**Exercise 1.5**

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

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

Let us assume that \(7\sqrt{5}\)

\(7\sqrt{5}\)

\(\sqrt{5}\)

We know thatÂ \(\sqrt{5}\)

Here we see thatÂ \(\sqrt{5}\)

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

Let us assume that \(6+\sqrt{2}\)

\(6+\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

Here we see thatÂ \(\sqrt{2}\)

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

**(iii)**Â **\(3-\sqrt{5}\)**

Let us assume that \(3-\sqrt{5}\)

\(3-\sqrt{5}\)

\(\sqrt{5}\)

\(\sqrt{5}\)

Here we see thatÂ \(\sqrt{5}\)

Hence \(3-\sqrt{5}\)

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

**Sol:**

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

Let us assume that \(2\sqrt{7}\)

\(2\sqrt{7}\)

\(\sqrt{7}\)

\(\sqrt{7}\)

Hence \(2\sqrt{7}\)

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

Let us assume that \(\frac{3}{2\sqrt{5}}\)

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

\(\sqrt{5}\)

\(\sqrt{5}\)

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

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

Let us assume that \(4+\sqrt{2}\)

\(4+\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

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

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

Let us assume that \(5\sqrt{2}\)

\(5\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

Hence \(5\sqrt{2}\)

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

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**Sol:**

Let us assume that \(2-\sqrt{3}\)

\(2-\sqrt{3}\)

\(\sqrt{3}\)

Here we see that \(\sqrt{3}\)

Hence \(2-\sqrt{3}\)

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

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**Sol:**

Let us assume that \(3+\sqrt{2}\)

\(3+\sqrt{2}\)

\(\sqrt{2}\)

\(\sqrt{2}\)

Here we see that \(\sqrt{2}\)

Hence \(3+\sqrt{2}\)

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

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**Sol:**

Let us assume that \(4-5\sqrt{2}\)

\(4-5\sqrt{2}\)

\(5\sqrt{2}\)

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

\(\sqrt{2}\)

This contradicts the fact that \(\sqrt{2}\)

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

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

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**Sol.**

Let us assume that \(5-2\sqrt{3}\)

\(5-2\sqrt{3}\)

\(2\sqrt{3}\)

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

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

This contradicts the fact that \(\sqrt{3}\)

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

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

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**Sol:**

Let us assume that \(2\sqrt{3}-1\)

\(2\sqrt{3}-1\)

\(2\sqrt{3}\)

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

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

This contradicts the fact that \(\sqrt{3}\)

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

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

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**Sol:**

Let us assume that \(2-3\sqrt{5}\)

\(2-3\sqrt{5}\)

\(3\sqrt{5}\)

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

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

This contradicts the fact that \(\sqrt{5}\)

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

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

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**Sol:**

Let us assume that \(\sqrt{5}+\sqrt{3}\)

\(\sqrt{5}+\sqrt{3}\)

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

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

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

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

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

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

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

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

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

Here we see that \(\sqrt{3}\)

HenceÂ \(\sqrt{5}+\sqrt{3}\)

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

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**Sol:**

Let us assume that \(\sqrt{3}+\sqrt{4}\)

\(\sqrt{3}+\sqrt{4}\)

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

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

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

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

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

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

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

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

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

Here we see thatÂ \(\sqrt{3}\)

HenceÂ \(\sqrt{3}+\sqrt{4}\)

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

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**Sol:**

Let us assume that \(\sqrt{p}\)

\(\sqrt{p}\)

\({p}\)

â‡’\({p}\)

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

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

\(\Rightarrow p|a\)

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

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

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

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

\(\Rightarrow p|b\)

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

This contradicts the fact that a and b are co primes

Hence \(\sqrt{p}\)

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

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**Sol:**

Let us assume that \(\sqrt{p}+\sqrt{q}\)

\(\sqrt{p}+\sqrt{q}\)

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

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

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

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

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

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

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

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

Here we see thatÂ \(\sqrt{q}\)

HenceÂ \(\sqrt{p}+\sqrt{q}\)