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
Question:-SL 1.1 Operations with numbers HL Paper 1
Integers divided by 2 leave a remainder 0 or 1; thus any integer has one of the forms
\(2n\) (even)
\(2n+1\) (odd)
Integers divided by 3 leave a remainder 0, 1 or 2; thus any integer has one of the forms
\(3n \) (multiples of 3)
\(3n+1\) (leave remainder 1)
\(3n+2\) (leave remainder 2)
(a) Prove that the sum of two even numbers is an even number. [1]
(b) Prove that the sum of two odd numbers is an even number. [1]
(c) Prove that the sum of two multiples of 3 is a multiple of 3. [1]
(d) Prove by using a counterexample that the sum of two numbers which are not multiples of 3 is not necessarily a multiple of 3. [1]
(e) Prove that the square of any integer cannot have the form \(3n+2\). [3]
▶️Answer/Explanation
(a)
Let the two even numbers be \( a \) and \( b \).
By the definition of even numbers, we can express them as:
\(a = 2m, \quad b = 2n\) where \( m \) and \( n \) are integers.
Compute the sum:
\(a + b = 2m + 2n\)
Factor out 2:
\(a + b = 2(m + n)\)
Since \( m + n \) is an integer (because the sum of two integers is an integer), we conclude that \( a + b \) is of the form \( 2k \), where \( k = m + n \).
(b)
Let the two odd numbers be \( a \) and \( b \).
By the definition of odd numbers, we can express them as:
\(a = 2m+1, \quad b = 2n+1\) where \( m \) and \( n \) are integers.
Compute the sum:
\(a + b = 2m+1 + 2n+1\)
Factor out 2:
\(a + b = 2(m + n+1)\)
Since \( m + n +1\) is an integer (because the sum of two integers is an integer), we conclude that \( a + b \) is of the form \( 2(k+1) \), where \( k = m + n \).
(c)
Let two numbers be \( a \) and \( b \), both of which are multiples of 3.
By definition of multiples of 3, we can write:
$a = 3m, \quad b = 3n$
where \( m \) and \( n \) are integers.
Compute the sum:
$a + b = 3m + 3n$
Factor out 3:
$a + b = 3(m + n)$
Since \( m + n \) is an integer, it follows that \( a + b \) is a multiple of 3.
(d)
Consider 4 and 7:
4 leaves remainder 1 when divided by 3.
7 leaves remainder 1 when divided by 3.
Compute the sum:
$4 + 7 = 11$
Since \( 11 \) leaves remainder 2 when divided by 3, it is not a multiple of 3.
(e)
1. Case 1: \( k = 3m \)
$k^2 = (3m)^2 = 9m^2 = 3(3m^2)$
Since this is a multiple of 3, it has the form \( 3n \).
2. Case 2: \( k = 3m+1 \)
$k^2 = (3m+1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1$
This is of the form \( 3n+1 \).
3. Case 3: \( k = 3m+2 \)
$k^2 = (3m+2)^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1$
This is also of the form \( 3n+1 \).
Since \( k^2 \) is always of the form \( 3n \) or \( 3n+1 \), it **can never be of the form \( 3n+2 \)**.
Question:
[Maximum mark: 13] [without GDC]
(a) Prove by using contradiction that the following numbers are irrational
(i) \(\sqrt{3}\) (ii) \(\sqrt[3]{2}\) (iii) log2 3 [9]
(b) Given the results in (a) prove also by contradiction that
(i) 31/2 + 3-1 is an irrational number.
(ii) log2 18 is an irrational number. [4]
▶️Answer/Explanation
(a)
(i)
Assume, for contradiction, that the result is not true, that is, $\sqrt[3]{3}$ is a rational number.
Then $\sqrt[3]{3} = \frac{a}{b}$ where $a, b$ are integers with no common factors (the fraction is simplified).
Then
$3 = \frac{a^3}{b^3} \implies a^3 = 3b^3.$
Thus $a^3$ is a multiple of 3 and $a$ is a multiple of 3.
Let $a = 3c$.
Then $(3c)^3 = 3b^3 \implies 27c^3 = 3b^3 \implies 9c^3 = b^3$.
Thus $b^3$ is a multiple of 3 and $b$ is a multiple of 3.
Contradiction since $a, b$ have no common factors.
Therefore, $\sqrt[3]{3}$ is irrational
(ii)
Assume, for contradiction, that $\sqrt[3]{2}$ is rational.
Let $\sqrt[3]{2} = \frac{a}{b}$ where $a, b$ are integers with no common factors.
Then
$2 = \frac{a^3}{b^3} \implies a^3 = 2b^3$
Thus $a^3$ is even, and thus $a$ is even.
Let $a = 2c$.
Then
$(2c)^3 = 2b^3 \implies 8c^3 = 2b^3 \implies 4c^3 = b^3.$
Thus $b^3$ is even, and thus $b$ is even.
Contradiction since $a, b$ have no common factors.
Therefore, $\sqrt[3]{2}$ is irrational.
(iii)
Assume, for contradiction, that $\log_2 3 = \frac{a}{b}$ (rational) where $a, b$ are positive integers
(with no common factors, but this is not necessary here).
Then
$2^{\frac{a}{b}} = 3 \implies 2^a = 3^b.$
But $2^a$ is even while $3^b$ is odd. Contradiction!
Therefore, $\log_2 3$ is irrational.
(b)
i)
1. Assume, for contradiction, that $ \sqrt{3} + \frac{1}{3} $ is rational.
Let $ \sqrt{3} + \frac{1}{3} = r $, where $ r \in \mathbb{Q} $ (rational).
2. Rearrange to isolate $ \sqrt{3} $:
$\sqrt{3} = r – \frac{1}{3}.$
3. Since $ r \in \mathbb{Q} $ and $ \frac{1}{3} \in \mathbb{Q} $, their difference $ r – \frac{1}{3} $ is also rational. Thus, $ \sqrt{3} $ would be rational.
4. This contradicts the well-known fact that $ \sqrt{3} $ is irrational (proven earlier).
Thus, $ \sqrt{3} + \frac{1}{3} $ is irrational.
(ii)
This part of the proof remains unchanged, as it depends on the properties of logarithms, not the specific form of the first question. Refer to the earlier proof:
1. Assume, for contradiction, that $ \log_2 18 $ is rational.
2. Let $ \log_2 18 = \frac{a}{b} $, where $ a, b \in \mathbb{Z} $ and $ \frac{a}{b} $ is in its simplest form.
3. Rewrite using the definition of logarithms:
$2^{\frac{a}{b}} = 18.$
4. Proceed with the proof to show a contradiction, as shown earlier.
$ \log_2 18 $ is irrational.
Question:
a) Prove that the number \(14 641\) is the fourth power of an integer in any base greater than \(6\).
b) For \(a,b \in \mathbb{Z}\) the relation \(aRb\) is defined if and only if \(\frac{a}{b} = {2^k}\) , \(k \in \mathbb{Z}\) .
(i) Prove that \(R\) is an equivalence relation.
(ii) List the equivalence classes of \(R\) on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
▶️Answer/Explanation
(a)
To prove that $ 14641 $ is the fourth power of an integer in any base $ b > 6 $, let’s follow the steps below:
Express $ 14641 $ as a fourth power
If $ 14641 $ is the fourth power of an integer, there must exist an integer $ k $ such that:
$k^4 = 14641$
To check this, we can take the fourth root of $ 14641 $ in base 10:
$k = \sqrt[4]{14641} = 11$
Thus, $ 14641 = 11^4 $.
Represent $ 14641 $ in a general base $ b > 6 $
For any base $ b > 6 $, the number $ 14641 $ can be expressed as:
$14641_b = 1 \cdot b^4 + 4 \cdot b^3 + 6 \cdot b^2 + 4 \cdot b^1 + 1 \cdot b^0$
Expanding this, we have:
$14641_b = b^4 + 4b^3 + 6b^2 + 4b + 1$
Verify $ 14641_b = (b + 1)^4 $
The binomial expansion of $ (b + 1)^4 $ is:
$(b + 1)^4 = b^4 + 4b^3 + 6b^2 + 4b + 1$
Notice that this matches the expansion of $ 14641_b $. Therefore, in any base $ b > 6 $, $ 14641 $ is the fourth power of $ b + 1 $.
Confirm $ b > 6 $ is necessary
If $ b \leq 6 $, the digits $ 4 $ and $ 6 $ in $ 14641_b $ are invalid because digits in base $ b $ must be less than $ b $. Hence, $ b > 6 $ is required for $ 14641 $ to be a valid number in base $ b $.
Conclusion:
For any base $ b > 6 $, $ 14641_b = (b + 1)^4 $. Thus, $ 14641 $ is the fourth power of an integer in any base $ b > 6 $.
(b)
(i)
1. Reflexivity:
For $ aRa $, we check whether $ \frac{a}{a} = 2^k $ for some $ k \in \mathbb{Z} $.
Since $ \frac{a}{a} = 1 = 2^0 $, with $ k = 0 \in \mathbb{Z} $, we have $ aRa $. Thus, $ R $ is reflexive.
2. Symmetry:
If $ aRb $, then $ \frac{a}{b} = 2^k $ for some $ k \in \mathbb{Z} $.
Taking the reciprocal, $ \frac{b}{a} = 2^{-k} $, where $ -k \in \mathbb{Z} $.
This implies $ bRa $. Thus, $ R $ is symmetric.
3. Transitivity:
If $ aRb $ and $ bRc $, then $ \frac{a}{b} = 2^k $ and $ \frac{b}{c} = 2^m $, where $ k, m \in \mathbb{Z} $.
Multiplying these equations, $ \frac{a}{c} = \frac{a}{b} \cdot \frac{b}{c} = 2^k \cdot 2^m = 2^{k+m} $, where $ k+m \in \mathbb{Z} $.
Thus, $ aRc $. Hence, $ R $ is transitive.
Since $ R $ satisfies reflexivity, symmetry, and transitivity, $ R $ is an equivalence relation.
(ii): List the equivalence classes of $ R $ on the set $ \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} $.
The equivalence class of an element $ a \in \mathbb{Z} $ under $ R $ is defined as:
$[a] = \{b \in \mathbb{Z} : bRa\}$
This means $ \frac{b}{a} = 2^k $ for some $ k \in \mathbb{Z} $, or equivalently $ b = a \cdot 2^k $.
To find the equivalence classes, we compute all multiples of each $ a $ by powers of 2 (within the given set $ \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} $).
1. Equivalence class of 1:
Multiples of 1 by powers of 2: $ 1 \cdot 2^k = 1, 2, 4, 8 $.
Thus, $ [1] = \{1, 2, 4, 8\} $.
2. Equivalence class of 2:
Multiples of 2 by powers of 2: $ 2 \cdot 2^k = 2, 4, 8 $.
Thus, $ [2] = \{2, 4, 8\} $.
3. Equivalence class of 3:
Multiples of 3 by powers of 2: $ 3 \cdot 2^k = 3, 6 $.
Thus, $ [3] = \{3, 6\} $.
4. Equivalence class of 4:
Multiples of 4 by powers of 2: $ 4 \cdot 2^k = 4, 8 $.
Thus, $ [4] = \{4, 8\} $.
5. Equivalence class of 5:
Multiples of 5 by powers of 2: $ 5 \cdot 2^k = 5, 10 $.
Thus, $ [5] = \{5, 10\} $.
6. Equivalence class of 6:
Multiples of 6 by powers of 2: $ 6 \cdot 2^k = 6 $ (since $ 12 > 10 $).
Thus, $ [6] = \{6\} $.
7. Equivalence class of 7:
Multiples of 7 by powers of 2: $ 7 \cdot 2^k = 7 $ (since $ 14 > 10 $).
Thus, $ [7] = \{7\} $.
8. Equivalence class of 8:
Multiples of 8 by powers of 2: $ 8 \cdot 2^k = 8 $.
Thus, $ [8] = \{8\} $.
9. Equivalence class of 9:
Multiples of 9 by powers of 2: $ 9 \cdot 2^k = 9 $ (since $ 18 > 10 $).
Thus, $ [9] = \{9\} $.
10. Equivalence class of 10:
Multiples of 10 by powers of 2: $ 10 \cdot 2^k = 10 $.
Thus, $ [10] = \{10\} $.
Final List of Equivalence Classes:
$
[1] = \{1, 2, 4, 8\}, \quad [2] = \{2, 4, 8\}, \quad [3] = \{3, 6\}, \quad [4] = \{4, 8\}, \quad [5] = \{5, 10\},
$
$
[6] = \{6\}, \quad [7] = \{7\}, \quad [8] = \{8\}, \quad [9] = \{9\}, \quad [10] = \{10\}.
$
Question:
Find the positive square root of the base 7 number \({(551662)_7}\), giving your answer as a base 7 number.
▶️Answer/Explanation
Convert the base 7 number to base 10
To convert $(551662)_7$ to base 10, we expand it as:
$551662_7 = 5 \cdot 7^5 + 5 \cdot 7^4 + 1 \cdot 7^3 + 6 \cdot 7^2 + 6 \cdot 7^1 + 2 \cdot 7^0$
Calculate each term:
$5 \cdot 7^5 = 5 \cdot 16807 = 84035$
$5 \cdot 7^4 = 5 \cdot 2401 = 12005$
$1 \cdot 7^3 = 1 \cdot 343 = 343$
$6 \cdot 7^2 = 6 \cdot 49 = 294$
$6 \cdot 7^1 = 6 \cdot 7 = 42$
$2 \cdot 7^0 = 2 \cdot 1 = 2$
Add them together:
$84035 + 12005 + 343 + 294 + 42 + 2 = 96721$
So, $(551662)_7 = 96721_{10}$.
Find the square root in base 10
The positive square root of $96721$ is:
$\sqrt{96721} = 311$
Convert the result back to base 7
To convert $311_{10}$ to base 7, we divide $311$ repeatedly by $7$ and record the remainders:
$311 \div 7 = 44$ remainder $3$
$44 \div 7 = 6$ remainder $2$
$6 \div 7 = 0$ remainder $6$
Reading the remainders from bottom to top, $311_{10} = 623_7$.
The positive square root of $(551662)_7$ is $(623)_7$.
Question:
[Maximum mark: 4] [without GDC]
Prove the following statement:
For all integers \(n\) , if n2+ 2n+5 is odd then \(n\) is even.
▶️Answer/Explanation
Let $n = 2k + 1$ (odd). Then
\(
n^2 + 2n + 5 = (2k + 1)^2 + 2(2k + 1) + 5 = 4k^2 + 4k + 1 + 4k + 2 + 7 = 4k^2 + 8k + 10
\)
\(
= 2(2k^2 + 4k + 5) \text{ which is even.}
\)
Assume that the result is not true. That is $n$ is odd. Then according to above condition,
\(
n^2 + 2n + 5 \text{ is even. Contradiction.}
\)
Therefore, $n$ is even.