Answer
384.6k+ views
Hint:
Here, we are required to prove that $\sqrt 3 $ is an irrational number. Thus, we will assume that it is a rational number such that it can be written in the fraction form where the denominator is not equal to zero. Also, we will try to prove the numbers in the numerator and denominator are co-prime numbers. Solving this further, we will observe that the numbers are not co-primes and this will contradict our assumption and hence, prove the given statement.
Complete step by step solution:
In order to prove that $\sqrt 3 $ is an irrational number, let us assume that $\sqrt 3 $ is a rational number.
Now, as we know, a rational number is a number which can be written in the form of $\dfrac{p}{q}$ where, $q \ne 0$ and $p$ and $q$ are co-prime numbers.
Hence, if $\sqrt 3 $ is a rational number, it can be written as:
$\sqrt 3 = \dfrac{p}{q}$
Multiplying both sides by $q$, we get,
$\sqrt 3 q = p$
Now, squaring both sides, we get
$ \Rightarrow 3{q^2} = {p^2}$………………………$\left( 1 \right)$
Hence, if 3 is a factor of ${p^2}$, then,
3 is also a factor of $p$
Thus, let $p = 3m$, where $m$ is any integer.
Now, squaring both sides, we get,
${p^2} = {\left( {3m} \right)^2} = 9{m^2}$
Substituting the value of ${p^2}$in $\left( 1 \right)$, we get,
$ \Rightarrow 3{q^2} = 9{m^2}$
Dividing both sides by 3, we get
$ \Rightarrow {q^2} = 3{m^2}$
Therefore, 3 is a factor of ${q^2}$.
Hence, 3 is a factor of $q$.
Now, from the above calculation, we can conclude that 3 is a factor of both $p$ and $q$
But, since we have assumed that $p$ and $q$ are co-prime numbers, they cannot have any factor common.
Thus, our assumption is wrong and $p$ and $q$ are not coprime and hence, this is contradicting our assumption that $\sqrt 3 $ is a rational number.
Therefore, $\sqrt 3 $ is an irrational number
Hence, proved.
Note:
A rational number is a number that is expressed as the ratio of two integers, i.e. in the form of $\dfrac{p}{q}$ where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating. An irrational number is a number that cannot be expressed as a fraction for any integers and irrational numbers have decimal expansions that neither terminate nor become periodic.
Here, we are required to prove that $\sqrt 3 $ is an irrational number. Thus, we will assume that it is a rational number such that it can be written in the fraction form where the denominator is not equal to zero. Also, we will try to prove the numbers in the numerator and denominator are co-prime numbers. Solving this further, we will observe that the numbers are not co-primes and this will contradict our assumption and hence, prove the given statement.
Complete step by step solution:
In order to prove that $\sqrt 3 $ is an irrational number, let us assume that $\sqrt 3 $ is a rational number.
Now, as we know, a rational number is a number which can be written in the form of $\dfrac{p}{q}$ where, $q \ne 0$ and $p$ and $q$ are co-prime numbers.
Hence, if $\sqrt 3 $ is a rational number, it can be written as:
$\sqrt 3 = \dfrac{p}{q}$
Multiplying both sides by $q$, we get,
$\sqrt 3 q = p$
Now, squaring both sides, we get
$ \Rightarrow 3{q^2} = {p^2}$………………………$\left( 1 \right)$
Hence, if 3 is a factor of ${p^2}$, then,
3 is also a factor of $p$
Thus, let $p = 3m$, where $m$ is any integer.
Now, squaring both sides, we get,
${p^2} = {\left( {3m} \right)^2} = 9{m^2}$
Substituting the value of ${p^2}$in $\left( 1 \right)$, we get,
$ \Rightarrow 3{q^2} = 9{m^2}$
Dividing both sides by 3, we get
$ \Rightarrow {q^2} = 3{m^2}$
Therefore, 3 is a factor of ${q^2}$.
Hence, 3 is a factor of $q$.
Now, from the above calculation, we can conclude that 3 is a factor of both $p$ and $q$
But, since we have assumed that $p$ and $q$ are co-prime numbers, they cannot have any factor common.
Thus, our assumption is wrong and $p$ and $q$ are not coprime and hence, this is contradicting our assumption that $\sqrt 3 $ is a rational number.
Therefore, $\sqrt 3 $ is an irrational number
Hence, proved.
Note:
A rational number is a number that is expressed as the ratio of two integers, i.e. in the form of $\dfrac{p}{q}$ where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating. An irrational number is a number that cannot be expressed as a fraction for any integers and irrational numbers have decimal expansions that neither terminate nor become periodic.
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