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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