Prove that $\sqrt{5}$ is irrational and hence prove that $\left( 2-\sqrt{5} \right)$ is also irrational.
ANSWER
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Hint: We can use the contradiction method to prove $\sqrt{5}$ as an irrational number i.e. suppose $\sqrt{5}$ is rational. Any rational number can be represented in the form of $\dfrac{p}{q}$, where q$\ne $0. So, represent $\sqrt{5}$ as $\dfrac{p}{q}$ and suppose p, q are co-primes. Now, use the property that if ${{n}^{2}}$ is a multiple of ‘a’ then ‘n’ is also a multiple of ‘a’. And hence contradict your assumption taken initially. And hence prove $\left( 2-\sqrt{5} \right)$ is irrational as well.
Complete Step-by-Step solution: Let us prove that $\sqrt{5}$ is an irrational number, by using the contradiction method. So, say that $\sqrt{5}$ is a rational number can be expressed in the form of $\dfrac{p}{q}$, where q $\ne $0. So, let $\sqrt{5}$ equals $\dfrac{p}{q}$. So, we get $\sqrt{5}\ =\ \dfrac{p}{q}$ Where p, q are co-prime integers i.e. they do not have any common factor except ‘1’. It means $\dfrac{p}{q}$ is the simplest form of fraction because p and q do not have any common factor. So, we have $\sqrt{5}\ =\ \dfrac{p}{q}$ $\left( q\ne 0 \right)$ On squaring both sides of the above equation, we get ${{\left( \sqrt{5} \right)}^{2}}\ =\ {{\left( \dfrac{p}{q} \right)}^{2}}$ $\dfrac{5}{1}\ =\ \dfrac{{{p}^{2}}}{{{q}^{2}}}$ On cross-multiplying the above equation, we get ${{p}^{2}}=\ \ 5{{q}^{2}}$ ………………………………………………………………………(i) Now, we can observe that ${{p}^{2}}$ is a multiple of ‘5’ because ${{p}^{2}}$ is expressed as $5\times {{q}^{2}}$. As we know that ‘n’ will be multiple of any number ‘a’ if ${{n}^{2}}$ is a multiple of ‘a’ which is the fundamental theorem of arithmetic. So, if ${{p}^{2}}$ is a multiple of ‘5’, then p will also be a multiple of 5. Hence, we can express ‘p’ as a multiple of in a following way as ${{p}^{2}}=\ {{\left( 5m \right)}^{2}}$ ……………………………………………………………………(ii) Where m is any positive integer. On squaring both sides of equation (ii), we get ${{p}^{2}}=\ {{\left( 5m \right)}^{2}}\ =\ 25{{m}^{2}}$ ……………………………………………………..(iii) Put value of ${{p}^{2}}$ as $5{{q}^{2}}$ from the equation (i) and to the equation (iii), we get $5{{q}^{2}}=\ 25{{m}^{2}}$ ${{q}^{2}}=\ 5{{m}^{2}}$ ……………………………………………………………………(iv) Now, we can observe that ${{q}^{2}}$ is a multiple of ‘5’ it means ‘q’ will also be a multiple of ‘5’. So, we get that ‘p’ and ‘q’ both have ‘5’ as a common factor from equation (i) and (iv), but we supposed earlier that p and q are co-primes which have ‘1’ as a common factor. So, it contradicts our assumption that $\dfrac{p}{q}$ supposed will not be a rational number. Hence, $\sqrt{5}$ is an irrational number. And now for the second query that, we have to prove ‘$2-\sqrt{5}$’ be an irrational number with the help of the condition that $\sqrt{5}$ is an irrational number. So, we can prove ‘$2-\sqrt{5}$’ is an irrational number by the contradiction method as well. So, let $2-\sqrt{5}$ is a rational number so we can represent it in the form of $\dfrac{p}{q}$, where $q\ne 0$. So, we have $\dfrac{p}{q}\ =\ 2-\sqrt{5}$ $\Rightarrow \dfrac{p}{q}-2\ =\ -\sqrt{5}$ $\Rightarrow 2-\dfrac{p}{q}\ =\ \sqrt{5}$ …………………………………………………(v) Now, we can observe that Left hand side of the above equation is rational term as we know that adding or subtracting any rational numbers will give a rational number but the right hand side of the equation is an irrational term $\sqrt{5}$ as we have already proved it in the first part of the problem. So, any rational number will not equal an irrational number. So, LHS $\ne $ RHS of the equation (v). Hence, our assumption was wrong and contradicts it. So, $2-\sqrt{5}$ is not a rational number. So, it is proved that $\sqrt{5}$ and $2-\sqrt{5}$ are rational numbers.
Note: It is obvious that $2-\sqrt{5}$ will be an irrational if $\sqrt{5}$ is an irrational, so one can write $2-\sqrt{5}$ directly a rational number as a sum of a rational (2) and irrational (-$\sqrt{5}$) will an irrational number. So, one may proceed this way as well. ‘p’ and ‘q’ are co-prime numbers that there are no common factors of ‘p’ and ‘q’ except ‘1’. It was the key point of the solution. One may get confused with the statement that p will also be a multiple of 5 if ${{p}^{2}}$ is also a multiple of 5. It is the arithmetic fundamental property in mathematics. Example: 25 is a multiple of 5 then $\sqrt{25}\ =\ 5$ is also a multiple of 5; 49 is a multiple of 7, then $\sqrt{49}\ =\ 7$ observed from the several examples as well. So, don’t be confused with this point.