Question

Show that √a +√b is an irrational number, if √a and √b are irrational

Answer 1

To find :-

If √a and √b are irrational then there sum is irrational too.

CONCEPT:-

This is a very basic question of class 9th.

SOLUTION:-

Let √a+√b =x (Assumption) be a rational nymber

Since it is given that,

√a and √b are irrational then,

squarring both sides.

=> (√a+√b)²=x² ------> all rational (assumption).

=> a+b+2√ab =x²

since,

it is very clear that is √a and √b are irrational then a and b are rationals.

Ulitmately now if we prove that √ab is irrational then we solve the problem.

so,

If for example ,

√2 and √3 are the values of a and be

then √ab= √2x3 = √6

its irrational.

so, therefore, √ab is irational ... therefore,

√a+√b is irrational.

this contradicts our assumptions.

ans

Answer 2

✤✤✤Question:✤✤✤

Show that √a + √b is an irrational number, if √a and √b are irrational.

✤✤✤Solution:✤✤✤

Let us assume that,

√a + √b is a rational number,, then there will be two co-prime integers p and q such that,

$\sqrt{a} + \sqrt{b} = \frac{p}{q} \\ \\ \sqrt{a} = \frac{p}{q} - \sqrt{b} \\ \\ (squaring \: both \: sides) \\ \\ { (\sqrt{a} )}^{2} = {( \frac{p}{q} - \sqrt{b} ) }^{2} \\ \\ a = \frac{ {p}^{2} }{ {q}^{2} } + b - \frac{2p \sqrt{b} }{q} \\ \\ \frac{2p \sqrt{b} }{q} = \frac{ {p}^{2} }{ {q}^{2} } + b - a \\ \\ \frac{2p \sqrt{b} }{q} = \frac{ {p}^{2} + b {q}^{2} - a{q}^{2} }{ {q}^{2} } \\ \\ \sqrt{b} = \frac{ {p}^{2} + b {q}^{2} - a {q}^{2} }{2pq}$

here,

RHS is totally rational

but LHS is irrational

it shows our that our assumption was wrong that

√a + √b is rational

hence,

by it contradicts our fact

and thus

√a + √b is an irrational number.

✤✤✤✤Proved✤✤✤✤