Question

Prove that $4+\sqrt{5}$ is irrational

Answer 1

Answer:-

first, prove that √5 is a irrational number

now,

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

Now,

Let us assume that 4+√5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒4+√5=p/q

=√5=p+4q/q

therefore,

p+4q/q is a rational number.

but wkt

√5 is a irrational number

so our assumption is wrong.

$\bf \therefore \: 4 + \sqrt{5} \: is \: a \: irrational \: number$

Answer 2

Let us assume that $\sqrt{5}$ is irrational,

Then,

$\sqrt{5}=\frac{p}{q}$  where p and q are co-prime and $q\neq0$

Squaring both sides,

$5=\frac{p^{2}}{q^{2}}$

$p^{2}=5q^{2}$ ....$(1)$

So, 5 is a factor of $p^{2}$

And in turn 5 is also a factor of $p$

Let,

$p=5m$  for some integer '$m$'

Squaring both sides,

$p^{2}=25m^{2}$

$5q^{2}=25m^{2}$     (from (1))

$q^{2}=5m^{2}$

So, 5 is a factor of $q^{2}$

And in turn 5 is also a factor of $q$

Thus, $p$ and $q$ have a common factor 5

But, this contradicts the fact that $p$ and $q$ are co-prime

Therefore, $\sqrt{5}$ is irrational

As the sum of a rational number and an irrational number is irrational,

$4+\sqrt{5}$ is irrational

Hence, proved

HOPE IT HELPS!

MARK AS BRAINLIEST!