Question

Prove that 3+√5 is an irrational number​

Answer 1

Proof:

Letus 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,

=>√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, it is contradictsour to our assumption.

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

3 + √5 is an irrational number

Hence proved.

Answer 2

To Prove:

✠ 3 + √5 is irrational number.

Proof:

Assume that 3 + √5 is a rational number of the form p/q where p & q are co-primes and q ≠ 0

➛ 3 + √5 = p/q

➛ √5 = p/q - 3

➛ √5 = (p-3q)/q

Here we may observe that the LHS is irrational but RHS is a rational number & this is a contradiction.

Hence our assumption is wrong!!

∴ 3 + √5 is irrational number.

More info:

• The square roots, cube roots, etc... of natural numbers are irrational numbers if their exact values cannot be obtained.

• A non-terminating and non-recurring decimal is an irrational number.

• The negative of an irrational number is always a rational.

• The sum of rational and an irrational number is always irrational.

• The product of a non-zero rational number and an irrational number is always irrational.

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