EXERCISE 1.2
Real Numbers • 3 Questions
Question 1
Hint available
Prove that 5 is irrational.
Key Idea
Proof by contradiction using the Fundamental Theorem of Arithmetic (unique prime factorisation).
Step-by-Step Solution
Given: We have to show that $\sqrt{5}$ cannot be expressed as a rational number.
To Prove: $\sqrt{5}$ is irrational.
Step 1 – Assume the contrary
Assume that $\sqrt{5}$ is rational. Then it can be written as a fraction of two integers having no common factor other than 1 (i.e., in lowest terms):
$$\sqrt{5}=\frac{p}{q},\qquad p,q\in\mathbb{Z},\; q
eq0,\; \gcd(p,q)=1.$$
Step 2 – Square both sides
Squaring gives
$$5 = \frac{p^{2}}{q^{2}} \;\Rightarrow\; p^{2}=5q^{2}.$$
Thus $p^{2}$ is a multiple of 5, which implies that $p$ itself must be a multiple of 5 (if a prime divides a square, it divides the base). Let $p=5k$ for some integer $k$.
Step 3 – Substitute back
Substituting $p=5k$ in $p^{2}=5q^{2}$:
$$ (5k)^{2}=5q^{2}\;\Rightarrow\;25k^{2}=5q^{2}\;\Rightarrow\;5k^{2}=q^{2}.$$
Hence $q^{2}$ is also a multiple of 5, so $q$ must be a multiple of 5.
Step 4 – Contradiction
Both $p$ and $q$ are multiples of 5, which means they have a common factor 5. This contradicts the assumption that $p$ and $q$ are coprime (i.e., the fraction is in lowest terms).
Conclusion
The assumption that $\sqrt{5}$ is rational leads to a contradiction. Therefore, $\sqrt{5}$ is irrational.
To Prove: $\sqrt{5}$ is irrational.
Step 1 – Assume the contrary
Assume that $\sqrt{5}$ is rational. Then it can be written as a fraction of two integers having no common factor other than 1 (i.e., in lowest terms):
$$\sqrt{5}=\frac{p}{q},\qquad p,q\in\mathbb{Z},\; q
eq0,\; \gcd(p,q)=1.$$
Step 2 – Square both sides
Squaring gives
$$5 = \frac{p^{2}}{q^{2}} \;\Rightarrow\; p^{2}=5q^{2}.$$
Thus $p^{2}$ is a multiple of 5, which implies that $p$ itself must be a multiple of 5 (if a prime divides a square, it divides the base). Let $p=5k$ for some integer $k$.
Step 3 – Substitute back
Substituting $p=5k$ in $p^{2}=5q^{2}$:
$$ (5k)^{2}=5q^{2}\;\Rightarrow\;25k^{2}=5q^{2}\;\Rightarrow\;5k^{2}=q^{2}.$$
Hence $q^{2}$ is also a multiple of 5, so $q$ must be a multiple of 5.
Step 4 – Contradiction
Both $p$ and $q$ are multiples of 5, which means they have a common factor 5. This contradicts the assumption that $p$ and $q$ are coprime (i.e., the fraction is in lowest terms).
Conclusion
The assumption that $\sqrt{5}$ is rational leads to a contradiction. Therefore, $\sqrt{5}$ is irrational.
Question 2
Hint available
Prove that 32 5 is irrational.
Key Idea
If \(\sqrt{n}\) is rational then \(n\) must be a perfect square. Also, the sum of a rational number and an irrational number is irrational.
Step-by-Step Solution
Given: \(\sqrt{32}+\sqrt{5}\).
To Prove: The number is irrational.
Step 1 – Assume the contrary
Assume that \(\sqrt{32}+\sqrt{5}\) is rational. Then there exist integers \(p,q\) (with \(q
eq0\) and \(\gcd(p,q)=1\)) such that
\[\sqrt{32}+\sqrt{5}=\frac{p}{q}.\]
Step 2 – Isolate one radical
\[\sqrt{5}=\frac{p}{q}-\sqrt{32}.\]
Since \(\sqrt{32}=4\sqrt{2}\) and \(4\sqrt{2}\) is irrational, the right‑hand side is the difference of a rational number and an irrational number; therefore it is irrational.
Step 3 – Square both sides
Square the equality in Step 1:
\[\left(\sqrt{32}+\sqrt{5}\right)^2 = \left(\frac{p}{q}\right)^2\]
\[32+5+2\sqrt{32\cdot5}=\frac{p^2}{q^2}\]
\[37+2\sqrt{160}=\frac{p^2}{q^2}.\]
Hence
\[2\sqrt{160}=\frac{p^2}{q^2}-37.\]
The right‑hand side is rational (difference of two rationals), so \(\sqrt{160}\) must be rational.
Step 4 – Reduce \(\sqrt{160}\)
\[\sqrt{160}=\sqrt{16\times10}=4\sqrt{10}.\]
Thus \(4\sqrt{10}\) is rational, which implies \(\sqrt{10}\) is rational.
Step 5 – Contradiction
If \(\sqrt{10}\) were rational, then 10 would be a perfect square, which is false. Hence our assumption that \(\sqrt{32}+\sqrt{5}\) is rational leads to a contradiction.
Conclusion
Therefore \(\sqrt{32}+\sqrt{5}\) is irrational.
To Prove: The number is irrational.
Step 1 – Assume the contrary
Assume that \(\sqrt{32}+\sqrt{5}\) is rational. Then there exist integers \(p,q\) (with \(q
eq0\) and \(\gcd(p,q)=1\)) such that
\[\sqrt{32}+\sqrt{5}=\frac{p}{q}.\]
Step 2 – Isolate one radical
\[\sqrt{5}=\frac{p}{q}-\sqrt{32}.\]
Since \(\sqrt{32}=4\sqrt{2}\) and \(4\sqrt{2}\) is irrational, the right‑hand side is the difference of a rational number and an irrational number; therefore it is irrational.
Step 3 – Square both sides
Square the equality in Step 1:
\[\left(\sqrt{32}+\sqrt{5}\right)^2 = \left(\frac{p}{q}\right)^2\]
\[32+5+2\sqrt{32\cdot5}=\frac{p^2}{q^2}\]
\[37+2\sqrt{160}=\frac{p^2}{q^2}.\]
Hence
\[2\sqrt{160}=\frac{p^2}{q^2}-37.\]
The right‑hand side is rational (difference of two rationals), so \(\sqrt{160}\) must be rational.
Step 4 – Reduce \(\sqrt{160}\)
\[\sqrt{160}=\sqrt{16\times10}=4\sqrt{10}.\]
Thus \(4\sqrt{10}\) is rational, which implies \(\sqrt{10}\) is rational.
Step 5 – Contradiction
If \(\sqrt{10}\) were rational, then 10 would be a perfect square, which is false. Hence our assumption that \(\sqrt{32}+\sqrt{5}\) is rational leads to a contradiction.
Conclusion
Therefore \(\sqrt{32}+\sqrt{5}\) is irrational.
Question 3
Hint available
Prove that the following are irrationals : (i) 1 2 (ii) 75 (iii) 62
Key Idea
Use proof by contradiction: assume the number is rational (expressible as \frac{p}{q} in lowest terms) and show that this leads to a contradiction using the Fundamental Theorem of Arithmetic (prime factorisation).
Step-by-Step Solution
(i) \(\sqrt{2}\)
Given: Assume \(\sqrt{2}=\frac{p}{q}\) where \(p,q\) are integers with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(2=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=2q^{2}\).
Step 2: Since \(p^{2}\) is even, \(p\) must be even. Write \(p=2k\).
Step 3: Substitute: \((2k)^{2}=2q^{2}\) ⇒ \(4k^{2}=2q^{2}\) ⇒ \(q^{2}=2k^{2}\).
Step 4: Hence \(q^{2}\) is even, so \(q\) is even.
Step 5: Both \(p\) and \(q\) are even, contradicting the assumption that they are coprime.
Conclusion: The assumption is false; therefore \(\sqrt{2}\) is irrational.
(ii) \(\sqrt{75}\)
Given: \(75=3\times5^{2}\) so \(\sqrt{75}=5\sqrt{3}\).
Step 1: Suppose \(\sqrt{75}\) is rational. Then \(\sqrt{3}=\frac{\sqrt{75}}{5}\) would also be rational.
Step 2: But \(\sqrt{3}\) is known to be irrational (proved similarly to part (i)).
Step 3: This contradiction shows the original assumption is false.
Conclusion: Hence \(\sqrt{75}\) is irrational.
(iii) \(\sqrt{62}\)
Given: Assume \(\sqrt{62}=\frac{p}{q}\) with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(62=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=62q^{2}=2\cdot31\,q^{2}\).
Step 2: In the prime factorisation of the right‑hand side, the primes 2 and 31 appear to the first power (odd exponent).
Step 3: The square of an integer (\(p^{2}\)) must have each prime factor with an even exponent (Fundamental Theorem of Arithmetic).
Step 4: Hence the equality cannot hold; the assumption that \(\sqrt{62}\) is rational leads to a contradiction.
Conclusion: Therefore \(\sqrt{62}\) is irrational.
Overall, each number leads to a contradiction when assumed rational, proving they are all irrational.
Given: Assume \(\sqrt{2}=\frac{p}{q}\) where \(p,q\) are integers with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(2=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=2q^{2}\).
Step 2: Since \(p^{2}\) is even, \(p\) must be even. Write \(p=2k\).
Step 3: Substitute: \((2k)^{2}=2q^{2}\) ⇒ \(4k^{2}=2q^{2}\) ⇒ \(q^{2}=2k^{2}\).
Step 4: Hence \(q^{2}\) is even, so \(q\) is even.
Step 5: Both \(p\) and \(q\) are even, contradicting the assumption that they are coprime.
Conclusion: The assumption is false; therefore \(\sqrt{2}\) is irrational.
(ii) \(\sqrt{75}\)
Given: \(75=3\times5^{2}\) so \(\sqrt{75}=5\sqrt{3}\).
Step 1: Suppose \(\sqrt{75}\) is rational. Then \(\sqrt{3}=\frac{\sqrt{75}}{5}\) would also be rational.
Step 2: But \(\sqrt{3}\) is known to be irrational (proved similarly to part (i)).
Step 3: This contradiction shows the original assumption is false.
Conclusion: Hence \(\sqrt{75}\) is irrational.
(iii) \(\sqrt{62}\)
Given: Assume \(\sqrt{62}=\frac{p}{q}\) with \(\gcd(p,q)=1\).
Step 1: Square both sides → \(62=\frac{p^{2}}{q^{2}}\) ⇒ \(p^{2}=62q^{2}=2\cdot31\,q^{2}\).
Step 2: In the prime factorisation of the right‑hand side, the primes 2 and 31 appear to the first power (odd exponent).
Step 3: The square of an integer (\(p^{2}\)) must have each prime factor with an even exponent (Fundamental Theorem of Arithmetic).
Step 4: Hence the equality cannot hold; the assumption that \(\sqrt{62}\) is rational leads to a contradiction.
Conclusion: Therefore \(\sqrt{62}\) is irrational.
Overall, each number leads to a contradiction when assumed rational, proving they are all irrational.