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

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that

5 \sqrt{3}

is an irrational.

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If possible, let

5 \sqrt{3}

is rational.

Let its simplest form be

5 \sqrt{3} = \frac{a}{b}

, where a and b are positive integers having no common factor other than 1.Then,

5 \sqrt{3} = \frac{a}{b}

\sqrt{3} = \frac{a}{5 b}

Since,

a

and

5 b

are non-zero integers, so

\frac{a}{5 b}

is rational.

Thus,

\sqrt{3}

is rational.

Let simplest form of

\sqrt{3}

\frac{a}{b}

Then, a and b are integers having no common factor other than 1, and b

\neq

Now,

\sqrt{3} = \frac{a}{b}

3 b^{2} = a^{2}

3 divides a

^{2}

. [3 divides 3b

^{2}

]

3

divides

a

Let

a = 3 c

for some integer

c

Therefore,

3 b^{2} = 9 c^{2}

b^{2} = 3 c^{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.