Question

Prove that √ x +√y is irrational, where x and y are primes.
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Answer 1

Answer:

Let us assume that √x and √y are irrational no.

⇒√x+√y = a/b where '' a and b'' are integers , b ≠0

⇒ √x = a/b-√y  -----(1)

now squaring both side we get

⇒ x = ( a/b -√y )² = a²/b² +y -2ab√y

on solving we get

2a/b√y = a²/b²+y -x

⇒ on solving the equation we get

√y = xb²+a²-b²y/2ab

Irrational = Rational

Thus it is not possible ,so our assumption is wrong

Hence √x+√y is irrational.

Answer 2

$\bold \red \star \: \large \underline{Solution:-}$

Let assume √x +√y is rational no.

so,

$\sqrt{x} + \sqrt{y} = \frac{a}{b} \: \: \: \: \: \: \: \: \: \: where \: a \: and \: b \: are \: co \: prime \\ \\ {( \sqrt{x} + \sqrt{y} ) }^{2} = \frac{{a }^{2} }{{b}^{2}} \\ \\ x + y + 2 \sqrt{xy} = \frac{{a }^{2} }{{b}^{2}} \\ \\ 2 \sqrt{xy} = \frac{{a }^{2} }{{b}^{2}} - x - y. \\ \\ 2 \sqrt{xy} = \frac{{a }^{2} - x {b}^{2} - y {b}^{2} }{{b}^{2}}. \\ \\ \sqrt{xy} = \frac{{a }^{2} - x {b}^{2} - y {b}^{2} }{{2b}^{2}}$

Now, We can see

$\sqrt{xy} \: is \: a \: irrational \: number \\ \: and \: \: \: \: \frac{{a }^{2} - x {b}^{2} - y {b}^{2} }{{2b}^{2}} \: \\ is \: rational \: number.$

$Irrational \: \: \cancel = \: \: Rational$

So, Our assumption is wrong,

$\sqrt{x} + \sqrt{y}$

is irrational no. proved