Question

Prove that square root 2 plus square root 3 is irrational

Answer 1

Answer:

Solution:

Let us suppose that √2+√3 is rational.

Let √2+√3=\frac{a}{b}

b

a

,

where a,b are integers and b≠0

Therefore,

\sqrt{2}=\frac{a}{b}-\sqrt{3}

2

=

b

a

−

3

On Squaring both sides , we get

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

b

2

a

2

+3−2×

b

a

×

3

Rearranging the terms ,

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

b

2a

×

3

=

b

2

a

2

+3−2

= \frac{a^{2}}{b^{2}}+1

b

2

a

2

+1

\sqrt{3}=\frac{a^{2}+b^{2}}{2ab}

3

=

2ab

a

2

+b

2

Since , a,b are integers ,

\frac{a^{2}+b^{2}}{2ab}

2ab

a

2

+b

2

is rational, and so √3 is rational.

This contradicts the fact √3 is irrational.

Hence, √2+√3 is irrational.

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