RD Sharma Solutions Class 10 Real Numbers Exercise 1.5

RD Sharma Solutions Class 10 Chapter 1 Exercise 1.5

RD Sharma Class 10 Solutions Chapter 1 Ex 1.5 PDF Free Download

Exercise 1.5

 

Q.1: Show that the following numbers are irrational.

 

 

(i) \(7\sqrt{5}\)

 

Let us assume that \(7\sqrt{5}\) is rational. Then, there exist positive co primes a and b such that

\(7\sqrt{5}\) = \(\frac{a}{b}\)

\(\sqrt{5}\)  = \(\frac{a}{7b}\)

We know that  \(\sqrt{5}\)   is an irrational number

Here we see that  \(\sqrt{5}\) is a rational number which is a contradiction.

 

(ii) \(6+\sqrt{2}\)

 

Let us assume that \(6+\sqrt{2}\) is rational. Then, there exist positive co primes a and b such that

\(6+\sqrt{2}\) = \(\frac{a}{b}\)

\(\sqrt{2}\)  = \(\frac{a}{b}-6\)

\(\sqrt{2}\)  = \(\frac{a-6b}{b}\)

Here we see that  \(\sqrt{2}\) is a rational number which is a contradiction as we know that \(\sqrt{2}\) is an irrational number

Hence \(6+\sqrt{2}\) is an irrational number

 

(iii) \(3-\sqrt{5}\)

 

Let us assume that \(3-\sqrt{5}\) is rational. Then, there exist positive co primes a and b such that

\(3-\sqrt{5}\) = \(\frac{a}{b}\)

\(\sqrt{5}\)  = \(3-\frac{a}{b}\)

\(\sqrt{5}\)  = \(\frac{3b-a}{b}\)

Here we see that  \(\sqrt{5}\) is a rational number which is a contradiction as we know that \(\sqrt{5}\) is an irrational number

Hence \(3-\sqrt{5}\) is an irrational number.

 

Q.2: Prove that the following numbers are irrationals.

 

Sol:

(i) \(\frac{2}{\sqrt[]{7}}\)

 

Let us assume that \(2\sqrt{7}\) is rational. Then, there exist positive co primes a and b such that

\(2\sqrt{7}\) = \(\frac{a}{b}\)

\(\sqrt{7}\)  = \(\frac{2b}{a}\)

\(\sqrt{7}\) is rational number which is a contradiction

Hence \(2\sqrt{7}\) is an irrational number

 

(ii) \(\frac{3}{2\sqrt{5}}\)

 

Let us assume that \(\frac{3}{2\sqrt{5}}\) is rational. Then, there exist positive co primes a and b such that

\(\frac{3}{2\sqrt{5}}\) = \(\frac{a}{b}\)

\(\sqrt{5}\)  = \(\frac{3b}{2a}\)

\(\sqrt{5}\) is rational number which is a contradiction

Hence \(\frac{3}{2\sqrt{5}}\) is irrational.

 

(iii) \(4+\sqrt{2}\)

 

Let us assume that \(4+\sqrt{2}\) is rational. Then, there exist positive co primes a and b such that

\(4+\sqrt{2}\) = \(\frac{a}{b}\)

\(\sqrt{2}\)  = \(\frac{a}{b}-4\)

\(\sqrt{2}\)  = \(\frac{a-4b}{b}\)

\(\sqrt{2}\) is rational number which is a contradiction

Hence \(4+\sqrt{2}\) is irrational.

 

(iv) \(5\sqrt{2}\)

 

Let us assume that \(5\sqrt{2}\) is rational. Then, there exist positive co primes a and b such that

\(5\sqrt{2}\) = \(\frac{a}{b}\)

\(\sqrt{2}\) = \(\frac{a}{b}-5\)

\(\sqrt{2}\) = \(\frac{a-5b}{b}\)

\(\sqrt{2}\) is rational number which is a contradiction

Hence \(5\sqrt{2}\)  is irrational

 

Q.3: Show that \(2-\sqrt{3}\) is an irrational number.

 

Sol:

Let us assume that \(2-\sqrt{3}\) is rational. Then, there exist positive co primes a and b such that

\(2-\sqrt{3}\) = \(\frac{a}{b}\)

\(\sqrt{3}\)  = \(2-\frac{a}{b}\)

Here we see that \(\sqrt{3}\) is a rational number which is a contradiction

Hence \(2-\sqrt{3}\)  is irrational

 

Q.4: Show that \(3+\sqrt{2}\) is an irrational number.

 

Sol:

Let us assume that \(3+\sqrt{2}\) is rational. Then, there exist positive co primes a and b such that

\(3+\sqrt{2}\) = \(\frac{a}{b}\)

\(\sqrt{2}\)  = \(\frac{a}{b}-3\)

\(\sqrt{2}\) = \(\frac{a-3b}{b}\)

Here we see that \(\sqrt{2}\) is a irrational number which is a contradiction

Hence \(3+\sqrt{2}\)  is irrational

 

Q.5: Prove that \(4-5\sqrt{2}\) is an irrational number.

 

Sol:

Let us assume that \(4-5\sqrt{2}\) is rational. Then, there exist positive co primes a and b such that

\(4-5\sqrt{2}\) = \(\frac{a}{b}\)

\(5\sqrt{2}\) = \(\frac{a}{b}-4\)

\(\sqrt{2}=\frac{\frac{a}{b}-4}{5}\)

\(\sqrt{2}\) = \(\frac{a-4b}{5b}\)

This contradicts the fact that \(\sqrt{2}\) is an irrational number

Hence \(4-5\sqrt{2}\) is irrational

 

Q.6: Show that \(5-2\sqrt{3}\) is an irrational number.

 

Sol.

Let us assume that \(5-2\sqrt{3}\) is rational. Then, there exist positive co primes a and b such that

\(5-2\sqrt{3}\) = \(\frac{a}{b}\)

\(2\sqrt{3}\) = \(\frac{a}{b}-5\)

\(\sqrt{3}=\frac{\frac{a}{b}-5}{2}\)

\(\sqrt{3}= \frac{a-5b}{2b}\)

This contradicts the fact that \(\sqrt{3}\) is an irrational number

Hence \(5-2\sqrt{3}\) is irrational

 

Q.7: Prove that \(2\sqrt{3}-1\) is an irrational number.

 

Sol:

Let us assume that \(2\sqrt{3}-1\) is rational. Then, there exist positive co primes a and b such that

\(2\sqrt{3}-1\) = \(\frac{a}{b}\)

\(2\sqrt{3}\)  = \(\frac{a}{b}+1\)

\(\sqrt{3}=\frac{\frac{a}{b}+1}{2}\)

\(\sqrt{3}= \frac{a+b}{2b}\)

This contradicts the fact that \(\sqrt{3}\) is an irrational number

Hence \(5-2\sqrt{3}\) is irrational

 

Q.8: Prove that \(2-3\sqrt{5}\) is an irrational number.

 

Sol:

Let us assume that \(2-3\sqrt{5}\) is rational. Then, there exist positive co primes a and b such that

\(2-3\sqrt{5}\) = \(\frac{a}{b}\)

\(3\sqrt{5}\) = \(\frac{a}{b}-2\)

\(3\sqrt{5} =\frac{\frac{a}{b}-2}{3}\)

\(\sqrt{5}=\frac{a-3b}{3b}\)

This contradicts the fact that \(\sqrt{5}\) is an irrational number

Hence \(2-3\sqrt{5}\) is irrational

 

Q.9: Prove that \(\sqrt{5}+\sqrt{3}\) is irrational.

 

Sol:

Let us assume that \(\sqrt{5}+\sqrt{3}\) is rational. Then, there exist positive co primes a and b such that

\(\sqrt{5}+\sqrt{3}\) = \(\frac{a}{b}\)

\(\sqrt{5}=\frac{a}{b}-\sqrt{3}\)

\(\left ( \sqrt{5} \right )^{2}=\left ( \frac{a}{b}-\sqrt{3} \right )^{3}\)

\(5=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}+3\)

\(\Rightarrow 5-3=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow 2=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow\left ( \frac{a}{b} \right )^{2}-2=\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow \frac{a^{2}-2b^{2}}{b^{2}}=\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow \left ( \frac{a^{2}-2b^{2}}{b^{2}} \right )\left ( \frac{b}{2a} \right )=\sqrt{3}\)

\(\Rightarrow \left ( \frac{a^{2}-2b^{2}}{2ab} \right )=\sqrt{3}\)

Here we see that \(\sqrt{3}\) is a rational number which is a contradiction as we know that \(\sqrt{3}\) is an irrational number

Hence \(\sqrt{5}+\sqrt{3}\)  is an irrational number

 

Q.10: Prove that \(\sqrt{3}+\sqrt{4}\) is irrational.

 

Sol:

Let us assume that \(\sqrt{3}+\sqrt{4}\) is rational. Then, there exist positive co primes a and b such that

\(\sqrt{3}+\sqrt{4}\) = \(\frac{a}{b}\)

\(\sqrt{4}=\frac{a}{b}-\sqrt{3}\)

\(\left ( \sqrt{4} \right )^{2}=\left ( \frac{a}{b}-\sqrt{3} \right )^{3}\)

\(4=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}+3\)

\(\Rightarrow 4-3=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow 1=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow\left ( \frac{a}{b} \right )^{2}-1=\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow \frac{a^{2}-b^{2}}{b^{2}}=\frac{2a\sqrt{3}}{b}\)

\(\Rightarrow \left ( \frac{a^{2}-b^{2}}{b^{2}} \right )\left ( \frac{b}{2a} \right )=\sqrt{3}\)

\(\Rightarrow \left ( \frac{a^{2}-b^{2}}{2ab} \right )=\sqrt{3}\)

Here we see that  \(\sqrt{3}\) is a rational number which is a contradiction as we know that \(\sqrt{3}\) is an irrational number

Hence \(\sqrt{3}+\sqrt{4}\)  is an irrational number

 

Q.11: Prove that for any prime positive integer p, \(\sqrt{p}\) is an irrational number.

 

Sol:

Let us assume that \(\sqrt{p}\) is rational. Then, there exist positive co primes a and b such that

\(\sqrt{p}\) = \(\frac{a}{b}\)

\({p}\) = \(\left (\frac{a}{b} \right )^{2}\)

\({p}\) =\(\frac{a^{2}}{b^{2}}\)

\(\Rightarrow p b^{2}=a^{2}\)

\(\Rightarrow p|a^{2}\)

\(\Rightarrow p|a\)

\(\Rightarrow a = pc for some positive integer c\)

\(\Rightarrow b^{2}p\) = \(a^{2}\)

\(\Rightarrow b^{2}p\) = \(p^{2}c^{2} \) ( ∵ a=pc )

\(\Rightarrow p|b^{2}\left ( since\;p|c^{2}p \right )\)

\(\Rightarrow p|b\)

\(\Rightarrow p|a\;and\;p|b\)

This contradicts the fact that a and b are co primes

Hence \(\sqrt{p}\) is irrational

 

Q.12: If p, q are prime positive integers, prove that \(\sqrt{p}+\sqrt{q}\) is an irrational number.

 

Sol:

Let us assume that \(\sqrt{p}+\sqrt{q}\) is rational. Then, there exist positive co primes a and b such that

\(\sqrt{p}+\sqrt{q}\) = \(\frac{a}{b}\)

\(\sqrt{p}=\frac{a}{b}-\sqrt{q}\)

\(\left ( \sqrt{p} \right )^{2}=\left ( \frac{a}{b}-\sqrt{q} \right )^{2}\)

\(p=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{q}}{b}+q\)

\(p-q=\left ( \frac{a}{b} \right )^{2}-\frac{2a\sqrt{q}}{b}\)

\(\left ( \frac{a}{b} \right )^{2}-\left ( p-q \right )=\frac{2a\sqrt{q}}{b}\)

\(\frac{a^{2}-b^{2}\left ( p-q \right )}{b^{2}}=\frac{2a\sqrt{q}}{b}\)

\(\left ( \frac{a^{2}-b^{2}\left ( p-q \right )}{b^{2}} \right )\left ( \frac{b}{2a} \right )=\sqrt{q}\)

\(\sqrt{q}=\frac{a^{2}-b^{2}\left ( p-q \right )}{2ab}\)

Here we see that  \(\sqrt{q}\) is a rational number which is a contradiction as we know that \(\sqrt{q}\) is an irrational number

Hence  \(\sqrt{p}+\sqrt{q}\)   is an irrational number

 

 


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