So I'm to teach myself proofs, and I'm struggling v a particular proof. Contradiction has actually been complicated for me because you kind of walk in blind concerning the statement C, for which you space trying to do a contradiction out of.

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

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

My Proof. intend for the sake of contradiction that cubeRoot(2) is rational. Climate cubeRoot(2) = a/b for part integers a,b. Letting both a and b be fully reduced, this suggests that they have opposite parity. For this reason let a it is in even and also b it is in odd. Then a = 2j and also b = 2k + 1, for part integers j,k. Substituting and cubing both sides us get, 2 = <(2j) / (2k + 1)>3 = 8 j3 / (2k + 1)3. Because 2 is an integer and 8 j3 / (2k + 1)3 is rational, the equation is false. QED

According come the book this is incorrect. The book starts as ns did, yet doesn't immediately substitute. Instead they cube both sides and also proceed as follows: 2 = a3 / b3 , climate 2 b3 = a3 which is even. This implies a is even so a = 2d. Substituting they get 2 b3 = (2d)3 = 8 d3. Separating they get, b3 = 4 d3 = 2 (2 d3 ), for this reason b3 is even, which indicates b is even. This is a contradiction since earlier in the proof the was declared that a and also b have actually opposite parity and also were fully reduced.

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


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

How does it indicate that?

Now that ns think around it, you're right, it doesn't. I was thinking, "well, if the portion is completely reduced, climate the parities should be opposite, otherwise just how else would certainly it be totally reduced." however I currently realize that's wrong, (for example 5/13 is fully reduced and both space odd).

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

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