Assume it is rational. Then there must be 2 coprime integers, p and q, such that p/q = ∛7
p^3 = 7q^3
Therefore p^3 is divisible by 7 and p is divisible by 7 as well.
Let p = 7k
(7k)^3 = 7q^3
343k^3 = 7q^3
49k^3 = q^3
Therefore q^3 is divisible by 7 and thus q is as well.
So both p and q are divisible by 7.
But by assumption these numbers are coprime so ∛7 is irrational.
A rational number can be written in fraction form. â7 cannot, so it is irrational.
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Assume it is rational. Then there must be 2 coprime integers, p and q, such that p/q = ∛7
p^3 = 7q^3
Therefore p^3 is divisible by 7 and p is divisible by 7 as well.
Let p = 7k
(7k)^3 = 7q^3
343k^3 = 7q^3
49k^3 = q^3
Therefore q^3 is divisible by 7 and thus q is as well.
So both p and q are divisible by 7.
But by assumption these numbers are coprime so ∛7 is irrational.
A rational number can be written in fraction form. â7 cannot, so it is irrational.