Question

Prove that the following are irrational.

(1) \[\dfrac{1}{\sqrt{2}}\]
(2) \[7\sqrt{5}\]
(3) \[6+\sqrt{2}\]

Answer 1

Hint: We can prove these by taking Contradicts i.e., given numbers are rational numbers and also the sum of rational and irrational numbers is an irrational number.

Complete step by step solution:
 (1) \[\dfrac{1}{\sqrt{2}}\]
Let the given number be a rational number.
Since it is a rational number. We can write it in \[\dfrac{p}{q}\] form.
\[\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}\] Where a and b are co primes.
Co-primes are the prime numbers which do not have a common root.
\[\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}\]
Squaring on both sides
\[\dfrac{1}{2}=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}\]
\[\Rightarrow \dfrac{\mathop{b}^{2}}{2}=\mathop{a}^{2}\] ............................……..(i)
So that we got 2 divides b2.
So it can also divides b ...........................…… (ii)
Let \[\dfrac{b}{2}=c\] where c in other numbers.
Squaring both sides
\[\dfrac{\mathop{b}^{2}}{4}=\mathop{c}^{2}\] …..................................(iii)
From equation(i) \[\dfrac{\mathop{b}^{2}}{4}=\mathop{c}^{2}\] we can substitute value in equation (iii)

\[\Rightarrow \dfrac{\mathop{a}^{2}}{2}=\mathop{c}^{2}\]
Therefore 2 also divides a2 .From 2 and 4 we got 2 divides both a and b. So 2 is a Cofactor of \[a,b\] . But as we assumed above a and b are Co primes.
This contradiction arises due to our assumption that\[\dfrac{1}{\sqrt{2}}\] is an irrational number.
Hence \[\dfrac{1}{\sqrt{2}}\] is an irrational number.
(2) \[7\sqrt{5}\]
Let us assume\[7\sqrt{5}\] is a rational number. So that it can be written in \[\dfrac{p}{q}\] form where p and q are co primes.
So we can let \[7\sqrt{5}=\dfrac{a}{b}\] where a and b are co primes.
\[\Rightarrow \sqrt{5}=\dfrac{a}{7b}\]
We know that is \[\sqrt{5}\] irrational number because we can’t write it in \[\dfrac{p}{q}\] form.
We got an irrational number = Rational number which is not Correct.
This contradiction raised due to because of our assumption that is \[7\sqrt{5}\] a rational number
Hence \[7\sqrt{5}\] is an irrational number.
(3) \[6+\sqrt{2}\]
Let us assume \[6+\sqrt{2}\] as a rational. So that it can be written in \[\dfrac{p}{q}\] form.
Let\[\left( 6+\sqrt{2} \right)=\dfrac{a}{b}\] where a and b are co prime number
On squaring on both sides
\[\Rightarrow \mathop{\left( 6+\sqrt{2} \right)}^{2}=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}\]
\[\Rightarrow {{(6)}^{2}}+2\times 6\times \sqrt{2}+{{\left( \sqrt{2} \right)}^{2}}=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}\] $\left\{ \because {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right\}$
\[\Rightarrow 36+12\sqrt{2}+2=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}\]
\[\Rightarrow 38+12\sqrt{2}=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}\]
\[\Rightarrow 12\sqrt{2}=\dfrac{\mathop{a}^{2}}{\mathop{b}^{2}}-38\]
\[\Rightarrow 12\sqrt{2}=\dfrac{{{a}^{2}}-38{{b}^{2}}}{{{b}^{2}}}\]
.\[\Rightarrow \sqrt{2}=\dfrac{{{a}^{2}}-38{{b}^{2}}}{12{{b}^{2}}}\]
From above equation Left hand side is\[\sqrt{2}\] which is an irrational number because we can’t write it in \[\dfrac{p}{q}\] form.
Right hand side of the equation is a rational number because it is in the required rational form.
So this contradict raised due to our assumption \[6+\sqrt{2}\] is rational.
Hence, \[6+\sqrt{2}\] is an irrational number.

Note: Co primes: These are the prime numbers who does not have a common factor
Eg: 3 and 5 are co primes because there is no common factor in 3 and 5.
From the fundamental theorem of Arithmetic when a prime no. p divides then p also divides a.