Question

prove that root 3+root 5is an irrational number​

Answer 1

To Prove:-

• 3 + √5 is an Irrational Number.

How To Prove:-

• The most easiest method to solve these type of questions is to first assume it as true.

• At last we will get a loophole which will contradicate the fact.

To Proof

Assume that 3 + √5 is Rational number equal to x

Where,. x is integer rational number

So,

→ 3 + √5 = x

• Squaring Both sides

→ ( 3 + √5 )² = ( x )²

→ (3)² + 2 × 3 × √5 + (√5)² = x²

→ 9 + 6√5 + 5 = x²

→ 14 + 6√5 = x²

→ 6√5 = x² - 14

→√5 = (x² - 14)/6

if x is Rational number then x² will be also a rational number.

Here,

x² is equal to √5

But, It Contradicate the fact that √5 is irrational number. So, our Assumption is wrong.

Hence, 3 + √5 is irrational Number

Answer 2

Question:-

Prove that $\sqrt{3}$ + $\sqrt{5}$ is an irrational number.

Answer:-

To prove,

• $\sqrt{3}$ + $\sqrt{5}$ is an irrational number

Solution,

Let,

$\sqrt{3}$ + $\sqrt{5}$ be rational number A

$\sqrt{3}$ = $\sqrt{5}$ - a

Squaring on both sides,

${\sqrt{3}^{2}}$ = ${\sqrt{5} -a^{2}}$

$\implies$ 3 = 5 + ${a}^{2}$ - 2a$\sqrt{5}$

$\implies$ 2a $\sqrt{5}$ = 2+ ${a}^{2}$

$\implies$ $\sqrt{5}$ = $\dfrac{2 + {a}^{2}}{2a}$

Now,

• Here $\dfrac{2+{a}^{2}}{2a}$ is rational.
• And even $\sqrt{5}$ is also rational because the rational numbers of right is also rational.
• But $\sqrt{5}$ is a irrational number.
• By this our assumption was wrong as the $\sqrt{3}$ + $\sqrt{5}$ is not a rational number.
• So $\sqrt{3}$ + $\sqrt{5}$ is an irrational number.

Know More:-

Irrational number:-

The number which cannot be written as fraction is known as Irrational number.

Ex:- $\sqrt{5}$ , $\sqrt{16}$ etc.

Rational numbers:-

Any number that can be written as a fraction is known as Rational number.

Ex:- $\dfrac{3}{4}$ , $\dfrac{7}{2}$ etc.

-Happies!!