# How can one prove that the cube root of 2 is irrational?

I heard that this can be done through a proof by contradiction, but what is the proof?

### 2 Answers

- testmathbpLv 41 decade agoFavorite Answer
yes

1. suppose is rational so it is= p/q where p and q are integers with no common divisor.

2. raise to exponent of 3 and we have 2 = (p/q)^3 or 2q^3=p^3

3. in LHS we have an even number , so in RHS q must be even. let say p= 2 r where r is an integer.

4. Substitute in 2 q^3=p^3 we have 2 q^3=(2r)^3 or

2 q^3=8 (r^3) or if we divide by 2, q^3=4 (r^3).

5. Now in the RHS we have an even number, so the LHS must be even or q = 2 s, where s is an integer.

6. From the last relation (q = 2 s) and p= 2 r (obtained above), we conclude that q and p have 2 as a common divisor.

7. The steps 1 and 6 are contradictory.

Cheers!!!

NOTE

I used in the steps 3, and 5. the fact "if n^3 has a divisor k prime number, then n has the same divisor k".

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