3:37 it's a quadratic in ab which is even easier I think.

1 & 3 imply 3

@tw5718

Жыл бұрын

Legendary

For step 2, you have a quadratic equation in ab so you can automatically solve it and we also know that a+b=ab.

@SkeletalBasis

Жыл бұрын

Nice. But you don’t actually need to solve it. Just observe that ab = 1 is not a solution. The art in problems like this is to never calculate the individual values of a and b.

If two imply the third, we may take all three to be true. You derive a contradiction in step 1, so by contradiction, we don't need to check the other cases.

@drpeyam

Жыл бұрын

But where’s the fun in that?

@colleen9493

Жыл бұрын

Oh wow I didn’t even think of that.

In the first set, didn't you just prove that if one set of two are true, the third won't be true, so it can't possibly have all three be true at once bc you'd always be able to go through the first argument and say a and b can't be real

This is a very interesting question! I used WolframAlpha to find a Groebner Basis for the ideal generated by a+b-t, a^2+b^2-u, and ab(1/a+1/b-v). (We multiplied by ab because Groebner bases apply only to polynomials, no fractions allowed.) What I got was 3 pieces: t^2*v - 2*t - u*v, 2b^2 - 2bt + t^2 - u, a + b - t. How we read this is that there’s one equation that relates t, u, and v together, one equation that lets you determine b from t, u, and v, and one equation that lets you determine a from all the other variables. So the main equation for our purposes is t^2*v = 2t + uv. Plugging in 1, 1, 1 doesn’t work, obviously, but it tells us that if we have values for u and v, there’s generically 2 values for t, but knowing t and one of u and v, there’s generically one value for the other. What about the problem of dividing by 0? We know that a and b must both be nonzero; plugging those into the last two equations, if t^2=u, then that’s a problem. The first equation then tells us that the problem case, plugging u=t^2 in for u, is when t=0. So we need to avoid the solution t=u=0; but t=v=0 and u is anything nonzero works! That gives a=-b, and u=2b^2. Anyway, the point is that Groebner bases are cool, and I think you should talk about them on your channel!!

Super entertaining!

If we proved that two of the equations contradict the third, have we not already proved that all of them cannot be true at the same time? Let me explain: it is one thing if "A and B" does not necessarily imply C, in that case it would still be possible for "A and B and C" to be true; but here we proved that "A and B" implies "not C", so it is impossible for "A and B and C" to be true. So, technically you could have stopped at the first option (exploring the other ones as you did is interesting, though).

I knew firsthand that there wouldn't be any implications. Even a common solution doesnt exist!

n = (a+b) n = 1/a + 1/b = (a+b)/ab = n/ab n = a^2 + b^2 = (a+b)^2 - 2ab = n^2 - 2 n = -1, 2 Which 2 implies the 3rd? n=2 ;)

👍👍👍 very relaxing video 6:20

Very nice video

Dammit Peyam... I was doing so well. Years of not losing...

Perhaps if A^N + B^N =1, then A^M+ B^M can never equal 1 for distinct integers N and M? I think you could show this with the binomial series of (A+B)^N and (A+B)^M

@rsa5991

Жыл бұрын

That's not true. All positive powers share trivial solutions (1, 0), (0, 1). Positive even powers also share (-1, 0), (0, -1). All negative odd powers share a limit solution (+0, -0). Now to the interesting part: negative and positive powers of different parity have a nontrivial solution! In the video, you already saw that N=2, M=-1 have a common point. But so does N=-2 and M=1 - and any two powers of different sign and parity.

You’re kinda done after the first round aren’t you? The first two imply the falsehood of the third, meaning they can’t all simultaneously be true. If any two imply the third, all three may simultaneously be true, but that can’t happen, so no two can imply the third. With that being said, watching the three rounds is still very interesting.

simple

One thing is certain: There are only 3C2 =3 attempts for this problem.

I lost the game.

Yes. I was got very fun with that.. Wtb high degree polynomial? 🤭

Oh , my gad you are a mazing thank you doctor

'ímply', not 'impies'. Nice video!

Am I first here 🌠🌠

What if we use your short where we got 1=-1 in the first round? ;)