Prove that \sqrt { 3 } +\sqrt { 8 }  is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rationalisation

Question

Prove that

\sqrt { 3 } +\sqrt { 8 }&nbsp;

is an irrational number.

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Let

\sqrt{8}

be a rational number

Then

\sqrt{8} = \frac{m}{n}

where m, n are integers and m, n are coprimes and

n \neq 0

\Rightarrow m = \sqrt{8} n

Squaring both sides we get

m^{2} = 8 n^{2}

\Rightarrow \frac{m^{2}}{8} = n^{2}

……………(iv)

\Rightarrow 8

divides

m^{2}

i.e.,

8

divides m

Then m can be written as

m = 8

k for some integer k.

Substituting value of m in (iv) we get

\Rightarrow \frac{(8 k)^{2}}{3} = n^{2}

\Rightarrow 8 k^{2} = n^{2}

\Rightarrow k^{2} = \frac{n^{2}}{8}

\Rightarrow 8

divides

n^{2}

i.e.,

8

divides n

Thus we get that

8

is a common factor of m and n but m and n are co-primes which is a contradiction to our assumption.

Hence

\sqrt{8}

is an irrational number.

Now consider

\sqrt{3} + \sqrt{8}

to be an rational number

Then

\sqrt{3} + \sqrt{8} = \frac{a}{b}

where a, b are integers, co-primes and

b \neq 0

\Rightarrow \sqrt{3} = \frac{a}{b} - \sqrt{8}

Squaring both sides we get

\Rightarrow 3 = \frac{a^{2}}{b^{2}} + (\sqrt{8})^{2} - 2 \frac{a}{b} \sqrt{8}

\Rightarrow 3 = \frac{a^{2}}{b^{2}} + 8 - \frac{2 a}{b} \sqrt{8}

\Rightarrow 3 - \frac{a^{2}}{b^{2}} - 8 = - \frac{2 a}{b} \sqrt{8}

\Rightarrow \frac{3 b^{2} - a^{2} - 8 b^{2}}{b^{2}} = \frac{- 2 a}{b} \sqrt{8}

\Rightarrow \sqrt{8} = \frac{(3 b^{2} - a^{2} - 8 b^{2})}{- 2 a b}

\Rightarrow \sqrt{8} = \frac{a^{2} + 8 b^{2} - 3 b^{2}}{2 a b}

\Rightarrow \sqrt{8} = \frac{a^{2} + 5 b^{2}}{2 a b}

Where

a^{2} + 5 b^{2}

is an integer and

2 a b

is also an integer, but

\sqrt{8}

is an irrational number, which is a contradiction.

Hence

\sqrt{3} + \sqrt{8}

is an irrational number.