Question

Prove that \[\sqrt{3}+\sqrt{8}\] is an irrational number.

Answer 1

Hint: In this question, we will prove that \[\sqrt{3}\] and \[\sqrt{8}\] are irrational separately by contradiction method. And after that, we will apply the property that the sum of two irrational numbers is always an irrational number.

Complete step-by-step answer:
In this question, we have to prove that \[\sqrt{3}+\sqrt{8}\] is an irrational number. For that, we will first prove that \[\sqrt{3}\] is irrational. Let \[\sqrt{3}\] be rational number and we know that a rational number can be expressed in the form of \[\dfrac{p}{q}\] where \[q\ne 0\] and p and q are coprime. So, let us write \[\sqrt{3}=\dfrac{p}{q}\] where p and q are co-prime and \[q\ne 0\].
Now, we will square both sides of the equation. So, we will get,
\[{{\left( \sqrt{3} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}}\]
\[3=\dfrac{{{p}^{2}}}{{{q}^{2}}}\]
Now, we will cross multiply the equation. So, we will get,
\[3{{q}^{2}}={{p}^{2}}\]
Here, we can see that 3 is a factor of \[{{p}^{2}}\]. So, we can say 3 is also a factor of p. So, we can write p = 3m. Now, by putting the value of p in the above equation, we will get,
\[3{{q}^{2}}={{\left( 3m \right)}^{2}}\]
\[3{{q}^{2}}=9{{m}^{2}}\]
\[{{q}^{2}}=3{{m}^{2}}\]
Here, we can see that 3 is a factor of \[{{q}^{2}}\] which implies 3 is also a factor of q. But we assumed that p and q are co-prime, which contradicts our assumption. Hence, our assumption is wrong. Therefore, we can say \[\sqrt{3}\] is an irrational number.
Now, we will prove that \[\sqrt{8}\] is irrational. We know that \[\sqrt{8}\] can be written as \[\sqrt{2\times 2\times 2}=2\sqrt{2}\]. So, if we will prove \[\sqrt{2}\] as irrational, then we will be able to prove that \[\sqrt{8}\] is irrational.
To prove \[\sqrt{2}\] as an irrational number, we will consider \[\sqrt{2}\] as a rational number and we know that a rational number can be expressed in terms of \[\dfrac{a}{b}\] where \[b\ne 0\] and a and b are co-prime. So, we can write \[\sqrt{2}=\dfrac{a}{b}\] where \[b\ne 0\] and a and b are co-prime.
Now we will square both the sides of the equations. So, we will get,
\[{{\left( \sqrt{2} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}\]
\[2=\dfrac{{{a}^{2}}}{{{b}^{2}}}\]
Now, we will multiply the equation, we get,
\[2{{b}^{2}}={{a}^{2}}\]
This means 2 is a factor of \[{{a}^{2}}\] or we can say 2 is a factor of a. So, we can write a = 2n. So by using it, we can write the above equation as,
\[2{{b}^{2}}={{\left( 2n \right)}^{2}}\]
\[2{{b}^{2}}=4{{n}^{2}}\]
\[{{b}^{2}}=2{{n}^{2}}\]
Here, we can see that 2 is a factor of \[{{b}^{2}}\], which implies 2 is also a factor of b. But we assumed that a and b are co-prime which contradicts our assumptions. Hence, our assumption is wrong. Therefore, we can say \[\sqrt{2}\] is an irrational number and we know that when a rational number is multiplied by an irrational number, we get an irrational number. So, we can say that \[2\sqrt{2}\] is an irrational number, which means \[\sqrt{8}\] is an irrational number. And we know that the sum of two irrational numbers is always an irrational number that means \[\sqrt{3}+\sqrt{8}\] is an irrational number.

Note: In this question, the most important point to remember is to prove \[\sqrt{3}\] and \[\sqrt{8}\] as irrational numbers separately. Also, we need to remember that to prove \[\sqrt{3}\] and \[\sqrt{8}\] as irrational, we will use the contradiction method by assuming them as rational, and then after expressing them in \[\dfrac{p}{q}\] form, we have to prove that p and q are not co-primes.