Question 91198
Prove that cube root of 7 is an IRRATIONAL number ?
Assume that cube rt 7 is rational.
Then (7)^(1/3) = a/b where a and b are integers and a/b is reduced to lowest terms.
Then a=b[7^(1/3)]
Since a is a multiple of b and a is an integer, b divides a.
Since b divides a, a = nb and n is an integer.
Therefore 7^(1/3) = a/b = nb/b, so a/b is not reduced to lowest terms.
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What led to this contradiction?
The assumption that 7^(1/3) was rational.
The assumption must be wrong.
Therefore 7^(1/3) if irrational.
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Cheers,
Stan H.