A Strong Induction Proof

blackpenredpen,
math for fun

[frozenmoon]

9:17 we all know you are very kind, but this is pure gold.

[grawuka]

Response to question asked at 1:53
Let a, b integer co-prime non 0 such that a/b + b/a is in Z
If a/b is in Z and isn't +/-1, b/a isn't in Z, thus a/b + b/a isn't in Z. Same for b/a.
So a/b and b/a are either +/-1 or not in Z, if they are not in Z by multiplying both side with b:
a + b^2/a = kb
kb - a = b^2/a, left side in Z but right side isn't because a and b are co-prime
Therefore there isn't any other solution other than a/b = +/-1

[insanesamp]

I'm actually curious about what went through your head at 8:59 lol

[Martynas-Pocius]

The glitch in the matrix: 8:59, and finally caught exception 9:17

[SciDiFuoco]

For the question asked at 1:53:
Just stick with r and solve the quadratic equation
r + 1/r = k
r² -kr + 1 = 0
Δ = k² - 4
We get a rational r if and only if Δ is a perfect square (Δ = d²). But note that b² is also a perfect square. We then get k² - d² = 4.

The only squares which difference is only 4 are with k = ±2 and d = 0. This can be proven considering the following cases:


For k=0, k=1 then and k=2 we can prove by hand. (we get respectively Δ=-4, Δ=-3 and Δ=0)


For k>=3 then:
If d = k-1 then k²-d² = k² - (k-1)² = 2k - 1 > 4 which is true for k>=3.
If d > k-1 then k²-d² <= 0.

If d < k-1 then k²-d² > k² -(k-1)² > 4.
This shows that k²-d² can never be 4 if k>=3



For k<=-3 it's the same but considering k+1 instead of k-1 and swapping some signs

[eliyasne]

2:20
It is *impossible*

Proof by contradiction:
Say there was a solution a, b such that a and b are relatively prime.
That would mean:
a/b+b/a €Z
aa/ab+bb/ab €Z
(a^2+b^2)/ab €Z
=> for some number n :
a^2+b^2=nab
Looking at the equation in terms of mod a :
a^2+b^2 Ξ 0 mod a
b^2 Ξ 0 mod a
=><= (a, b are supposed to be relatively prime)
[|||]

[gaaraofddarkness]

8:59 me when i go to speak to my crush...

[topakyou]

8:59 I though my neighbor turned off their wifi

[eurotrash]

8:59 me throughout the entire duration of my discrete math final exam

[nahkaimurrao]

Thank you, it was bugging me that the Fibonacci closed form is basically infinite pairs of irrational numbers adding to be integers, but this proves it.

[iceIceCold]

No one:
My cat when it sees a bird outside the window: 8:59

[xxxprawn]

12:40 the quarantine is turning bprp into pierre de fermat

[VaradMahashabde]

8:58 I watch things at 2x speed and that made me go "How am I at normal speed? Is this slowed?!"

[ivarangquist]

1:53 the expression: a/b + b/a can be expanded into (a^2 + b^2) / ab. Assuming a and b are coprime, the top part of the fraction isn’t divisible by either a nor b. Therefore, the expression cannot be an integer.

[pinedelgado]

THANK YOU LOTS for your wonderful and amazing math videos!! You're one of my ALL-TIME FAVORITE YouTube personalities and hosts and I LOVE, LOVE, LOVE your channel!!!! Keep up the grand work!!!! :) :)

[nozua]

8:58 I never thought I'd laugh this hard at a bprp video

[lucadamian]

9:12 when there is the answer given in a test and you thougt you had ist correct before you checked the solution...

[yoelfeinmesser]

This is a question from the 2014/15 British maths olympiad 1 (albeit half of it)

[franciscocirelli]

Once you prove r belongs to the integers you can prove r can only be one or minus one(r^n + (1/r^n) = r^2n + 1/r^n. As this results in an integer, r^n divides r^2n + 1. Since r^n divides r^2n, r^n must divide 1, meaning r = 1 or r = -1)

[ngalenor]

If we can assume true for k-1 we can assume true for k+1 and promblem