Question

prove that root 2 plus root3 is an irrational

Answer 1

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Answer 2

$\fontsize{18}{10}{\textup{\textbf{The proof is done below.}}}$

Step-by-step explanation:

Let us assume that $\sqrt2+\sqrt3$ is rational.

Then, according to the definition of rational numbers, there exists p and q (co-prime to each other, integers and q≠0) such that

$\sqrt2+\sqrt3=\dfrac{p}{q}\\\\\Rightarrow \sqrt2=\dfrac{p}{q}-\sqrt3\\\\\Rightarrow 2=\left(\dfrac{p}{q}-\sqrt3\right)^2~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Squaring both sides}]\\\\\\\Rightarrow 2=\dfrac{p^2}{q^2}-2\sqrt3\dfrac{p}{q}+3\\\\\\\Rightarrow 2\sqrt3\dfrac{p}{q}=\dfrac{p^2}{q^2}+1\\\\\\\Rightarrow \sqrt3=\dfrac{p}{2q}+\dfrac{q}{2p}\\\\\\\Rightarrow \sqrt3=\dfrac{p^2-q^2}{2pq}.$

Since p and q are integers, so the fraction on the right-hand side of the above equation is rational.

This implies that √3 is also rational, which is a contradiction.

Thus, our assumption is wrong and hence $\sqrt2+\sqrt3$ is rational.

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Question : Prove that root 3-root 2 is irrational.

Link : https://brainly.in/question/1224229.