Is #sqrt(2)# a rational number?
2 Answers
See explanation.
Explanation:
The number
Let's assume that
If we raise this equation to the second power we get:
But this equality is not possible for integer
This is a contradiction which leads to the conclusion that the assumption is false, so
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.
Here is a basic proof by contradiction, just for fun.
Proof.
Suppose
#=>a=b*sqrt2#
#=>a^2=2b^2#
Therefore
Then
So,
#=>2k^2=b^2#
Therefore,
But, then