Question

Hello guyz

Question --> Prove that root 3 + root 5 is an irrational number.​

Answer 1

Answer :-

Step-by-step explanation :-

Let √3 + √5 is a rational number .

°•° √3 + √5 = p/q where q ≠ 0 and p and q are co-prime number .

==> √3 = p/q - √5 .

Now, squaring both side .

==> (√3)² = ( p/q - √5 )² .

==> 3 = p²/q² + ( √5) ² - 2(p/q)(√5) .

==> √5 × 2p/q = p²/q² + 5 - 3 .

==> √5 = ( p² + 2q² )/q² × q/2p .

==> √5 = ( p² + 2q² )/2pq .

$\therefore$ p , q are integers then ( p² + 2q² )/2pq is a rational number.

Then, √5 is also a rational number .

But this contradicts the fact that √5 is an irrational number.

So,our assumption is false.

Therefore, √3 + √5 is an irrational number.

Answer 2

✴ola!!✴

⬇Solution⬇

Root 3

Let root 3 = a/b where , (a,b = 1,b is not equal to 0)

squaring ,

3b^2 = a^2

3 divides L.H.S. of (1)

Therefore , 3 divides R.H.S of (1)

therefore , 3 divides a^2 i.e., 3 divide a

therefore a = 3k

3 divide b^2 i.e.,

3 divides b

therefore , a and b both are multiple of 3 which is a contradiction . [ because, a and b has no common factor]

Therefore , root 3 is a irrational number.

Root 5

Let root 5 = a/b , where (a,b) = 1 , b is not equal to 0

squaring,

5 b^2 = a^2

5 divides L.H.S of (1)

Therefore, 5 divides R.H.S of (1)

Therefore , 5 divides a^2 , i.e., 5 divides a

therefore, a=5k

From (1) 5b^2 = 25k^2,

i.e.,5 = 5k^2

5 divides b,

therefore, a and b both are multiple of 5 which is contradiction.[because, a and b has no common factor]

Therefore , Root 5 is a rational number.

Root 3 and Root 5 is an irrational number .

Therefore , Sum of irrational number is irrational.✔✔

Hope it helps u✌

tysm❤❤