Question

prove that cube root of 26 is irational

Answer 1

Suppose cube root of 26 is rational.

Then there exist with such that ,

now 2 divides 26 so 2 divides   and so 2 divides .

2  divides , since 2 is a prime so 2 divides implies 2 divides .

for some .

Substituting value of in we get ,

........

Since p is divisible by 2 and so 2 does not divides q .

Now 2 does not divides q ,

8 does not divides

But 8 divides .

So 8 divides but 8 does not divides so they cannot be equal .

Which is a contradiction to .

So our assumption  cube root of 26 is rational is wrong .

Hence  cube root of 26 is irrational.

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