IBDP Maths MAA HL – SL 1.1 Operations with numbers HL Paper 1- Exam Style Questions
IB MATHEMATICS AA HL – Practice Questions- Paper1-All Topics
Topic : SL 1.1 Operations with numbers HL Paper 1
Topic 1- Number and algebra- Weightage : 15 %
All Questions for Topic : SL 1.1 – Operations with numbers in the form a × 10k where 1 ≤ a < 10 and k is an integer
▶️ Answer/Explanation
(a)
Even numbers: \( a = 2m \), \( b = 2n \), \( m, n \) integers.
Sum: \( a + b = 2m + 2n = 2(m + n) \), even.
(b)
Odd numbers: \( a = 2m + 1 \), \( b = 2n + 1 \), \( m, n \) integers.
Sum: \( a + b = 2m + 2n + 2 = 2(m + n + 1) \), even.
(c)
Multiples of 3: \( a = 3m \), \( b = 3n \), \( m, n \) integers.
Sum: \( a + b = 3m + 3n = 3(m + n) \), multiple of 3.
(d)
Non-multiples: 4 (\( 3 \cdot 1 + 1 \)), 7 (\( 3 \cdot 2 + 1 \)).
Sum: \( 4 + 7 = 11 \) (\( 3 \cdot 3 + 2 \)), not a multiple of 3.
(e)
Integer \( k \): \( 3m \), \( 3m + 1 \), or \( 3m + 2 \).
1. \( k = 3m \): \( k^2 = 9m^2 = 3(3m^2) \), form \( 3n \).
2. \( k = 3m + 1 \): \( k^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1 \), form \( 3n + 1 \).
3. \( k = 3m + 2 \): \( k^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1 \), form \( 3n + 1 \).
\( k^2 \) never \( 3n + 2 \).
▶️ Answer/Explanation
(a)
(i) \(\sqrt{3}\)
Assume \(\sqrt{3} = \frac{a}{b}\), \(a, b \in \mathbb{Z}\), no common factors.
Then \(3 = \frac{a^2}{b^2} \implies a^2 = 3b^2\).
\(a^2\) multiple of 3, so \(a\) multiple of 3. Let \(a = 3c\).
Then \(9c^2 = 3b^2 \implies 3c^2 = b^2\).
\(b^2\) multiple of 3, so \(b\) multiple of 3. Contradiction.
Thus, \(\sqrt{3}\) irrational.
(ii) \(\sqrt[3]{2}\)
Assume \(\sqrt[3]{2} = \frac{a}{b}\), \(a, b \in \mathbb{Z}\), no common factors.
Then \(2 = \frac{a^3}{b^3} \implies a^3 = 2b^3\).
\(a^3\) even, so \(a\) even. Let \(a = 2c\).
Then \(8c^3 = 2b^3 \implies 4c^3 = b^3\).
\(b^3\) even, so \(b\) even. Contradiction.
Thus, \(\sqrt[3]{2}\) irrational.
(iii) \(\log_2 3\)
Assume \(\log_2 3 = \frac{a}{b}\), \(a, b \in \mathbb{Z}^+\).
Then \(2^{\frac{a}{b}} = 3 \implies 2^a = 3^b\).
\(2^a\) even, \(3^b\) odd. Contradiction.
Thus, \(\log_2 3\) irrational.
(b)
(i) \(3^{1/2} + 3^{-1}\)
Assume \( \sqrt{3} + \frac{1}{3} = r \), \( r \in \mathbb{Q} \).
Then \(\sqrt{3} = r – \frac{1}{3}\).
Since \( r, \frac{1}{3} \in \mathbb{Q} \), \( r – \frac{1}{3} \in \mathbb{Q} \).
But \(\sqrt{3}\) irrational (from (a)(i)). Contradiction.
Thus, \( \sqrt{3} + \frac{1}{3} \) irrational.
(ii) \(\log_2 18\)
Assume \(\log_2 18 = \frac{a}{b}\), \( a, b \in \mathbb{Z}^+ \).
Then \( 2^{\frac{a}{b}} = 18 = 2 \cdot 3^2 \implies 2^{\frac{a}{b} – 1} = 3^2 \).
Thus, \(\frac{a}{b} – 1 = \log_2 9 = 2 \log_2 3\).
So, \(\log_2 3 = \frac{a – b}{2b}\), rational.
But \(\log_2 3\) irrational (from (a)(iii)). Contradiction.
Thus, \(\log_2 18\) irrational.