Prove by contradiction that the cuberoot of 7 is irrational.
Similar to prove that the equation 7m^3=n^3
1 Answer | Add Yours
Suppose r is the cube root of 7 and it is rational, and in lowest terms.
So I can write `r = n/m` where m and n are relatively prime.
Then `(n/m)^3 = 7` so `n^3 = 7m^3` . Since 7 is prime then n must be a factor of 7. So I can write n=7p. And our equation is `7^3p^3 = 7m^3`. This reduces to 7p^3=m^3, and I must infer that m is a multiple of 7 also from the same reasoning as n being a multiple of 7. But my assumption was that m and n are relatively prime so they cannot both be multiples of 7 and relatively prime. This is a contradiction, and therefore r is not a rational number.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes