Question

prove that 3+√5 is irrational number​

Answer 1

$\huge\underline\mathbb{\red S\pink{O}\purple{L} \blue{UT} \orange{I}\green{ON :}}$

Let us assume that 3 + √5 is a rational number.

$⟹3 + \sqrt{5} = \frac{a}{b}$

• [ a & b are co - primes ]

$⟹ \sqrt{5} = \frac{a}{b} - 3$

$⟹ \sqrt{5} = \frac{a - 3b}{b}$

Here, {(a - 3b) ÷ b} is a rational number.

But we know that √5 is a irrational number.

So, {(a - 3b) ÷ b} is also a irrational number.

So, our assumption is wrong. 3 + √5 is a irrational number.

Hence, proved.

$\underline{\boxed{\bf{\purple{∴ “ \: 3 + \sqrt{5} \: ”\;is\;an\;irrational\;number.}}}}$

Step-by-step explanation:

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Answer 2

Answer:

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