Prove that root 3 + root 7 is irrational
​

√3+√7 is irrational.

Step-by-step explanation:

Let us assume that √3+√7 is rational.

That is , we can find coprimes a and b (b≠0) such that \sqrt{3}+\sqrt{7}=\sqrt{a}{b}

3

+

7

=

a

b

Therefore,

\sqrt{7}=\frac{a}{b}-\sqrt{3}

7

=

b

a

−

3

Squaring on both sides ,we get

7=\frac{a^{2}}{b^{2}}+3-2\times \frac{a}{b}\times \sqrt{3}7=

b

2

a

2

+3−2×

b

a

×

3

Rearranging the terms ,

\begin{gathered}2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}}{b^{2}}+3-7\\=\frac{a^{2}}{b^{2}}-4\end{gathered}

2×

b

a

×

3

=

b

2

a

2

+3−7

=

b

2

a

2

−4

\implies 2\times \frac{a}{b}\times \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}⟹2×

b

a

×

3

=

b

2

a

2

−4b

2

\begin{gathered}\implies \sqrt{3}=\frac{a^{2}-4b^{2}}{b^{2}}\times \frac{b}{2a}\\=\frac{a^{2}-4b^{2}}{2ab}\end{gathered}

⟹

3

=

b

2

a

2

−4b

2

×

2a

b

=

2ab

a

2

−4b

2

Since, a and b are integers , \frac{(a^{2}-4b^{2})}{2ab}

2ab

(a

2

−4b

2

)

is rational ,and so √3 also rational.

But this contradicts the fact that √3 is irrational.

This contradiction has arisen because of our incorrect assumption that √3+√7 is rational.

Hence, √3+√7 is irrational.

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step-by-step explanation:

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