\displaystyle \textbf{Question 1. }\text{Show that the following numbers are irrational.}
\displaystyle (i)\ \frac{1}{\sqrt{2}}\qquad (ii)\ 7\sqrt{5}\qquad (iii)\ 6+\sqrt{2}\qquad (iv)\ 3-\sqrt{5}\qquad (v)\ \frac{1}{\sqrt{5}}\qquad [\mathrm{CBSE}\ 2025]
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let us assume that }\frac{1}{\sqrt{2}}\text{ is rational.}
\displaystyle \therefore \frac{1}{\sqrt{2}}=r,\text{ where }r\text{ is rational and }r\ne0.
\displaystyle \therefore \sqrt{2}=\frac{1}{r}
\displaystyle \text{This implies that }\sqrt{2}\text{ is rational, which is a contradiction.}
\displaystyle \therefore \frac{1}{\sqrt{2}}\text{ is irrational.}
\\
\displaystyle (ii)\ \text{Let us assume that }7\sqrt{5}\text{ is rational.}
\displaystyle \therefore 7\sqrt{5}=r,\text{ where }r\text{ is rational.}
\displaystyle \sqrt{5}=\frac{r}{7}
\displaystyle \text{This implies that }\sqrt{5}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 7\sqrt{5}\text{ is irrational.}
\\
\displaystyle (iii)\ \text{Let us assume that }6+\sqrt{2}\text{ is rational.}
\displaystyle \therefore 6+\sqrt{2}=r,\text{ where }r\text{ is rational.}
\displaystyle \sqrt{2}=r-6
\displaystyle \text{This implies that }\sqrt{2}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 6+\sqrt{2}\text{ is irrational.}
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\displaystyle (iv)\ \text{Let us assume that }3-\sqrt{5}\text{ is rational.}
\displaystyle \therefore 3-\sqrt{5}=r,\text{ where }r\text{ is rational.}
\displaystyle \sqrt{5}=3-r
\displaystyle \text{This implies that }\sqrt{5}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 3-\sqrt{5}\text{ is irrational.}
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\displaystyle (v)\ \text{Let us assume that }\frac{1}{\sqrt{5}}\text{ is rational.}
\displaystyle \therefore \frac{1}{\sqrt{5}}=r,\text{ where }r\text{ is rational and }r\ne0.
\displaystyle \therefore \sqrt{5}=\frac{1}{r}
\displaystyle \text{This implies that }\sqrt{5}\text{ is rational, which is a contradiction.}
\displaystyle \therefore \frac{1}{\sqrt{5}}\text{ is irrational.}
\\

\displaystyle \textbf{Question 2. }\text{Prove that the following numbers are irrational:}
\displaystyle (i)\ \frac{2}{\sqrt{7}}\qquad (ii)\ \frac{3}{2\sqrt{5}}\qquad (iii)\ 4+\sqrt{2}\qquad (iv)\ 5\sqrt{2}\qquad [\mathrm{CBSE}\ 2009]
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let us assume that }\frac{2}{\sqrt{7}}\text{ is rational.}
\displaystyle \therefore \frac{2}{\sqrt{7}}=r,\text{ where }r\text{ is rational.}
\displaystyle \therefore \sqrt{7}=\frac{2}{r}
\displaystyle \text{This implies that }\sqrt{7}\text{ is rational, which is a contradiction.}
\displaystyle \therefore \frac{2}{\sqrt{7}}\text{ is irrational.}
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\displaystyle (ii)\ \text{Let us assume that }\frac{3}{2\sqrt{5}}\text{ is rational.}
\displaystyle \therefore \frac{3}{2\sqrt{5}}=r,\text{ where }r\text{ is rational.}
\displaystyle \therefore \sqrt{5}=\frac{3}{2r}
\displaystyle \text{This implies that }\sqrt{5}\text{ is rational, which is a contradiction.}
\displaystyle \therefore \frac{3}{2\sqrt{5}}\text{ is irrational.}
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\displaystyle (iii)\ \text{Let us assume that }4+\sqrt{2}\text{ is rational.}
\displaystyle \therefore 4+\sqrt{2}=r,\text{ where }r\text{ is rational.}
\displaystyle \therefore \sqrt{2}=r-4
\displaystyle \text{This implies that }\sqrt{2}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 4+\sqrt{2}\text{ is irrational.}
\\
\displaystyle (iv)\ \text{Let us assume that }5\sqrt{2}\text{ is rational.}
\displaystyle \therefore 5\sqrt{2}=r,\text{ where }r\text{ is rational.}
\displaystyle \therefore \sqrt{2}=\frac{r}{5}
\displaystyle \text{This implies that }\sqrt{2}\text{ is rational, which is a contradiction.}
\displaystyle \therefore 5\sqrt{2}\text{ is irrational.}
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\displaystyle \textbf{Question 3. }\text{Show that }3+\sqrt{2}\text{ is an irrational number.}\qquad [\mathrm{CBSE}\ 2009]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }3+\sqrt{2}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 3+\sqrt{2}=\frac{a}{b}
\displaystyle \Rightarrow \sqrt{2}=\frac{a}{b}-3
\displaystyle \Rightarrow \sqrt{2}=\frac{a-3b}{b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a-3b}{b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{2}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{2}\text{ is irrational.}
\displaystyle \therefore 3+\sqrt{2}\text{ is an irrational number.}
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\displaystyle \textbf{Question 4. }\text{Prove that }4-5\sqrt{2}\text{ is an irrational number.}\qquad [\mathrm{CBSE}\ 2010]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }4-5\sqrt{2}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 4-5\sqrt{2}=\frac{a}{b}
\displaystyle \Rightarrow 5\sqrt{2}=4-\frac{a}{b}
\displaystyle \Rightarrow 5\sqrt{2}=\frac{4b-a}{b}
\displaystyle \Rightarrow \sqrt{2}=\frac{4b-a}{5b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{4b-a}{5b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{2}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{2}\text{ is irrational.}
\displaystyle \therefore 4-5\sqrt{2}\text{ is an irrational number.}
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\displaystyle \textbf{Question 5. }\text{Prove that }2\sqrt{3}-1\text{ is an irrational number.}\qquad [\mathrm{CBSE}\ 2010]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }2\sqrt{3}-1\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 2\sqrt{3}-1=\frac{a}{b}
\displaystyle \Rightarrow 2\sqrt{3}=\frac{a}{b}+1
\displaystyle \Rightarrow 2\sqrt{3}=\frac{a+b}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{a+b}{2b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a+b}{2b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 2\sqrt{3}-1\text{ is an irrational number.}
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\displaystyle \textbf{Question 6. }\text{Prove that }2-3\sqrt{5}\text{ is an irrational number.}\qquad [\mathrm{CBSE}\ 2010]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }2-3\sqrt{5}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 2-3\sqrt{5}=\frac{a}{b}
\displaystyle \Rightarrow 3\sqrt{5}=2-\frac{a}{b}
\displaystyle \Rightarrow 3\sqrt{5}=\frac{2b-a}{b}
\displaystyle \Rightarrow \sqrt{5}=\frac{2b-a}{3b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{2b-a}{3b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{5}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{5}\text{ is irrational.}
\displaystyle \therefore 2-3\sqrt{5}\text{ is an irrational number.}
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\displaystyle \textbf{Question 7. }\text{Prove that }\sqrt{5}+\sqrt{3}\text{ is irrational.}
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }\sqrt{5}+\sqrt{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle \sqrt{5}+\sqrt{3}=\frac{a}{b}
\displaystyle \Rightarrow \sqrt{5}=\frac{a}{b}-\sqrt{3}
\displaystyle \text{Squaring both sides,}
\displaystyle 5=\frac{a^{2}}{b^{2}}+3-\frac{2a\sqrt{3}}{b}
\displaystyle \Rightarrow \frac{2a\sqrt{3}}{b}=\frac{a^{2}}{b^{2}}-2
\displaystyle \Rightarrow \sqrt{3}=\frac{a^{2}-2b^{2}}{2ab}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a^{2}-2b^{2}}{2ab}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore \sqrt{5}+\sqrt{3}\text{ is irrational.}
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\displaystyle \textbf{Question 8. }\text{Given that }\sqrt{2}\text{ is irrational, prove that }5+3\sqrt{2} \\ \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2018]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }5+3\sqrt{2}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 5+3\sqrt{2}=\frac{a}{b}
\displaystyle \Rightarrow 3\sqrt{2}=\frac{a}{b}-5
\displaystyle \Rightarrow 3\sqrt{2}=\frac{a-5b}{b}
\displaystyle \Rightarrow \sqrt{2}=\frac{a-5b}{3b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a-5b}{3b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{2}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{2}\text{ is irrational.}
\displaystyle \therefore 5+3\sqrt{2}\text{ is an irrational number.}
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\displaystyle \textbf{Question 9. }\text{Prove that }\frac{2+\sqrt{3}}{5}\text{ is an irrational number, given that }\sqrt{3}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2019]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }\frac{2+\sqrt{3}}{5}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle \frac{2+\sqrt{3}}{5}=\frac{a}{b}
\displaystyle \Rightarrow 2+\sqrt{3}=\frac{5a}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{5a}{b}-2
\displaystyle \Rightarrow \sqrt{3}=\frac{5a-2b}{b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{5a-2b}{b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore \frac{2+\sqrt{3}}{5}\text{ is an irrational number.}
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\displaystyle \textbf{Question 10. }\text{Prove that }2+5\sqrt{3}\text{ is an irrational number, given that }\sqrt{3}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2019]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }2+5\sqrt{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 2+5\sqrt{3}=\frac{a}{b}
\displaystyle \Rightarrow 5\sqrt{3}=\frac{a}{b}-2
\displaystyle \Rightarrow 5\sqrt{3}=\frac{a-2b}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{a-2b}{5b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a-2b}{5b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 2+5\sqrt{3}\text{ is an irrational number.}
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\displaystyle \textbf{Question 11. }\text{Prove that }\sqrt{2}+\sqrt{3}\text{ is irrational.}
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }\sqrt{2}+\sqrt{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle \sqrt{2}+\sqrt{3}=\frac{a}{b}
\displaystyle \Rightarrow \sqrt{2}=\frac{a}{b}-\sqrt{3}
\displaystyle \text{Squaring both sides,}
\displaystyle 2=\frac{a^{2}}{b^{2}}+3-\frac{2a\sqrt{3}}{b}
\displaystyle \Rightarrow \frac{2a\sqrt{3}}{b}=\frac{a^{2}}{b^{2}}+1
\displaystyle \Rightarrow \sqrt{3}=\frac{a^{2}+b^{2}}{2ab}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a^{2}+b^{2}}{2ab}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore \sqrt{2}+\sqrt{3}\text{ is irrational.}
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\displaystyle \textbf{Question 12. }\text{Prove that }2+\sqrt{3}\text{ is an irrational number, given that }\sqrt{3}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2023]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }2+\sqrt{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 2+\sqrt{3}=\frac{a}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{a}{b}-2
\displaystyle \Rightarrow \sqrt{3}=\frac{a-2b}{b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a-2b}{b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 2+\sqrt{3}\text{ is an irrational number.}
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\displaystyle \textbf{Question 13. }\text{Prove that }5-2\sqrt{3}\text{ is an irrational number. It is given that }\sqrt{3}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2024]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }5-2\sqrt{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 5-2\sqrt{3}=\frac{a}{b}
\displaystyle \Rightarrow 2\sqrt{3}=5-\frac{a}{b}
\displaystyle \Rightarrow 2\sqrt{3}=\frac{5b-a}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{5b-a}{2b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{5b-a}{2b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 5-2\sqrt{3}\text{ is an irrational number.}
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\displaystyle \textbf{Question 14. }\text{Prove that }\frac{2-\sqrt{3}}{5}\text{ is an irrational number, given }\sqrt{3} \\ \text{is an irrational number.}\  [\mathrm{CBSE}\ 2024]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }\frac{2-\sqrt{3}}{5}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle \frac{2-\sqrt{3}}{5}=\frac{a}{b}
\displaystyle \Rightarrow 2-\sqrt{3}=\frac{5a}{b}
\displaystyle \Rightarrow \sqrt{3}=2-\frac{5a}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{2b-5a}{b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{2b-5a}{b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore \frac{2-\sqrt{3}}{5}\text{ is an irrational number.}
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\displaystyle \textbf{Question 15. }\text{Prove that }(\sqrt{2}+\sqrt{3})^{2}\text{ is an irrational number, given that }\sqrt{6}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2024]
\displaystyle \text{Answer:}
\displaystyle (\sqrt{2}+\sqrt{3})^{2}=2+3+2\sqrt{6}
\displaystyle =5+2\sqrt{6}
\displaystyle \text{Let us assume on the contrary that }5+2\sqrt{6}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 5+2\sqrt{6}=\frac{a}{b}
\displaystyle \Rightarrow 2\sqrt{6}=\frac{a}{b}-5
\displaystyle \Rightarrow 2\sqrt{6}=\frac{a-5b}{b}
\displaystyle \Rightarrow \sqrt{6}=\frac{a-5b}{2b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{a-5b}{2b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{6}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{6}\text{ is irrational.}
\displaystyle \therefore (\sqrt{2}+\sqrt{3})^{2}\text{ is an irrational number.}
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\displaystyle \textbf{Question 16. }\text{Prove that }5\sqrt{3}+\frac{2}{3}\text{ is an irrational number given that }\sqrt{3}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2025]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }5\sqrt{3}+\frac{2}{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 5\sqrt{3}+\frac{2}{3}=\frac{a}{b}
\displaystyle \Rightarrow 5\sqrt{3}=\frac{a}{b}-\frac{2}{3}
\displaystyle \Rightarrow 5\sqrt{3}=\frac{3a-2b}{3b}
\displaystyle \Rightarrow \sqrt{3}=\frac{3a-2b}{15b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{3a-2b}{15b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{3}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{3}\text{ is irrational.}
\displaystyle \therefore 5\sqrt{3}+\frac{2}{3}\text{ is an irrational number.}
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\displaystyle \textbf{Question 17. }\text{Prove that }\left(4\sqrt{2}+\frac{5}{3}\right)\text{ is an irrational number given that }\sqrt{2}
\displaystyle \text{is an irrational number.}\qquad [\mathrm{CBSE}\ 2025]
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume on the contrary that }4\sqrt{2}+\frac{5}{3}\text{ is rational.}
\displaystyle \text{Then, there exist co-prime positive integers }a\text{ and }b\text{ such that}
\displaystyle 4\sqrt{2}+\frac{5}{3}=\frac{a}{b}
\displaystyle \Rightarrow 4\sqrt{2}=\frac{a}{b}-\frac{5}{3}
\displaystyle \Rightarrow 4\sqrt{2}=\frac{3a-5b}{3b}
\displaystyle \Rightarrow \sqrt{2}=\frac{3a-5b}{12b}
\displaystyle \text{Since }a,b\text{ are integers, }\frac{3a-5b}{12b}\text{ is a rational number.}
\displaystyle \therefore \sqrt{2}\text{ is rational.}
\displaystyle \text{This contradicts the fact that }\sqrt{2}\text{ is irrational.}
\displaystyle \therefore 4\sqrt{2}+\frac{5}{3}\text{ is an irrational number.}
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\displaystyle \textbf{Question 18. }\text{Prove that for any prime positive integer }p,\sqrt{p}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume that }\sqrt{p}\text{ is rational.}
\displaystyle \therefore \sqrt{p}=\frac{a}{b},\text{ where }a\text{ and }b\text{ are coprime positive integers and }b\ne0.
\displaystyle \therefore p=\frac{a^{2}}{b^{2}}
\displaystyle \therefore a^{2}=pb^{2}
\displaystyle \therefore p\text{ divides }a^{2}.
\displaystyle \therefore p\text{ divides }a.
\displaystyle \text{Let }a=pk,\text{ for some integer }k.
\displaystyle \therefore a^{2}=p^{2}k^{2}
\displaystyle \therefore p^{2}k^{2}=pb^{2}
\displaystyle \therefore b^{2}=pk^{2}
\displaystyle \therefore p\text{ divides }b^{2}.
\displaystyle \therefore p\text{ divides }b.
\displaystyle \text{Thus, }p\text{ divides both }a\text{ and }b,\text{ which contradicts that }a\text{ and }b\text{ are coprime.}
\displaystyle \therefore \sqrt{p}\text{ is an irrational number.}
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\displaystyle \textbf{Question 19. }\text{If }p,q\text{ are prime positive integers, prove that }\sqrt{p}+\sqrt{q}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let us assume that }\sqrt{p}+\sqrt{q}\text{ is rational.}
\displaystyle \therefore \sqrt{p}+\sqrt{q}=r,\text{ where }r\text{ is rational.}
\displaystyle \sqrt{p}=r-\sqrt{q}
\displaystyle \text{Squaring both sides,}
\displaystyle p=r^{2}+q-2r\sqrt{q}
\displaystyle 2r\sqrt{q}=r^{2}+q-p
\displaystyle \sqrt{q}=\frac{r^{2}+q-p}{2r}
\displaystyle \text{This implies that }\sqrt{q}\text{ is rational, which is a contradiction.}
\displaystyle \therefore \sqrt{p}+\sqrt{q}\text{ is an irrational number.}
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