k [/spoiler] of such numbers

let <span class="math"> a_0=1 [/spoiler] and <span class="math"> b_0=1 [/spoiler], and <span class="math"> a_{k+1}=a_k+b_k [/spoiler], <span class="math"> b_{k+1}=2a_k+b_k [/spoiler]
I say that (1):there is a <span class="math"> n_k [/spoiler] such that <span class="math"> (a_kb_k)^2=\frac{n_k(n_k+1)}{2} [/spoiler]
(then <span class="math"> n_k [/spoiler] works)

To prove (1), let's have an induction about k: there is a <span class="math"> n_k [/spoiler] such that either
(2):<span class="math"> a_k^2=\frac{n_k}{2} [/spoiler] and <span class="math"> b_k^2=n_k+1 [/spoiler], either
(3):<span class="math"> a_k^2=\frac{n_k+1}{2} [/spoiler] and <span class="math"> b_k^2=n_k [/spoiler]

First, (2)=>(1) and (3)=>1 is kinda obvious
Then, <span class="math"> a_0=1 [/spoiler] and <span class="math"> b_0=1 [/spoiler] satisfy (3) with <span class="math"> n_0=1 [/spoiler]

And the induction:
First case, if <span class="math"> a_k [/spoiler] and <span class="math"> b_k [/spoiler] satisfy (3) with <span class="math"> n_k [/spoiler],
then we have
<span class="math"> b_{k+1}^2=(2a_k+b_k)^2=4a_k^2+b_k^2+4a_kb_k=4\frac{n_k+1}{2}+n_k+4a_kb_k [/spoiler]
<span class="math"> b_{k+1}^2=2n_k+2+n_k+4a_kb_k=n_{k+1}+1 [/spoiler] if we call <span class="math"> n_{k+1}=3n_k+1+4a_kb_k [/spoiler]
<span class="math"> a_{k+1}^2=(a_k+b_k)^2=a_k^2+b_k^2+2a_kb_k=\frac{n_k+1}{2}+n_k+2a_kb_k=\frac{3n_k+1+4a_kb_k}{2}=\frac{n_{k+1}}{2} [/spoiler]
and we have property (2) for k+1 (...)

[Anonymous No.6754027]

>>6754023

(...) Second case, if <span class="math"> a_k [/spoiler] and <span class="math"> b_k [/spoiler] satisfy (2) with <span class="math"> n_k [/spoiler],
<span class="math"> b_{k+1}^2=(2a_k+b_k)^2=4a_k^2+b_k^2+4a_kb_k=4\frac{n_k}{2}+n_k+1+4a_kb_k [/spoiler]
<span class="math"> b_{k+1}^2=2n_k+n_k+1+4a_kb_k=3n_k+1+4a_kb_k [/spoiler], let's call that again <span class="math"> n_{k+1}=3n_k+1+4a_kb_k [/spoiler]
<span class="math"> a_{k+1}^2=(a_k+b_k)^2=a_k^2+b_k^2+2a_kb_k=\frac{n_k}{2}+n_k+1+2a_kb_k=\frac{3n_k+2+4a_kb_k}{2}=\frac{n_{k+1}+1}{2}} [/spoiler]
and we have (3) for k+1

Finally, we have an increasing <span class="math"> n_k [/spoiler] such that <span class="math"> (a_kb_k)^2=\frac{n_k(n_k+1)}{2} [/spoiler]
which is a square of a natural number <span class="math"> (a_kb_k)[/spoiler], so the pairs <span class="math"> (a_kb_k), n[/spoiler] do the job.

[Anonymous No.6754031]

>>6754027
*<span class="math"> a_{k+1}^2=(a_k+b_k)^2=a_k^2+b_k^2+2a_kb_k=\frac{n_k}{2}+n_k+1+2a_kb_k=\frac{3n_k+2+4a_kb_k}{2}=\frac{n_{k+1}+1}{2} [/spoiler]

[Anonymous No.6754045]

I love algebra. All you fucking analysisfags are just like surgeons. All you do is figuratively cut and sew. I'll take elegance, breadth, and structure all day 'eryday.

[Anonymous No.6754050]

prove that any NxN checker board with one corner removed can be covered by 3 square "L" shaped pieces

[Anonymous No.6754089]

>>6754050
i don't get the "L shaped pieces". can i have a picture?

[Anonymous No.6754277]

>>6752661
>you need measure theory to appreciate the inclusion-exclusion principle

Bitch it's useful for probability even, not just measures in general

[Anonymous No.6754294]

>>6749084
>metrizable topological spaces are Hausdorff spaces.
>Inclusion exclusion. I was 16.

everybody look this guy is special and needs recognition for his aptitude

[Anonymous No.6754330]

>>6754294
Neither of those are impressive though.

[Anonymous No.6754463]

>>6754050
Yeah, I'm not sure I get what you mean either. Couldn't it be covered by 1 very fat L shaped piece, i.e.: a checker board with one corner removed is L shaped.

[Anonymous No.6754617]

i derived x^2 = (sigma(z=1 to n) 2z-1)

i realized by adding every odd number it became a perfect square root, made the formula for it and found that it was on wiki, i was stunned