Question

✩Question:-
Prove that $\sqrt{p}$+$\sqrt{q}$ is irrational, where p and q are primes

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•Step wise explanation needed.​

Answer 1

Answer:

First, we'll assume that

p

and

q

is rational , where p and q are distinct primes

p

+

q

=x, where x is rational

Rational numbers are closed under multiplication, so if we square both sides, we still get rational numbers on both sides.

(

p

+

q

)

2

=x

2

p+2

pq

+q=x

2

2

pq

=x

2

−p−q

pq

=

2

(x

2

−p−q)

Now, x, x

2

, p, q, & 2 are all rational, and rational numbers are closed under subtraction and division.

So,

2

(x

2

−p−q)

is rational.

But since p and q are both primes, then pq is not a perfect square and therefore

pq

is not rational. But this is contradiction. Original assumption must be wrong.

So,

p

and

q

is irrational, where p and q are distinct primes.

Step-by-step explanation:

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Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$ $\huge\bold\blue{iS}$ $\huge\bold{iN}$ $\huge{\underline{\underline{\mathrm{\green{AttaChMenT}}}}}$

Let √p + √q

be rational

⇒ √p + √q

[where a, b are co-primes and integers]

Squaring both sides

⇒(√p+q²)=(a/b)²

⇒p+q+2√pq=(a/b)²

⇒ 2√pq = a^2/b^2 -p-q

⇒2√pq=a²-pb²-qb²/b²

⇒√p =a²-pb²-qb²/2b²√q

….I

∴ from Eq. I our assumption contradicts here, because p is rational

Hence, √p + √q is irrational number.

$\huge{\underline{\underline{\mathrm{\green{thanku}}}}}$

three answers attached. hope you understand.