Question

Prove that $7\sqrt 5 $​ is an irrational number.

Answer 1

Hint: In these types of questions remember rational numbers can be represented in form of $\dfrac{a}{b}$ where b$ \ne $0 also use the method of contradiction to prove that $7\sqrt 5 $ cannot be represented in the form of $\dfrac{a}{b}$.

Complete step-by-step answer:
We know that all the real numbers which do not belong to rational numbers family are irrational numbers.
Let us assume that $7\sqrt 5 $​ is a rational number
As we know any rational number can be written in form of $\dfrac{a}{b}$ where b​$ \ne $0
so$7\sqrt 5 $ can be written in the form of $\dfrac{a}{b}$ ​ where b​$ \ne $0 .
⟹$7\sqrt 5 $=$\dfrac{a}{b}$​
⟹$\sqrt 5 $​=$\dfrac{a}{{7b}}$​
But as we all know that$\sqrt 5 $​ is an irrational number and $\dfrac{a}{{7b}}$​ is rational.
So as we know that if rational numbers are multiplied or divided by any irrational number or vice versa, then the resultant number will always be an Irrational number. Here also we got that there is a rational and an irrational number. both the numbers are being multiplied.
So, the number $7\sqrt 5 $​ is an irrational number.


Note: The term rational number can be explained as; a rational number is a number which can be represented as a quotient or a fraction p / q of two integers, a number p and a non-zero denominator q as q will be equal to 1 therefore every integer will be equal to a rational number.