Question

Prove that 2 - 3√5 is an irrational number.

Answer 1

so, assume that 2+3root 5 is rational.

2+3root5=p/q,where p and q are co-primes and q not equals to 0.

3root 5=p/q-2.

root 5=p/q-2/3.

A rational number never equals to an irrational number.

We assume that p and q are co-primes and q not equals to 0.

so, our assumption is wrong.

2+3root 5 is an irrational number

Answer 2

Rational numbers

Any number which can be expressed in the form $\dfrac {p}{q}$ for two integers $p,\ q$ (may or may not be coprimes) such that $q\neq0$ are called rational numbers.

E.g.: 2, 7, 5/3, 0, -2/5, 0.333..., etc.

Irrational numbers

Unlike rational numbers, any number which cannot be expressed in the form $\dfrac {p}{q}$ for two integers $p,\ q$ (may or may not be coprimes) such that $q\neq0$ are called irrational numbers.

E.g.: √2, √3, π, e, etc.

=================================

We're given to prove that 2 - 3√5 is irrational. We prove the statement by the method of contradiction.

We first assume that 2 - 3√5 is a rational number. Let me call it as 'x'. So we have,

$x=2-3\sqrt5$

Now we are going to express √5 in terms of x from this equation. So,

$x=2-3\sqrt5\\\\2-x=3\sqrt5\\\\\dfrac {2-x}{3}=\sqrt5$

Now, consider the final equation. Here the LHS of the equation is rational since x is assumed to be rational, but the RHS, √5, is an irrational number. In this equation we see that √5, being an irrational number, is expressed in fractional form, which is not possible.

So we have arrived at the contradiction, and hence proved that 2 - 3√5 is an irrational number.

#answerwithquality

#BAL