Question

prove that root 2 + root 5 is irrational

Answer 1

Let √2+√5 be a rational number.

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

√2+√5 = p/q

Squaring on both sides,

(√2+√5)² = (p/q)²

√2²+√5²+2(√5)(√2) = p²/q²

2+5+2√10 = p²/q²

7+2√10 = p²/q²

2√10 = p²/q² - 7

√10 = (p²-7q²)/2q

p,q are integers then (p²-7q²)/2q is a rational number.

Then √10 is also a rational number.

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

.°. Our supposition is false.

√2+√5 is an irrational number.

Hence proved.

Answer 2

Answer:

To prove,

√2 + √5 is irrational.

Recall the concepts,

Rational numbers are the numbers that can be expressed in the form $\frac{p}{q},$ where p and q are integers and q≠0

Solution:

Required to prove that √2 + √5 is irrational.

Let us assume that √2 + √5 is rational, then there exists two coprime integers such that 'a' and 'b' such that

√2 + √5 = $\frac{a}{b}$

√2   = $\frac{a}{b}$ - √5

Squaring on both sides

(√2)²   = ($\frac{a}{b}$ - √5)²

2 = $\frac{a^2}{b^2}$ + 5 - 2√5$\frac{a}{b}$

2√5$\frac{a}{b}$ = 2 = $\frac{a^2}{b^2}$ + 5 -2 = $\frac{a^2}{b^2}$ + 3

√5  = $\frac{a^2+3b^2} {2ab}$

Since a and b are integers, we can say that a ²+3b² and 2ab are also integers.

From this, we can say that √5 is rational, as it is expressed in the form $\frac{p}{q}$ , where p and q are integers.

But we know that √5 is irrational, hence 'our assumption that √2 + √5 is rational' is wrong.

Hence √2 + √5 is irrational.

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