Question

Prove that√ 2 + √ 5 is irrational​

Answer 1

Answer:

√2+√5 is a irrational number.

Step-by-step explanation:

Let √2 + √5 be rational number

A rational number can be written in the form of p/q where p,q are integers.

√2+ √5 = p/q --(where, q ≠ 0)

Squaring on both sides,

(√2+√5)^2 = (p/q)^2

(√2)^2 + (√5)^2 +2 (√5) × (√2) = p^2 /q^2

2+5+2√10 = p2/q ----( since, (√a)^2 = a ; and √a × √b =√a×b)

7+2 √10 = p^2/q^2

2-√10 = p^2/q^2 - 7

√10 = (p^2-7q^2) / 2q

p,q are integers then (p2-7q)/2q is a rational number.

Then √10 is also a rational number.

But this contradicts the fact that √10 is an irrational number.

therefore, Our supposition is false.

√2+√5 is an irrational number.

Hence proved.

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hope it helps....

Answer 2

Answer:

√2+√5 is irrational

Step-by-step explanation:

Let us assume,to the contradiction that √2+√5 is rational

√2+√5=a/b [where a and b are co-primes and $b\neq0$]

      √2=a/b-√5

squaring on bothsides,

(√2)²=(a/b-√5)²

                 ↓

            It is in the form of $(a-b)^{2}$ and we know that $(a-b)^{2} =a^{2} +b^{2}+2ab$

$a^{2}=(a/b)^{2} \\b^{2}=(\sqrt 5)^{2}$

(√2)²=(a/b)²+(√5)²-2×a/b×√5

     2=(a/b)²+5-2a/b×√5

2a/b×√5=a²/b²+5-2

2a/b×√5=a²/b²+3

2a/b×√5=a²+3b²/b²

         √5=a²+3b²/b×2a/b

         √5=a²+3b²/2ab

As a and b are integers,a²+2b²/2ab is rational,so √5 is irrational.But this contradicats the fact that √5 is irrational.This contradiction has arisen because of our correct assumption √2+√5 is rational.

∴Please mark it as brainlist answer