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Prove by contradiction that the cuberoot of 7 is irrational.Similar to prove that the...

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svjr | Student, Undergraduate | Honors

Posted November 21, 2011 at 2:28 AM via web

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Prove by contradiction that the cuberoot of 7 is irrational.

Similar to prove that the equation 7m^3=n^3

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beckden | High School Teacher | (Level 1) Educator

Posted November 21, 2011 at 6:51 AM (Answer #1)

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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.

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