Question

2) Prove that 3 + √5 is an irrational number.​

Answer 1

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✳ Prove that 3 + √5 is an irrational number.

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➡ According to the question,

Considering, 3 + √5 is a rational number.

Now,

3 + √5 = (a ÷ b)

[Here a and b are co-prime numbers]

√5 = [(a ÷ b) - 3]

√5 = [(a - 3b) ÷ b]

Here, {(a - 3b) ÷ b} is a rational number.

But we know that √5 is a irrational number.

So, {(a - 3b) ÷ b} is also a irrational number.

So, our assumption is wrong.

3 + √5 is a irrational number.

Hence, proved.

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Answer 2

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Rational numbers:

• Rational numbers are the numbers that can be written in the form of p/q where p and q are integers and q is not equal to zero.
• Example 5/7, 1/4, 9/5 etc.

Given:

• We have been given a number 3 + √5.

To Prove:

• We need to prove that 3 + √5 is an irrational number.

Proof:

Let us assume that 3 + √5 is a rational number.

Therefore, 3 + √5 can be written in the form of p/q where p and q are coprime.

=> 3+√5 = a/b

=> √5 = a - 3b/b

√5 is an irrational number. Irrational number can never be equal to a rational number.

Hence, our assumption is wrong.

Hence 3 + √5 is an irrational number.