Question

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

Answer 1

Answer:

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Step-by-step explanation:

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

Now,

3+5=ba [Here a and b are co-prime numbers]

5=[(ba)−3]

5=[(ba−3b)]

Here, [(ba−3b)] is a rational number.

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But we know that 5 is an irrational number.

So, [(ba−3b)] is also a irrational number.

So, our assumption is wrong.

3+5 is an irrational number.

Answer 2

Answer:

Answer:

Answer:

V

Given 3 + √5

To prove:3 + √5 is an irrational number.

Proof:

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

So it can be written in the form a/b

3 + √5 = a/b

Here a and b are coprime numbers and b

#0

Solving

3 + √5 = a/b

we get,

V

=>√5 = a/b -3

<

=>√5 = (a-3b)/b

=>√5 = (a-3b)/b

This shows (a-3b)/b is a rational number.

But we know that √5 is an irrational number, which contradicts our assumption.

Our assumption 3 + √5 is a rational. number is incorrect.

3 + √5 is an irrational number

Hence proved.

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Step-by-step explanation:

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