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Prove by contradiction that the cuberoot of 7 is irrational.Similar to prove that the...
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High School Teacher
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.
Posted by beckden on November 21, 2011 at 6:51 AM (Answer #1)
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