Question

prove that 3+√5 is irrational​

Answer 1

Answer:

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

Here, {(a - 3b) ÷ b} is a rational number. ... → a and b both are co-prime numbers and 5 divide both of them. So, 3 + √5 is a irrational number.

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Answer 2

$\large\bf\underline {To \: prove:-}$

• 3+√5 is irrational

$\huge\bf\underline{Solution:-}$

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

then,

$\longmapsto\rm\:3 + \sqrt{5} = \frac{a}{b}$

where, a and b are co - primes and b ≠ 0

$\longmapsto\rm\:3 + \sqrt{5} = \frac{a}{b} \\ \\ \longmapsto\rm\: \sqrt{5} = \frac{a}{b} - 3 \\ \\ \longmapsto\rm\: \sqrt{5} = \frac{a - 3b}{b}$

So a and b are intigers , and then $\rm\: \frac{a - 3b}{b}$ is rational.

Then √5 is Also rational .

but we know that√5 is irrational , So this contradiction is arissen because of our wrong assumption.

therefore , 3 + √5 is irrational.

hence Proved.