Question

Prove that 5-3√5 is an irrational number.​

Answer 1

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let us assume 5 - 3√5 to be a rational number

so we can say that 5 -3√5 = a/b , where a and b are co-prime integers

5 - 3√5 = a / b

3√5 = 5b / a

√5 = 5b / 3a

where 5,3,a and b are non negative rational numbers.

so √5 must be a rational number

but this contradicts the fact that√5 is irrational

so our assumption is wrong

and thus, 5 - 3√5 is irrational number.

• hope it helps..

If so..

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Answer 2

Answer:

First prove √5 as irrational.

Step-by-step explanation:

If possible let √5 be rational.

Let √5 be = p/q,

where 1. p and q are integers

2. q is not equal to 0

3. p and q are co-prime

Therefore,

√5=p/q

5=p²/q² (on squaring both sides)

p²=5q² (5×some integer q²=p²)--------(1)

Therefore, 5 is a factor of p²

Thus, 5 is a factor of p.---------(2)

Now,

let p² = 5k(5×some integer k)

(5k)²=5q²(from 1 we get)

25k²=5q²

5k²=q²

Thus, 5 is a factor of q²

thus,. 5 is a factor of q.--------(3)

(2) and (3) are contradicting our claim that √5 is rational because 5 is a factor of both p and q and thus we cannot express √5 as p/q or rational. So √5 is irrational.

Now,

let 5-3√5 be rational = p/q where,

1. p and q are integers

2. q is not equal to 0

3. p and q are co-prime

5-3√5=p/q

√5= -3(p/q-5)--------(4)

√5 irrational= -3(p/q-5) rational

(4) is contradicting our claim that 5-3√5 is rational because a irrational no. can never be equal to rational no. So 5-3√5 is irrational.