Question

Prove that $\sqrt{7}$ is an irrational number.

Answer 1

Hint: First we need to assume the number to be rational and we need to find a way to contradict our assumption. Use definition of rational number, If rational number = $\dfrac{p}{q}$ . p, q must not have common factors.

Complete step-by-step solution -
First, we will prove that given variable: $\sqrt{7}$ is irrational
Assume $\sqrt{7}$ is rational
By basic algebraic concepts, we get: A number which can be written the form $\dfrac{p}{q}$ with p and q being in lowest terms is called rational number
By applying above concept, we get:
$\sqrt{7}=\dfrac{p}{q}$
Here we have p, q in lowest terms, that means there are no common factors between them.
By squaring on both sides, we get:
$7=\dfrac{{{p}^{2}}}{{{q}^{2}}}$
By multiplying ${{q}^{2}}$ on both sides we get: $7{{q}^{2}}={{p}^{2}}$
Since, the left-hand side and right-hand side are equal if 7 divides left hand side it must also divide right hand side.
So, we can write p = 7k, where k is any integer except zero.
By substituting this, we get:
$\begin{align}
  & 7{{q}^{2}}={{\left( 7k \right)}^{2}} \\
 & {{q}^{2}}=7{{k}^{2}} \\
\end{align}$
Since the left-hand side and right-hand side are equal if 7 divides right hand side it must also divide left hand side.
So, we can write q = 7k, where k is any integer except zero.
By above conditions, we can say:
7 divides both a and b.
So, we get:
7 is a common factor of a, b.
By definition of rational number, a number is said to be rational if and only if p, q are in lowest terms, there should not be any common factors.
But here we have a common factor 7. So, it contradicts our assumption. So, our assumption is wrong.
Therefore $\sqrt{7}$ is irrational.
Hence proved.

Note: The idea of proving 7 is a common factor for both numerator and denominator by cross multiplying is a crucial step. Do carefully, or else you might reach the wrong conclusion.