Question

Prove that root 3 - root 7 is irrational

Answer 1

root 3=1.414....which is irrational number and root 7=2.645....which is also an irrational number.So root 3 and root 7 are irrational numbers

Answer 2

Step-by-step explanation:

Rational number is given in fraction form say p/q where q and p are integers.

Let us prove the statement by contradiction.

Let us assume that $\sqrt{3}-\sqrt{2}$ is rational.

That implies \sqrt{3}-\sqrt{2} can be expressed in the form of $\frac{p}{q}$

$\sqrt{3}-\sqrt{2}=\frac{p}{q}$

Squaring on both sides,

$3+2-2 \sqrt{6}=\frac{p^{2}}{q^{2}}$

$5-2 \sqrt{6}=\frac{p^{2}}{q^{2}}$

$5-\frac{p^{2}}{q^{2}}=2 \sqrt{6}$

Again squaring on both sides,

$25+\frac{p^{4}}{q^{4}}-10 \frac{p^{2}}{q^{2}}=24$

$\Rightarrow \frac{p^{4}}{q^{4}}-10 \frac{p^{2}}{q^{2}}=-1$

A fraction cannot be equal to an integer.

This is a contradiction

$\sqrt{3}-\sqrt{2}$ cannot be a rational.

Therefore, $\sqrt{3}-\sqrt{2}$ is irrational.