Answer
Verified
409.2k+ views
Hint: 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.
Complete step-by-step answer:
Let us prove that $\sqrt{5}$ is an irrational number, by using the contradiction method.
So, say that $\sqrt{5}$ is a rational number and as we know any rational number can be expressed in form of $\dfrac{p}{q}$ with $q\ne 0$ .So let $\sqrt{5}$ equals to $\dfrac{p}{q}$
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 simple 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
$\begin{align}
& {{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}} \\
& \dfrac{5}{1}=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
\end{align}$
On cross-multiplying the above equation, we get
${{p}^{2}}=5{{q}^{2}}......\left( i \right)$
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 5 in a following way as
$p=5m......\left( ii \right)$
Where m is any positive integer.
On squaring both sides of equation (ii), we get
${{p}^{2}}={{\left( 5m \right)}^{2}}=25{{m}^{2}}........\left( iii \right)$
Put value of ${{p}^{2}}$ as $5{{q}^{2}}$ from the equation (i) to the equation equation(iii) we get
$\begin{align}
& 5{{q}^{2}}=25{{m}^{2}} \\
& {{q}^{2}}=5{{m}^{2}}........\left( iv \right) \\
\end{align}$
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.
Note: ‘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 multiple of 5. It is the Arithmetic fundamental property in mathematics.
Example : 25 is multiple of 5 then $\sqrt{25}=5$ is also a multiple of 5; 49 is a multiple of 7, then $\sqrt{49}=7$ is also a multiple of 7. So, it can be observed from the several examples as well. So, don’t be confused with this point.
Complete step-by-step answer:
Let us prove that $\sqrt{5}$ is an irrational number, by using the contradiction method.
So, say that $\sqrt{5}$ is a rational number and as we know any rational number can be expressed in form of $\dfrac{p}{q}$ with $q\ne 0$ .So let $\sqrt{5}$ equals to $\dfrac{p}{q}$
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 simple 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
$\begin{align}
& {{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{p}{q} \right)}^{2}} \\
& \dfrac{5}{1}=\dfrac{{{p}^{2}}}{{{q}^{2}}} \\
\end{align}$
On cross-multiplying the above equation, we get
${{p}^{2}}=5{{q}^{2}}......\left( i \right)$
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 5 in a following way as
$p=5m......\left( ii \right)$
Where m is any positive integer.
On squaring both sides of equation (ii), we get
${{p}^{2}}={{\left( 5m \right)}^{2}}=25{{m}^{2}}........\left( iii \right)$
Put value of ${{p}^{2}}$ as $5{{q}^{2}}$ from the equation (i) to the equation equation(iii) we get
$\begin{align}
& 5{{q}^{2}}=25{{m}^{2}} \\
& {{q}^{2}}=5{{m}^{2}}........\left( iv \right) \\
\end{align}$
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.
Note: ‘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 multiple of 5. It is the Arithmetic fundamental property in mathematics.
Example : 25 is multiple of 5 then $\sqrt{25}=5$ is also a multiple of 5; 49 is a multiple of 7, then $\sqrt{49}=7$ is also a multiple of 7. So, it can be observed from the several examples as well. So, don’t be confused with this point.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE