Ex.1.3 Q3 Real Numbers Solution - NCERT Maths Class 10

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Question

Prove that the following are irrationals:

(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}      (ii) \begin{align}7\sqrt 5 \end{align}     (iii)  \begin{align}6 +\sqrt 2 \end{align}

Video Solution
Real Numbers
Ex 1.3 | Question 3

Text Solution

What is unknown?

(i) \begin{align}\frac{1}{{\sqrt 2 }}\end{align}      (ii) \begin{align}7\sqrt 5 \end{align}     (iii)  \begin{align}6 +\sqrt 2 \end{align} are Irrationals

Steps:

(i) \begin{align} \frac{1}{{\sqrt 2 }}\end{align}

Let us assume, to the contrary \begin{align} \frac{1}{{\sqrt 2 }}\end{align} that is a rational number.

\begin{align} \frac{1}{{\sqrt 2 }} = \frac{p}{q}\end{align}

Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.

\begin{align}\frac{1}{{\sqrt 2 }}& = \frac{a}{b} \\\sqrt 2 {\text{a}} &= {\text{b}}\\\sqrt 2 &= \frac{b}{a}\end{align}
(where $$a$$ and $$b$$ are co - primes and have no common factor otherthan $$1$$)

Since, $$b$$ and $$a$$ are integers, \begin{align} \frac{b}{a}\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is rational.

But we know that \begin{align} \sqrt 2 \end{align} is irrational this contradicts the fact that \begin{align} \sqrt 2 \end{align} is rational.

So, our assumption was wrong. Therefore, \begin{align} \frac{1}{{\sqrt 2 }}\end{align} is a rational number.

(ii) \begin{align} 7\sqrt 5 \end{align}

Let us assume, to the contrary that \begin{align} 7\sqrt 5 \end{align} is a rational number.

\begin{align} 7\sqrt 5 = \frac{p}{q}\end{align}

Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.

\begin{align} 7\sqrt 5 &= \frac{a}{b} \\7\sqrt 5 {\text{b}} &= {\text{a}}\\\sqrt 5 &= \frac{a}{{7b}}\end{align}

(where $$a$$ and  $$b$$ are co - primes and have no common factor other than $$1$$)

Since, $$a,\; 7$$ and $$b$$ are integers. So, \begin{align} \frac{a}{{7b}}\end{align} is rational number and so, \begin{align} \sqrt 5 \end{align} is rational. But this contradict the fact that \begin{align} \sqrt 5 \end{align}

So, our assumption was wrong. Therefore, \begin{align} 7\sqrt 5 \end{align} is a rational number.

(iìi) \begin{align} 6{\text{ }} + \sqrt 2 \end{align}

Let us assume, to the contrary that \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.

\begin{align} 6{\text{ }} + \sqrt 2 = \frac{p}{q}\end{align}

Let $$p$$ and $$q$$ have common factors, so by cancelling them we will get \begin{align} \frac{a}{b},\end{align} where $$a$$ and $$b$$ are co-primes.

\begin{align} 6 + \sqrt 2 &= \frac{a}{b}\\\sqrt 2 &= \frac{a}{b} - 6 \end{align}

(where $$a$$ and $$b$$ are co - primes and have no common factor other than $$1$$)

Since, $$a, \,b$$ and $$6$$ are integers. So, \begin{align} \frac{a}{b} - 6\end{align} is rational number and so, \begin{align} \sqrt 2 \end{align} is also a rational number.

But this contradicts the fact that \begin{align} \sqrt 2 \end{align} is irrational. So, our assumption was wrong.

Therefore, \begin{align} 6{\text{ }} + \sqrt 2 \end{align} is a rational number.

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