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?
- testmathbpLv 41 decade agoFavorite Answer
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.
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".