Question

Prove that √5 is irrational. Hence show that 3√5 is also irrational.

Answer 1

Answer:

Step-by-step explanation:

To prove : 3√5 is irrational number.

Proof :

Let us assume that √5 is rational

Then √5 = a/b

(a and b are co primes, with only 1 common factor and b≠0)

⇒ √5 = a/b

(cross multiply)

⇒ a = √5b

⇒ a² = 5b² -------> α

⇒ 5/a²

(by theorem if p divides q then p can also divide q²)

⇒ 5/a ----> 1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations α and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b-------> 2

We know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that √5 is a rational number

∴ Our assumption is wrong

∴ √5 is irrational number

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

Now,

3√5 = p/q

[Here p and q are co-prime numbers]

√5 = [p/3q]

√5 = [p/3q]

Here, {p/3q} is a rational number.

But we know that √5 is a irrational number but p/3q is a rational number. We know that LHS = RHS.

So, { p/3q} is also a irrational number.

So, our assumption is wrong.

3√5 is a irrational number.

Hence, proved.

Answer 2

Answer:

Step-by-step explanation:

Let us assume that $\sqrt{5}$ is rational

Then $\sqrt{5}$ = a/b

(a and b are co primes, with only 1 common factor and b≠0)

$\sqrt{5}$ = a/b (cross multiply)

⇒ a = $\sqrt{5}$b

⇒ a² = 5b² -------> α

⇒ 5/a² (by theorem if p divides q then p can also divide q²)

⇒ 5/a ----> 1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations α and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b-------> 2

We know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that $\sqrt{5}$ is a rational number

∴ Our assumption is wrong

∴ $\sqrt{5}$ is irrational number

Let us assume that 3 + $\sqrt{5}$ is a rational number.

Now,

3$\sqrt{5}$ = p/q

[Here p and q are co-prime numbers]

$\sqrt{5}$ = [p/3q]

$\sqrt{5}$ = [p/3q]

Here, {p/3q} is a rational number.

But we know that $\sqrt{5}$ is a irrational number but p/3q is a rational number. We know that LHS = RHS.

So, { p/3q} is also a irrational number.

So, our assumption is wrong.

3$\sqrt{5}$ is a irrational number.

Hence, proved.