Question

[tex]prove \: that \: \sqrt{3} + \sqrt{5} \: is \: an \: irrational \: number.

Answer 1

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Hey Mate !!

Here's your answer !!

To Prove: √3 + √5  is irrational.

Proof: Let us assume √3 + √5 to be rational.

If they are rational, then certainly they must be of the form p / q.

But here q ≠ 0 and p,q have a HCF of 1.

So √3 + √5 = p / q

Squaring on both sides :

= ( √3 + √5 )² = ( p / q )² 

= (√3 )² + ( √5 )² + 2 × √3 × √5  = p² / q²

= 3 + 5 + 2√3 × √ 5 = p² / q²

= 8 + 2√3 × √5 = p² / q²

= 2√3 × √5 = p² / q² - 8

= 2√3 × √5 = ( p² - 8q² ) / q²

= √3 × √5 = ( p² - 8q² ) / q² × 2

= √15 = ( p² - 8q² ) / q² × 2

= ( p² - 8q² ) / q² × 2 is a rational number. 

=> √15 is a rational number.

But we know that √15 is irrational. 

Hence a contradiction has arisen.  Hence √3 + √5  is irrational.

Hope this helps !!

Cheers !!
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# Kalpesh :)

Answer 2

Hey friend, Harish here.

Here is your answer:

To prove:

√3 + √5 is irrational

Assumption.

Let us assume √3 + √5 to be rational and equal to a.

Proof:

As √3 + √5 is rational and equals to a then, a² must also be rational.

a² = (√3 + √5)² = 5 + 3 + 2×√(5×3) = 8 + √15.

As 8 + √15 is rational it must be of the form p/q, where q≠o & p,q are coprimes.

Then,

$8 + 2 \sqrt{15} = \frac{p}{q}$

→ $2 \sqrt{15}= \frac{p}{q} -8$

→ $2 \sqrt{15}= \frac{p-8q}{q}$

→ $\sqrt{15} = \frac{p-8q}{2q}$

We know that,

$\sqrt{15} \ is \ irrational.$

$\frac{p-8q}{2q} \ is\ rational$

Rational ≠ Irrational.

Therefore we contradict the statement that 8 + 2√15 is rational.

Hence √3 + √5 is irrational.
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Hope my answer is helpful to you.