Question

Is `sqrt(2)` a rational number?

Answer 1

See explanation.

Explanation:

The number `sqrt(2)` is irrational. The proof can be as follows:

Let's assume that `sqrt(2)` is rational. Then we have:

`sqrt(2)=p/q` where `p,q in ZZ`

If we raise this equation to the second power we get:

`2=(p^2)/(q^2)`

`2q^2=p^2`

But this equality is not possible for integer `p` and `q`. If we analyze the prime factorizarion of both sides we can see that number `2` appears even number of times on the right hand side and odd number of times on the left side.

This is a contradiction which leads to the conclusion that the assumption is false, so `sqrt(2)` is an irrational number

Answer 2

The square root of 2 is not rational, i.e. it is irrational.

Explanation:

We can think of a rational number as that which can be expressed as a fraction or ratio of two integers.

i.e. a number is rational if there exists integers `p` and `q` such that the number can be expressed by the quotient `p/q`

The square root of 2 cannot be expressed as the quotient of two integers, and therefore is called an irrational number.

Irrational numbers do not terminate or repeat, and cannot be represented by a finite number of digits.

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Here is a basic proof by contradiction, just for fun.

Proof.

Suppose `sqrt2` is rational. Then there exists integers `a` and `b` such that `a/b=sqrt2`, and `gcd(a,b)=1`.

`=>a=b*sqrt2`

`=>a^2=2b^2`

Therefore `a^2` is even (a multiple of 2), and so `a` must also be even. Then there exists an integer `k` such that `a=2k`.

Then `a^2=(2k)^2=2b^2`.

So, `4k^2=2b^2`.

`=>2k^2=b^2`

Therefore, `b^2` is even, and so `b` must also be even.

But, then `gcd(a,b)!=1`, a contradiction.

`:. sqrt2` is irrational.