# how to solve this- prove that cube root of 2 cannot be represented in the form p + q,p,q are rational nos (q>0 andis not a perfect square)

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### 1 Answer

Here is one way to look at the problem (although, it doesn't use the extra information about q)

If p and q are both rational numbers, then so is p+q

So if you can show that

`root(3)(2)` is not rational, then you are done

You can do this in the "standard way" that you can show that `sqrt(2)` is not rational:

(see: http://en.wikipedia.org/wiki/Square_root_of_2#Proof_by_infinite_descent )

Suppose (by contradition) that `root(3)/(2)` were rational.

Then you could write:

`root(3)(2) = (a)/(b)` where a and b are natural numbers (whole numbers), and the fraction `(a)/(b)` is in lowest terms (it is "irreducible")

Cube both sides:

`2=(a^3)/(b^3)`

So `2b^3=a^3`

The left hand side is even, because it is divisible by 2. So the right hand side must be even as well. But then `a` must be even. (If you cube an odd number, the result is an odd number.) If `a` is even, we can write it as `a=2k` , where `k` is a whole number.

Then `a^3= (2k)^3 = 8k^3`

So:

`2b^3=8k^3`

`b^3=4k^3`

Now the right hand side is even. So the left must be even as well. But then `b` must be even (since an odd cubed is odd).

But then `a` and `b` are both even, and we said they were in lowest terms.

So `root(3)(2)` can't be written as ANY rational, and so it can't be written as a sum of rationals.