EXAMPLES
Real Numbers • 7 Questions
Question 1
Hint available
Consider the numbers 4 n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero.
Key Idea
A number ends with the digit zero iff it is divisible by 10. Since $10 = 2 \times 5$, a number must contain both the prime factors 2 and 5. The power $4^n$ contains only the prime factor 2.
Step-by-Step Solution
Given: $4^n$ where $n \in \mathbb{N}$.
To Find: Whether there exists $n$ such that $4^n$ ends with 0 (i.e., $4^n$ is divisible by 10).
Step 1: Express $4^n$ in terms of its prime factors.
\[4^n = (2^2)^n = 2^{2n}.\]
Thus $4^n$ contains only the prime factor 2.
Step 2: For a number to be divisible by 10, it must contain the factor 5 as well (since $10 = 2 \times 5$).
Step 3: Observe that $2^{2n}$ has no factor 5 for any natural $n$ because the only prime factor present is 2.
Therefore $4^n$ can never be divisible by 5, and consequently it can never be divisible by 10.
Conclusion: No natural number $n$ makes $4^n$ end with the digit zero.
To Find: Whether there exists $n$ such that $4^n$ ends with 0 (i.e., $4^n$ is divisible by 10).
Step 1: Express $4^n$ in terms of its prime factors.
\[4^n = (2^2)^n = 2^{2n}.\]
Thus $4^n$ contains only the prime factor 2.
Step 2: For a number to be divisible by 10, it must contain the factor 5 as well (since $10 = 2 \times 5$).
Step 3: Observe that $2^{2n}$ has no factor 5 for any natural $n$ because the only prime factor present is 2.
Therefore $4^n$ can never be divisible by 5, and consequently it can never be divisible by 10.
Conclusion: No natural number $n$ makes $4^n$ end with the digit zero.
Question 2
Hint available
Find the LCM and HCF of 6 and 20 by the prime factorisation method.
Key Idea
Prime factorisation of numbers; HCF is the product of common prime factors with the smallest exponents, and LCM is the product of all prime factors taken with the highest exponents.
Step-by-Step Solution
Given: Two numbers 6 and 20.
Step 1: Write the prime factorisation of each number.
$$6 = 2 \times 3$$
$$20 = 2^{2} \times 5$$
Step 2: Identify the common prime factors.
- Common prime factor: 2 (lowest power = $2^{1}$).
Hence, HCF = $2^{1} = 2$.
Step 3: For LCM, take each prime factor appearing in either number with the highest power.
- From 6: $2^{1},\;3^{1}$
- From 20: $2^{2},\;5^{1}$
Highest powers: $2^{2},\;3^{1},\;5^{1}$.
Thus, LCM = $2^{2} \times 3 \times 5 = 4 \times 3 \times 5 = 60$.
Conclusion: HCF(6,20) = 2 and LCM(6,20) = 60.
Step 1: Write the prime factorisation of each number.
$$6 = 2 \times 3$$
$$20 = 2^{2} \times 5$$
Step 2: Identify the common prime factors.
- Common prime factor: 2 (lowest power = $2^{1}$).
Hence, HCF = $2^{1} = 2$.
Step 3: For LCM, take each prime factor appearing in either number with the highest power.
- From 6: $2^{1},\;3^{1}$
- From 20: $2^{2},\;5^{1}$
Highest powers: $2^{2},\;3^{1},\;5^{1}$.
Thus, LCM = $2^{2} \times 3 \times 5 = 4 \times 3 \times 5 = 60$.
Conclusion: HCF(6,20) = 2 and LCM(6,20) = 60.
Question 3
Hint available
Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM.
Key Idea
Prime factorisation method: HCF is the product of common prime factors raised to the smallest powers; LCM is the product of all prime factors raised to the highest powers.
Step-by-Step Solution
Given: Two numbers 96 and 404.
Step 1 – Prime factorisation
\[96 = 2 \times 48 = 2 \times 2 \times 24 = 2 \times 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 2 \times 6 = 2^5 \times 3\]
\[404 = 2 \times 202 = 2 \times 2 \times 101 = 2^2 \times 101\]
(Here 101 is a prime number.)
Step 2 – Find HCF
Identify the common prime factors and take the smallest exponent of each:
- Common prime factor: \(2\)
- Smallest exponent of \(2\) in the two factorizations = \(2\) (since \(96 = 2^5\) and \(404 = 2^2\)).
Thus,
\[\text{HCF} = 2^{2} = 4\]
Step 3 – Find LCM
Take each prime factor appearing in either number with the highest exponent:
- \(2\) : highest exponent = \(5\) (from 96)
- \(3\) : appears only in 96 → exponent = \(1\)
- \(101\) : appears only in 404 → exponent = \(1\)
Hence,
\[\text{LCM} = 2^{5} \times 3^{1} \times 101^{1} = 32 \times 3 \times 101 = 96 \times 101 = 9696\]
Step 4 – Verification (optional)
Using the relation \(\text{LCM} \times \text{HCF} = a \times b\):
\[9696 \times 4 = 96 \times 404 = 38784\]
Both sides are equal, confirming the results.
Conclusion
- HCF of 96 and 404 = 4
- LCM of 96 and 404 = 9696
Step 1 – Prime factorisation
\[96 = 2 \times 48 = 2 \times 2 \times 24 = 2 \times 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 2 \times 6 = 2^5 \times 3\]
\[404 = 2 \times 202 = 2 \times 2 \times 101 = 2^2 \times 101\]
(Here 101 is a prime number.)
Step 2 – Find HCF
Identify the common prime factors and take the smallest exponent of each:
- Common prime factor: \(2\)
- Smallest exponent of \(2\) in the two factorizations = \(2\) (since \(96 = 2^5\) and \(404 = 2^2\)).
Thus,
\[\text{HCF} = 2^{2} = 4\]
Step 3 – Find LCM
Take each prime factor appearing in either number with the highest exponent:
- \(2\) : highest exponent = \(5\) (from 96)
- \(3\) : appears only in 96 → exponent = \(1\)
- \(101\) : appears only in 404 → exponent = \(1\)
Hence,
\[\text{LCM} = 2^{5} \times 3^{1} \times 101^{1} = 32 \times 3 \times 101 = 96 \times 101 = 9696\]
Step 4 – Verification (optional)
Using the relation \(\text{LCM} \times \text{HCF} = a \times b\):
\[9696 \times 4 = 96 \times 404 = 38784\]
Both sides are equal, confirming the results.
Conclusion
- HCF of 96 and 404 = 4
- LCM of 96 and 404 = 9696
Question 4
Hint available
Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method.
Key Idea
Use the prime factorisation method: HCF is the product of the lowest powers of common prime factors; LCM is the product of the highest powers of all prime factors appearing in the numbers.
Step-by-Step Solution
Given: Numbers are 6, 72 and 120.
To Find: HCF and LCM using prime factorisation.
Step 1 – Prime factorise each number
$$\begin{aligned}
6 &= 2 \times 3 \\
72 &= 2^3 \times 3^2 \\
120 &= 2^3 \times 3 \times 5
\end{aligned}$$
Step 2 – Identify the common prime factors
The primes common to all three numbers are 2 and 3.
Step 3 – Determine the lowest powers of the common primes
- For 2: lowest exponent = $\min\{1,3,3\}=1$
- For 3: lowest exponent = $\min\{1,2,1\}=1$
Step 4 – Compute the HCF
$$\text{HCF}=2^{1}\times 3^{1}=2\times 3=6$$
Step 5 – Determine the highest powers of all primes appearing
- For 2: highest exponent = $\max\{1,3,3\}=3$
- For 3: highest exponent = $\max\{1,2,1\}=2$
- For 5: highest exponent = $\max\{0,0,1\}=1$ (appears only in 120)
Step 6 – Compute the LCM
$$\text{LCM}=2^{3}\times 3^{2}\times 5^{1}=8\times 9\times 5=360$$
Conclusion: HCF = 6 and LCM = 360.
To Find: HCF and LCM using prime factorisation.
Step 1 – Prime factorise each number
$$\begin{aligned}
6 &= 2 \times 3 \\
72 &= 2^3 \times 3^2 \\
120 &= 2^3 \times 3 \times 5
\end{aligned}$$
Step 2 – Identify the common prime factors
The primes common to all three numbers are 2 and 3.
Step 3 – Determine the lowest powers of the common primes
- For 2: lowest exponent = $\min\{1,3,3\}=1$
- For 3: lowest exponent = $\min\{1,2,1\}=1$
Step 4 – Compute the HCF
$$\text{HCF}=2^{1}\times 3^{1}=2\times 3=6$$
Step 5 – Determine the highest powers of all primes appearing
- For 2: highest exponent = $\max\{1,3,3\}=3$
- For 3: highest exponent = $\max\{1,2,1\}=2$
- For 5: highest exponent = $\max\{0,0,1\}=1$ (appears only in 120)
Step 6 – Compute the LCM
$$\text{LCM}=2^{3}\times 3^{2}\times 5^{1}=8\times 9\times 5=360$$
Conclusion: HCF = 6 and LCM = 360.
Question 5
Hint available
Prove that 3 is irrational.
Key Idea
Proof by contradiction using the Fundamental Theorem of Arithmetic (unique prime factorisation) and the fact that in a reduced fraction the powers of any prime factor must be even for a perfect square.
Step-by-Step Solution
Given: We have to show that $\sqrt{3}$ cannot be expressed as a rational number.
To Prove: $\sqrt{3}$ is irrational.
Step 1 – Assume the contrary
Assume that $\sqrt{3}$ is rational. Then there exist two integers $p$ and $q$ (with $q
eq 0$) such that
$$\sqrt{3}=\frac{p}{q}$$
where the fraction $\frac{p}{q}$ is in its lowest terms, i.e., $\gcd(p,q)=1$.
Step 2 – Square both sides
Squaring the equality gives
$$3 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad p^{2}=3q^{2}.$$
Thus $p^{2}$ is a multiple of $3$.
Step 3 – Use the property of prime divisibility
If a prime divides a square, it must divide the number itself. Hence, from $3\mid p^{2}$ we conclude $3\mid p$.
Let $p = 3k$ for some integer $k$.
Step 4 – Substitute back
Substituting $p = 3k$ in $p^{2}=3q^{2}$ gives
$$ (3k)^{2}=3q^{2} \quad \Rightarrow \quad 9k^{2}=3q^{2} \quad \Rightarrow \quad 3k^{2}=q^{2}.$$
Thus $q^{2}$ is also a multiple of $3$, which implies $3\mid q$.
Step 5 – Arrive at a contradiction
We have shown that both $p$ and $q$ are divisible by $3$. This contradicts the initial assumption that $p$ and $q$ have no common factor (i.e., $\gcd(p,q)=1$).
Conclusion
The assumption that $\sqrt{3}$ is rational leads to a contradiction. Hence the assumption is false and $\sqrt{3}$ is irrational.
(If the question literally asks to prove that the number 3 itself is irrational, the answer is that 3 is a rational number because it can be expressed as $\frac{3}{1}$. The intended statement in the textbook is to prove that $\sqrt{3}$ is irrational.)
To Prove: $\sqrt{3}$ is irrational.
Step 1 – Assume the contrary
Assume that $\sqrt{3}$ is rational. Then there exist two integers $p$ and $q$ (with $q
eq 0$) such that
$$\sqrt{3}=\frac{p}{q}$$
where the fraction $\frac{p}{q}$ is in its lowest terms, i.e., $\gcd(p,q)=1$.
Step 2 – Square both sides
Squaring the equality gives
$$3 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad p^{2}=3q^{2}.$$
Thus $p^{2}$ is a multiple of $3$.
Step 3 – Use the property of prime divisibility
If a prime divides a square, it must divide the number itself. Hence, from $3\mid p^{2}$ we conclude $3\mid p$.
Let $p = 3k$ for some integer $k$.
Step 4 – Substitute back
Substituting $p = 3k$ in $p^{2}=3q^{2}$ gives
$$ (3k)^{2}=3q^{2} \quad \Rightarrow \quad 9k^{2}=3q^{2} \quad \Rightarrow \quad 3k^{2}=q^{2}.$$
Thus $q^{2}$ is also a multiple of $3$, which implies $3\mid q$.
Step 5 – Arrive at a contradiction
We have shown that both $p$ and $q$ are divisible by $3$. This contradicts the initial assumption that $p$ and $q$ have no common factor (i.e., $\gcd(p,q)=1$).
Conclusion
The assumption that $\sqrt{3}$ is rational leads to a contradiction. Hence the assumption is false and $\sqrt{3}$ is irrational.
(If the question literally asks to prove that the number 3 itself is irrational, the answer is that 3 is a rational number because it can be expressed as $\frac{3}{1}$. The intended statement in the textbook is to prove that $\sqrt{3}$ is irrational.)
Question 6
Hint available
Show that 5– 3 is irrational.
Key Idea
Proof by contradiction: assume the number is rational, express it as a fraction of two coprime integers, and derive an impossible integer equation using the properties of squares of integers.
Step-by-Step Solution
Given: \(\sqrt{5}-\sqrt{3}\)
To Prove: \(\sqrt{5}-\sqrt{3}\) is irrational.
Step 1: Assume the contrary, i.e., \(\sqrt{5}-\sqrt{3}=\dfrac{p}{q}\) where \(p,q\in\mathbb{Z},\;q
eq0\) and \(\gcd(p,q)=1\).
Step 2: Rearrange the equation
\[\sqrt{5}=\sqrt{3}+\dfrac{p}{q}.\]
Step 3: Square both sides
\[5 = \left(\sqrt{3}+\dfrac{p}{q}\right)^{2}=3+\dfrac{p^{2}}{q^{2}}+2\sqrt{3}\,\dfrac{p}{q}.\]
Step 4: Isolate the term containing \(\sqrt{3}\)
\[2\sqrt{3}\,\dfrac{p}{q}=5-3-\dfrac{p^{2}}{q^{2}}=2-\dfrac{p^{2}}{q^{2}}.\]
Step 5: Multiply by \(q^{2}\) to clear denominators
\[2\sqrt{3}\,p q = 2q^{2}-p^{2}.\]
Step 6: The right‑hand side is an integer; therefore \(2\sqrt{3}\,p q\) must be an integer. Since \(p,q\) are integers, this implies \(\sqrt{3}\) is rational, which is impossible because \(\sqrt{3}\) is known to be irrational (proved earlier in the chapter).
Step 7: The contradiction arises from the assumption that \(\sqrt{5}-\sqrt{3}\) is rational. Hence the assumption is false.
Conclusion: \(\sqrt{5}-\sqrt{3}\) is irrational.
To Prove: \(\sqrt{5}-\sqrt{3}\) is irrational.
Step 1: Assume the contrary, i.e., \(\sqrt{5}-\sqrt{3}=\dfrac{p}{q}\) where \(p,q\in\mathbb{Z},\;q
eq0\) and \(\gcd(p,q)=1\).
Step 2: Rearrange the equation
\[\sqrt{5}=\sqrt{3}+\dfrac{p}{q}.\]
Step 3: Square both sides
\[5 = \left(\sqrt{3}+\dfrac{p}{q}\right)^{2}=3+\dfrac{p^{2}}{q^{2}}+2\sqrt{3}\,\dfrac{p}{q}.\]
Step 4: Isolate the term containing \(\sqrt{3}\)
\[2\sqrt{3}\,\dfrac{p}{q}=5-3-\dfrac{p^{2}}{q^{2}}=2-\dfrac{p^{2}}{q^{2}}.\]
Step 5: Multiply by \(q^{2}\) to clear denominators
\[2\sqrt{3}\,p q = 2q^{2}-p^{2}.\]
Step 6: The right‑hand side is an integer; therefore \(2\sqrt{3}\,p q\) must be an integer. Since \(p,q\) are integers, this implies \(\sqrt{3}\) is rational, which is impossible because \(\sqrt{3}\) is known to be irrational (proved earlier in the chapter).
Step 7: The contradiction arises from the assumption that \(\sqrt{5}-\sqrt{3}\) is rational. Hence the assumption is false.
Conclusion: \(\sqrt{5}-\sqrt{3}\) is irrational.
Question 7
Hint available
Show that 32 is irrational.
Key Idea
Proof by contradiction using the definition of rational numbers and the fact that the square of an even integer is divisible by 4.
Step-by-Step Solution
Given: \(\sqrt{32}\) \(=\) \(\sqrt{2^5}\).
To Prove: \(\sqrt{32}\) cannot be expressed as a ratio of two integers with no common factor (i.e., it is irrational).
Step 1: Assume the contrary, that \(\sqrt{32}\) is rational. Then there exist coprime integers \(p\) and \(q\) (\(q
eq 0\)) such that
$$\sqrt{32}=\frac{p}{q}$$
Step 2: Square both sides to eliminate the square root.
$$32 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad 32 q^{2}=p^{2}.$$
Step 3: Write 32 as a power of 2: \(32 = 2^{5}\). Hence
$$2^{5} q^{2}=p^{2}.$$
Step 4: From the equation, \(p^{2}\) is divisible by \(2\); therefore \(p\) must be even. Let \(p=2k\) for some integer \(k\).
Step 5: Substitute \(p=2k\) into the equation:
$$2^{5} q^{2} = (2k)^{2}=4k^{2} \quad \Rightarrow \quad 2^{5} q^{2}=4k^{2}.$$
Divide both sides by 4 (i.e., \(2^{2}\)):
$$2^{3} q^{2}=k^{2} \quad \Rightarrow \quad 8 q^{2}=k^{2}.$$
Step 6: The right‑hand side \(k^{2}\) is again divisible by 2, so \(k\) is even. Hence \(k=2m\) for some integer \(m\).
Step 7: Substituting back, we get
$$8 q^{2}= (2m)^{2}=4m^{2} \quad \Rightarrow \quad 2 q^{2}=m^{2}.$$
Thus \(m^{2}\) is even, implying \(m\) is even, and consequently \(q\) is also even.
Step 8: We have shown that both \(p\) and \(q\) are even, which means they have a common factor 2. This contradicts the initial assumption that \(p\) and \(q\) are coprime.
Conclusion: The assumption that \(\sqrt{32}\) is rational leads to a contradiction. Hence \(\sqrt{32}\) is irrational.
To Prove: \(\sqrt{32}\) cannot be expressed as a ratio of two integers with no common factor (i.e., it is irrational).
Step 1: Assume the contrary, that \(\sqrt{32}\) is rational. Then there exist coprime integers \(p\) and \(q\) (\(q
eq 0\)) such that
$$\sqrt{32}=\frac{p}{q}$$
Step 2: Square both sides to eliminate the square root.
$$32 = \frac{p^{2}}{q^{2}} \quad \Rightarrow \quad 32 q^{2}=p^{2}.$$
Step 3: Write 32 as a power of 2: \(32 = 2^{5}\). Hence
$$2^{5} q^{2}=p^{2}.$$
Step 4: From the equation, \(p^{2}\) is divisible by \(2\); therefore \(p\) must be even. Let \(p=2k\) for some integer \(k\).
Step 5: Substitute \(p=2k\) into the equation:
$$2^{5} q^{2} = (2k)^{2}=4k^{2} \quad \Rightarrow \quad 2^{5} q^{2}=4k^{2}.$$
Divide both sides by 4 (i.e., \(2^{2}\)):
$$2^{3} q^{2}=k^{2} \quad \Rightarrow \quad 8 q^{2}=k^{2}.$$
Step 6: The right‑hand side \(k^{2}\) is again divisible by 2, so \(k\) is even. Hence \(k=2m\) for some integer \(m\).
Step 7: Substituting back, we get
$$8 q^{2}= (2m)^{2}=4m^{2} \quad \Rightarrow \quad 2 q^{2}=m^{2}.$$
Thus \(m^{2}\) is even, implying \(m\) is even, and consequently \(q\) is also even.
Step 8: We have shown that both \(p\) and \(q\) are even, which means they have a common factor 2. This contradicts the initial assumption that \(p\) and \(q\) are coprime.
Conclusion: The assumption that \(\sqrt{32}\) is rational leads to a contradiction. Hence \(\sqrt{32}\) is irrational.