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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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