Question

# Prove that $$5\sqrt { 3 }$$ is an irrational.

Solution

## If possible, let $$5\sqrt{3}$$ is rational.Let its simplest form be $$5\sqrt{3}=\dfrac{a}{b}$$, where a and b are positive integers having no common factor other than 1.Then,$$5\sqrt{3}=\dfrac{a}{b}$$$$\sqrt{3}=\dfrac{a}{5b}$$Since, $$a$$ and $$5b$$ are non-zero integers, so $$\dfrac{a}{5b}$$ is rational.Thus, $$\sqrt{3}$$ is rational.Let simplest form of $$\sqrt{3}$$ be $$\dfrac{a}{b}$$.Then, a and b are integers having no common factor other than 1, and b$$\neq$$0.Now,$$\sqrt{3}=\dfrac{a}{b}$$$$3b^{2}=a^{2}$$3 divides a$$^{2}$$.     [3 divides 3b$$^{2}$$]$$3$$ divides $$a$$.Let $$a = 3c$$ for some integer $$c$$.Therefore,$$3b^{2}=9c^{2}$$$$b^{2}=3c^{2}$$$$3$$ divides b$$^{2}$$.      [3 divides 3c$$^{2}$$]$$3$$ divides $$b$$.Thus, $$3$$ is a common factor of $$a$$ and $$b$$.But this contradicts the fact that $$a$$ nad $$b$$ have no common factor other than 1.Thus, $$\sqrt{3}$$ is irrational.Hence, $$5\sqrt{3}$$ is irrational.Mathematics

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