Question

Prove that \[\sqrt{3}\] is an irrational number. Hence, show that \[7+2\sqrt{3}\] is also an irrational number.

Answer 1

Hint: Take \[\sqrt{3}=\dfrac{p}{q}\] where p, q are coprime then square both the sides and do cross multiplication. After that compare with divisible by 3.

Complete step-by-step answer:

In the question we are asked to prove \[\sqrt{3}\] is an irrational number.
If we assume \[\sqrt{3}\] as a rational number and then prove our assumption is wrong, then our work will be done.
As we know that rational number is in the form \[\dfrac{p}{q}\] where p, q are co-prime numbers of H.C.F. as 1.
So, we can represent \[\sqrt{3}\] as a rational number by writing,
\[\sqrt{3}=\dfrac{p}{q}\]
Now, by squaring both sides we get,
\[3=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
Now, on cross multiplication we get,
\[3{{q}^{2}}={{p}^{2}}\]
So, we can say that 3 divides \[3{{q}^{2}}\] then, 3 should also divide \[{{p}^{2}}\].
If 3 is divided by \[{{p}^{2}}\] then 3 should p. So, we can write
\[p=3r\] where r is an integer.
Now, we will substitute \[p=3r\] so, we can substitute p as 3r in equation \[3{{q}^{2}}={{p}^{2}}\] so we get,
\[3{{q}^{2}}={{\left( 3r \right)}^{2}}\]
Which can be simplified as,
\[{{q}^{2}}=3{{r}^{2}}\]
In this equation we can say that 3 divides \[3{{r}^{2}}\] so 3 divides \[{{q}^{2}}\]. Hence, we can also say that 3 divides q also.
In the above relations we saw p is divisible by 3 and q is also divisible by 3. So, there exist a common factor 3 between p and q but we assumed that p, q were co-prime. Hence, our assumption gets contradicted.
Thus \[\sqrt{3}\] is a rational number.
In the second part we are asked to prove that \[7+2\sqrt{3}\] is an irrational number.
As we know that \[\sqrt{3}\] is irrational as we proved earlier so,
\[2\times \sqrt{3}\] is also irrational because the product of rational and irrational numbers is always irrational. And \[7+2\sqrt{3}\] will be irrational as 7 is a rational number and \[2\sqrt{3}\] is an irrational number and sum of a rational and an irrational number is always irrational.
Hence proved, \[\sqrt{3}\] and \[7+2\sqrt{3}\] both are irrational .

Note: At first take \[\sqrt{3}=\dfrac{a}{b}\] then take a, b as odd numbers which are distinct so that they are coprime, if they were even then 2 might be the common factor. After that square then and find the relation which is a very tedious process.