So I'm to teach myself proofs, and I'm struggling with a specific proof. Contradiction has been challenging for me because you sort of walk in blind concerning the explain C, because that which you room trying to do a contradiction out of.

You are watching: Prove the cube root of 2 is irrational

Proposition: Prove that the cube source of 2 is irrational.

My Proof. expect for the services of contradiction that cubeRoot(2) is rational. Climate cubeRoot(2) = a/b for part integers a,b. Letting both a and b be totally reduced, this implies that they have opposite parity. Therefore let a it is in even and b it is in odd. Climate a = 2j and b = 2k + 1, for some integers j,k. Substituting and also cubing both sides we get, 2 = <(2j) / (2k + 1)>3 = 8 j3 / (2k + 1)3. Since 2 is an integer and 8 j3 / (2k + 1)3 is rational, the equation is false. QED

According to the publication this is incorrect. The book begins as i did, however doesn't automatically substitute. Instead they cube both sides and also proceed together follows: 2 = a3 / b3 , then 2 b3 = a3 i beg your pardon is even. This means a is even so a = 2d. Substituting they gain 2 b3 = (2d)3 = 8 d3. Splitting they get, b3 = 4 d3 = 2 (2 d3 ), therefore b3 is even, which indicates b is even. This is a contradiction due to the fact that earlier in the proof it was proclaimed that a and also b have opposite parity and were totally reduced.

Why is my way invalid? Or is it? Isn't the the situation that once you cube both political parties (even without substituting) the you would have actually 2 = a3 / b3, which is saying that an integer is equal to a rational number? i beg your pardon is false. What have I implicitly assumed the is incorrect?


Letting both a and also b be completely reduced, this means that they have actually opposite parity.

How go it imply that?

Now that i think about it, you're right, it doesn't. I was thinking, \"well, if the portion is fully reduced, then the parities must be opposite, otherwise just how else would it be fully reduced.\" yet I now realize that's wrong, (for example 5/13 is totally reduced and both room odd).

Your proof doesn't fairly work. First, it's incomplete; your proof through contradiction actually has two cases: a is also & b is odd and also a is odd & b is even. This is fairly minor, though, and both situations are usually identical, however you need to at least cite it. Second, girlfriend haven't really displayed that 2 = 8j3/(2k+1)3 leader to a contradiction. 2 is an integer, yet it is additionally rational, so that isn't quite enough. You must show an ext specifically the 8j3/(2k+1)3 is not one integer.

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