Question

10. Show that 7 is not the cube of a rational number​

Answer 1

Answer:

Assume that cube root 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. 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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