Question

PROVE THAT 7-3 ROOT 5 IS IRRATIONAL

Answer 1

Step-by-step explanation:

⇒ Assume that 7 - 3√5 is rational. So let x = 7 - 3√5, for a rational number x.

⇒ In x = 7 - 3√5, both sides are rational as assumed.

$\bullet\ x=7-3\sqrt{5} \\ \\ \bullet\ 3\sqrt{5}=7-x \\ \\ \bullet\ \sqrt{5}=\frac{7-x}{3}$

⇒ Here, as both sides of x = 7 - 3√5 are rational, then so should be √5 = (7 - x)/3.  But this contradicts our earlier assumption that 7 - 3√5 is rational, because the RHS of √5 = (7 - x)/3 is rational while the LHS is irrational.

⇒ Hence proved!!!

Answer 2

Let us assume that √5 is rational
So,
√5 = a/b. where a and b are integers such that they're coprime to each other.

So,
a^2 = 5b^2
So,
a^2/5 = b^2
which means 5 divides p^2
which indirectly also means, 5 divides p

Hence, we can say
a/5 = c where c is some other integer
So,

a = 5c
a^2 = 25c^2
5b^2 = 25c^2
b^2/5 = c^2
which means 5 divides b^2 and again indirectly that 5 divides b
This means, 5 divides both a and b
Hence, 5 is a factor of both a and b.
Therefore, a and b are not coprime
This contradicts our assumption and thus our assumption is false
Therefore, √5 is irrational number.

Now,
Let us assume that 7-3√5 is rational.
So,
7-3√5 = p/q where p and q are integers coprime to each other.
So,
(7-p/q) = 3√5
So,
√5 = (7-p/q)/3
Now, here at the Left hand side we have an irrational number.
and on the right hand side we have a rational number because even p and q are integers.
This contradicts the equality
and hence our assumption is incorrect , and ultimately hence, 7-3√5 is also an irrational number

Hope this helps you!