Question

Show that (√3+√5)^2 is an irrational number.
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Answer 1

Answer:-

To Prove:-

• $\tt(\sqrt{3}+\sqrt{5})^2$ is a irrational number.

ID Used:-

• $\tt (a+b)^2 = a^2+b^2+2ab.$

So Now,

Let us first simplify $\bf(\sqrt3+\sqrt5)^2.$

$\\ \tt\mapsto( \sqrt{3} + \sqrt{5} ) {}^{2}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt= ( \sqrt{3}) {}^{2} + ( \sqrt{5}) {}^{2} + 2( \sqrt{3})( \sqrt{5}). \\ \\ \tt = 3 + 5 + 2 \sqrt{15}. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \tt = 7 + 2 \sqrt{15} . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now,

Let $\bf 7+\sqrt{15}$ be a rational number.

$\\ \tt\mapsto7 + 2\sqrt{15} = \dfrac{p}{q},$

$\\ \textrm{Where p and q are integers and q \neq 0.}$

$\\ \tt\mapsto 2 \sqrt{15} = \dfrac{p}{q} - 7.$

$\\ \tt\mapsto2 \sqrt{15} = \dfrac{p - 7q}{q} \: \: \rm(by \: \: taking \: \: lcm.)$

$\\ \tt\mapsto \sqrt{15} = \frac{p - 7q}{2q} = \frac{integer}{integer}.$

$\\ \tt\mapsto \sqrt{15} = \dfrac{integer}{integer} .$

▪︎ And we know that every rational number is in the form of p/q where p and q are integers and q is not equal to 0.

▪︎ And these all conditions are satisfied by $\bf\sqrt{15}$ by the above simplifications.

▪︎ So we can say that $\bf\sqrt{15}$ is a rational number.

▪︎ But it contradicts the fact that is it an irrational number.

▪︎ This contradiction has arisen because of our wrong assumption.

▪︎ So our assumption that $\bf(\sqrt{3}+\sqrt{5})^2$ Was wrong.

▪︎ And hence we can conclude that it is an irrational number.

...Hence Proved...