Answer

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Hint: We use the method of contradiction to solve this problem. One can start by assuming that $3+\sqrt{5}$ is rational number, such that,

$x=3+\sqrt{5}$

(Where, x is a rational number)

Subsequently, we will prove that this assumption is incorrect, which would then prove that $3+\sqrt{5}$ is not rational.

Complete step-by-step answer:

We have,

$x=3+\sqrt{5}$

Squaring LHS and RHS, we get,

$\begin{align}

& {{x}^{2}}={{\left( 3+\sqrt{5} \right)}^{2}} \\

& {{x}^{2}}=9+5+2(3)(\sqrt{5}) \\

& {{x}^{2}}=14+6(\sqrt{5}) \\

& \dfrac{{{x}^{2}}-14}{6}=\sqrt{5} \\

\end{align}$

Now, LHS should be a rational number, since x was rational (by assumption). Further, all other terms on LHS are integers, so any arithmetic operations among these numbers would only yield a rational number. However, RHS is an irrational number, since $\sqrt{5}$ is an irrational number. According to equality (LHS=RHS), that is even RHS should have been a rational number, however, this is not true. This contradicts our earlier made assumption. Hence, by method of contradiction, x is an irrational method.

In case, we have to prove that $\sqrt{5}$ is an irrational number, we can again prove this by method of contradiction. Let’s assume,

$\dfrac{a}{b}=\sqrt{5}$

(Where, a and b are integers with b$\ne $0 such that a and b are co-prime)

Now, to prove, we have, by squaring on both sides,

\[\begin{align}

& {{\left( \dfrac{a}{b} \right)}^{2}}=5 \\

& \dfrac{{{a}^{2}}}{5}={{b}^{2}}--(1) \\

\end{align}\]

Thus, 5 should divide a. Now, a is a multiple of 5. Let, a=5m (where, m is an integer).

a = 5m

Squaring both sides, we have,

${{a}^{2}}=25{{m}^{2}}$ -- (2)

From (1) and (2), we have,

\[\dfrac{25{{m}^{2}}}{5}={{b}^{2}}\]

\[{{b}^{2}}=5{{m}^{2}}\]

Thus, now even b is a multiple of 5. However, earlier we assumed that a and b are co-prime. Thus, our assumption was incorrect. Hence, by method of contradiction, $\sqrt{5}$ is an irrational number.

Hence, finally, we can conclude that 3+$\sqrt{5}$ is an irrational number.

Note: Generally, for questions in which one has to prove a given expression to be irrational, it is always a good idea to prove it by contradiction (as done in the above problem). After that, one should solve until the point where it can be proved that the equation cannot be true with the assumption made in the problem. For example, in this problem, we solved the algebraic expression till we reached a point where LHS was rational and RHS was irrational and then could use a method of contradiction. Another way to solve this problem is by concluding that since we are adding a rational number to an irrational number, the sum should be an irrational number.

$x=3+\sqrt{5}$

(Where, x is a rational number)

Subsequently, we will prove that this assumption is incorrect, which would then prove that $3+\sqrt{5}$ is not rational.

Complete step-by-step answer:

We have,

$x=3+\sqrt{5}$

Squaring LHS and RHS, we get,

$\begin{align}

& {{x}^{2}}={{\left( 3+\sqrt{5} \right)}^{2}} \\

& {{x}^{2}}=9+5+2(3)(\sqrt{5}) \\

& {{x}^{2}}=14+6(\sqrt{5}) \\

& \dfrac{{{x}^{2}}-14}{6}=\sqrt{5} \\

\end{align}$

Now, LHS should be a rational number, since x was rational (by assumption). Further, all other terms on LHS are integers, so any arithmetic operations among these numbers would only yield a rational number. However, RHS is an irrational number, since $\sqrt{5}$ is an irrational number. According to equality (LHS=RHS), that is even RHS should have been a rational number, however, this is not true. This contradicts our earlier made assumption. Hence, by method of contradiction, x is an irrational method.

In case, we have to prove that $\sqrt{5}$ is an irrational number, we can again prove this by method of contradiction. Let’s assume,

$\dfrac{a}{b}=\sqrt{5}$

(Where, a and b are integers with b$\ne $0 such that a and b are co-prime)

Now, to prove, we have, by squaring on both sides,

\[\begin{align}

& {{\left( \dfrac{a}{b} \right)}^{2}}=5 \\

& \dfrac{{{a}^{2}}}{5}={{b}^{2}}--(1) \\

\end{align}\]

Thus, 5 should divide a. Now, a is a multiple of 5. Let, a=5m (where, m is an integer).

a = 5m

Squaring both sides, we have,

${{a}^{2}}=25{{m}^{2}}$ -- (2)

From (1) and (2), we have,

\[\dfrac{25{{m}^{2}}}{5}={{b}^{2}}\]

\[{{b}^{2}}=5{{m}^{2}}\]

Thus, now even b is a multiple of 5. However, earlier we assumed that a and b are co-prime. Thus, our assumption was incorrect. Hence, by method of contradiction, $\sqrt{5}$ is an irrational number.

Hence, finally, we can conclude that 3+$\sqrt{5}$ is an irrational number.

Note: Generally, for questions in which one has to prove a given expression to be irrational, it is always a good idea to prove it by contradiction (as done in the above problem). After that, one should solve until the point where it can be proved that the equation cannot be true with the assumption made in the problem. For example, in this problem, we solved the algebraic expression till we reached a point where LHS was rational and RHS was irrational and then could use a method of contradiction. Another way to solve this problem is by concluding that since we are adding a rational number to an irrational number, the sum should be an irrational number.

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