Is #sqrt(2)# a rational number?

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

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.

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.