Question

prove that √3+√5 is an irrational number​

Answer 1

Answer:

Let us assume that 3 + √5 is a rational number. ... So, {(a - 3b)/b} should also be an irrational number. Hence, it is a contradiction to our assumption. Thus, 3 + √5 is an irrational number.

Step-by-step explanation:

3 + √5 = a/b

[Here a and b are co-prime numbers, where b ≠ 0]

√5 = a/b - 3

√5 = (a - 3b)/b

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

But we know that √5 is an irrational number.

So, {(a - 3b)/b} should also be an irrational number.

Hence, it is a contradiction to our assumption.

Thus, 3 + √5 is an irrational number.

Hence proved, 3 + √5 is an irrational number