Question

Prove that $\sqrt 5$ is irrational and hence, $3 + \sqrt 5$ is irrational.

Answer 1

We will assume the contrary that $\sqrt 5$ is rational and bring on a contradiction to some actual fact, which will prove our assumption to be wrong and then, we will use the fact that the sum of a rational and an irrational number results into an irrational number.

Complete step-by-step answer:
Let us assume that on the contrary, if possible $\sqrt 5$ is rational.
Now, since the definition of rational numbers says that we can write any rational number in the form of $\dfrac{p}{q}$, where p and q are co-prime integers and $q \ne 0$.
So, $\sqrt 5 = \dfrac{p}{q}$ for some p and q coprime integers and $q \ne 0$.
Squaring both sides, we will get:-
$\Rightarrow 5 = {\left( {\dfrac{p}{q}} \right)^2}$ for some p and q coprime integers and $q \ne 0$.
We can write this as:-
$\Rightarrow 5 = \dfrac{{{p^2}}}{{{q^2}}}$ for some p and q coprime integers and $q \ne 0$.
Cross multiplying it to get the following expression:-
$\Rightarrow 5{q^2} = {p^2}$ for some p and q coprime integers and $q \ne 0……….(1)$.
Hence, ${p^2}$ is a multiple of 5. (By Euclid division lemma)
[According to Euclid's Division Lemma, if we have two positive integers a and b, then there would be whole numbers q and r that satisfy the equation: a = bq + r, where $0 \leqslant r < b$, a is the dividend and b is the divisor]
Hence, $p$ is a multiple of 5. (By Fundamental Theorem of Arithmetic)………..(2)
[the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors]
So, p = 5n.
Squaring it, we will get:-
$\Rightarrow {p^2} = 25{n^2}$
Putting this in (1), we will get:-
$\Rightarrow 5{q^2} = 25{n^2}$
On simplifying it, we will get:-
$\Rightarrow {q^2} = 5{n^2}$.
Now, applying the same facts on this, we will get that q is a multiple of 5. …………..(3)
By (2) and (3), p and q are both multiples of 5 and hence, not coprime.
This is a contradiction to our previous assumption.
Hence, our assumption was wrong and $\sqrt 5$ is irrational.
Now, we know that the sum of a rational and an irrational number always gives us an irrational number.
Now, that $\sqrt 5$ is irrational and 3 is rational. Therefore, $3 + \sqrt 5$ is irrational.

Note: The students must note that we used a fundamental theorem to get that p is a multiple of 5. Let us prove that $p$ is a multiple of 5 when ${p^2}$ is a multiple of 5 by contradiction:
Let if possible, p is not a multiple of 5. So, p = 5n + k, where $1 \leqslant k \leqslant 4$.
Now, squaring it, we will get:-
$\Rightarrow {p^2} = {\left( {5n + k} \right)^2} = 25{n^2} + {k^2} + 2 \times 5n \times k = 25{n^2} + {k^2} + 10kn$ , where $1 \leqslant k \leqslant 4$.
Rearranging the terms, we will get:-
$\Rightarrow {p^2} = 5(5{n^2} + 2kn) + {k^2}$, where $1 \leqslant k \leqslant 4$.
We see that ${k^2} = 1,4,9,16$ for $1 \leqslant k \leqslant 4$ , none of which is a multiple of 5.
Hence, ${p^2}$ is not a multiple of 5, which is a contradiction to what was given to us.
Hence, $p$ is a multiple of 5.