Answer
In order to prove that $\sqrt{2}+\sqrt{3}$ is irrational, we can use proof by contradiction
Work Step by Step
We have to show that
$\sqrt{2}+\sqrt{3}$
This can be proved by using $Proof$ $by$ $Contradiction$
Suppose, for the sake of contradiction that $\sqrt{2}+\sqrt{3}$ is not irrational.
Since $\sqrt{2}+\sqrt{3}$ Is not irrational, $\sqrt{2}+\sqrt{3}$ is rational. By the definition of rational, there exist integers a and $b \neq 0$ such that:
$\sqrt{2}+\sqrt{3}=\frac{a}{b}$
Square each side of the previous equation:
$(\sqrt{2}+\sqrt{3})^{2}=\left(\frac{a}{b}\right)^{2}$
Use the property
$(a+b)^{2}=a^{2}+2 a b+b^{2}$
Simplify:
$2+2 \sqrt{6}+3=\frac{a^{2}}{b^{2}}$
Combine like terms:
$5+2 \sqrt{6}=\frac{a^{2}}{b^{2}}$
Subtract 5 from each side:
$2 \sqrt{6}=\frac{a^{2}}{b^{2}}-5$
Write the difference as one fraction:
$2 \sqrt{6}=\frac{a^{2}-5 b^{2}}{b^{2}}$
Divide each side of the equation by 2:
$\sqrt{6}=\frac{a^{2}-5 b^{2}}{2 b^{2}}$
As we know that $a$ and $b$ are integers, $a^{2}-5 b^{2}$ and 2$b^{2}$ are also integers. Moreover $2b^{2}$ is nonzero as $b$ is nonzero.
By the definition of rational, $\sqrt{6}$ is then rational. However $6$ is not a perfect square and thus the given statement tells us this then $\sqrt{6}$ that is irrational (as $\sqrt{n}$ is irrational when $n$ is not a perfect square), which means that we have derived a contradiction.
Thus our supposition that “$\sqrt{2}+\sqrt{3}$ is not irrational” is false which means that $\sqrt{2}+\sqrt{3}$ is irrational. This implies that supposition has been proven.
