Question

If \[7\] is a prime number then how to prove that \[\sqrt{7}\] is irrational?

Answer 1

Hint: We can prove it by contradictory method that is we will assume that \[\sqrt{7}\] is rational number and try to write it as \[\dfrac{p}{q}\] form where \[q \ne 0\] and conclude that our assumption is wrong and hence proved.

Complete step by step answer:
In this question we have given \[7\]is a prime number and let us assume that \[\sqrt{7}\] is a rational number.
As we know that any rational number can be represented in \[\dfrac{p}{q}\]form ; \[q\ne 0\].
So, we can say that there exist coprime positive integers \[a\]and \[b\]such that \[\sqrt{7}\,=\dfrac{a}{b}\]
Now multiply by \[b\] on both sides we get, \[\sqrt{7}b=a\].
On squaring both sides we will get,
Which implies that \[{{a}^{2}}=7{{b}^{2}}\]
Therefore, it means that \[{{a}^{2}}\] is divided by \[7\].
And hence, we can say that $a$ is divide by \[7\]
Now, we will substitute the value of \[a\] in equation \[\left( 1 \right)\] and we get, \[49{{m}^{2}}\,=\,7{{b}^{2}}\]
Which implies that \[{{b}^{2}}=7{{m}^{2}}\]
If further implies that \[{{b}^{2}}\] is also divisible by \[7\].
And Hence, we can say that \[b\] is also divisible by \[7\].
Therefore, it implies that \[a\]and \[b\] have at least one common factor that is \[7\].
But this is contradiction to our assumption that \[a\] and \[b\] are, co-prime,
Thus, our assumption that \[\sqrt{7}\] is rational number is wrong
Which implies that, \[\sqrt{7}\] is an irrational number.

Note: The irrational numbers are all the real numbers which are not rational numbers. That is irrational numbers are those numbers that cannot be expressed as the ratio of two integers.
Also, we can say that irrational numbers have decimal expansions that neither terminate nor become periodic.
Among irrational number are the ratio \[\pi \] of \[a\] circle’s circumference to its diameter, Euler’s number \[e\], the golden ratio \[\theta \], and the square root of two natural numbers, other than of perfect squares are irrational.