Recent content by MandelbrotI

p was supposed to stand for prime... And no, no questions anymore.

Yes, I tried to say, that because of a/b is in it's smallest form, this automatically means a and b don't share any factors. But I can't think of a condition, the variable in the square root has to satisfy.

I am aware, that you can't generalise something with just one example being given. However, I assumed this in the first place, because I have already seen this way of proving the irrationality of primes with sqrt(2), sqrt(3) and others. At the start of this proof, we assume that sqrt(2) (or...

I appreciate your answer, and understand, that there are either certain guidelines a and b have to follow, or the proof is incorrect. In my initial statement, I left out an important detail. a and b can't share the same factors. This could lead to the proof being correct, as all counterexamples...

Fine. I still don't understand, which part of any proof here is correct. I came with the question whether my proof is correct. Seemingly it is not. I tried another approach, still no conclusions. Both of these statements are used in various proofs on the internet. You tell me, they are correct...

Yes, only for prime numbers.

Then the proofs on the internet are wrong. Thanks, the first insight I gained so far.

It is the reasoning of multiple "proofs" on the internet, including the video I posted.

Yes, the proof uses contradiction. But this contradiction only works, because, when p divides a2, p also divides a. So p divides a and b -> contradiction. But for that to happen, the statement that when x divides y2, x divides y, also has to be true. Only because of this you can say, that they...

I see that problem. The thing is, this statement is used by many people to proof that the square root of 2 is irrational. sqr(2) = a/b 2 = a^2/b^2 2*b^2 = a^2 At this point, it says: 2 divides a^2, so 2 divides a -> a = 2* some variable. I assumed this is correct. Of course, this is true as a...

Do you have a proof for that? Maybe this works: a*x=b^2 With the rule that a and b have to be integers, b can be split up into primes: b=p1*p2*...pn Using that for b^2: a*x=p1*p1*p2*p2...pn*pn We see that a has to divide one of these prime factors. And all of these factors can be found in...

I was trying to find a proof concerning this statement, but I couldn't find anything. I have tried to come up with my own solution: If a divides b^2, that means that : a*x=b^2 a*(x/b) = b This shows that b is divisible by a. Is this correct?