Prove that \sqrt{5} is irrrational

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

zeros of a polynomial

Question

Prove that

\sqrt{5}

is irrrational

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Assume that

\sqrt{5}

is rational.

Hence,

\sqrt{5}

can be written in the form of

\frac{a}{b}

where

a

and

b

are co-prime and

b \neq 0

Hence

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

\Rightarrow a = b \sqrt{5}

\Rightarrow a^{2} = 5 b^{2}

by squaring on both sides

\Rightarrow \frac{a^{2}}{5} = b^{2}

Hence

5

divides

a^{2}

By theorem,if

p

is a prime number,and

p

divided

a^{2}

, then

p

divides

a

, where

a

is a positive number.

So,

5

shall divide

a

also. ......

(1)

Hence, we can say

\frac{a}{5} = c

where

c

is some integer.

So,

a = 5 c

Now we know that

a^{2} = 5 b^{2}

Put

a = 5 c

we get

25 c^{2} = 5 b^{2}

\Rightarrow b^{2} = 5 c^{2}

\Rightarrow \frac{b^{2}}{5} = c^{2}

Hence,

5

divides

b^{2}

So,

5

divides

b

also. ......

(2)

(1)

and

(2)

5

divides both

a

and

b

Hence

5

is a factor of

a

and

b

So,

a

and

b

have a factor

5

∴, a

and

b

are not co-prime.

Hence, our assumption is wrong.

∴

by contradiction,

\sqrt{5}

is irrational.