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]
▶️Answer/Explanation
Ans: \(a = 2m\) and \(b = 2n\) Then \(a + b = 2m + 2n = 2(m + n)\) even
(b) Prove that the sum of two odd numbers is an even number. [1]
▶️Answer/Explanation
Ans: \(a = 2m +1\) and \(b = 2n +1\) Then \(a + b = 2m +1 + 2n +1 = 2(m + n +1)\) even
(c) Prove that the sum of two multiples of 3 is a multiple of 3. [1]
▶️Answer/Explanation
Ans: \(a = 3m\) and \(b = 3n\) Then \(a + b = 3m + 3n = 3(m + n)\) multiple of 3
(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]
▶️Answer/Explanation
Ans: \(a = 4\) and \(b = 7\) gives \(a + b =11\) which is not necessarily a multiple of 3.
(e) Prove that the square of any integer cannot have the form \(3n+2\). [3]
▶️Answer/Explanation
Ans: If \(a=3m\Rightarrow a^{2}=9m^{2}=3(3m^{2})\) it has the form 3n
If \(a=3m+1\Rightarrow a^{2}=9m^{2}+6m+1=3(3m^{2}+3m)+1\) it has the form 3n+1
If \(a=3m+2\Rightarrow a^{2}=9m^{2}+12m+4m=3(3m^{2}+4m+1)+1\) it has the form 3n+1
So it never has 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]
▶️Answer/Explanation
Ans: Suppose that \(\sqrt{3}=\frac{a}{b}\) (rational) where \(a,b\) have no common factors.
Then \(a=\sqrt{3b}\Leftrightarrow a^{2}=3b^{2}\). Thus 3 divides a2 and thus 3 divides \(a\) . Let \(a = 3c\)
Then \(\left ( 3c \right )^{2}=3b^{2}\Leftrightarrow 9c^{2}=3b^{2}\Leftrightarrow 3c^{2}=b^{2}\).
Thus 3 divides b2 and thus 3 divides \(b\) .
Contradiction since \(a,b\) have no common factors
Therefore, \(\sqrt{3}\) is irrational.
(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
Ans: Suppose that \(\sqrt[3]{2}=\frac{a}{b}\) (rational) where \(a,b\) have no common factors.
Then \(a=\sqrt[3]{2b}\Leftrightarrow a^{3}=2b^{3}\). Thus a3 is even and thus \(a\) is even. Let \(a = 2c\)
Then \(\left ( 2c \right )^{3}=2b^{3}\Leftrightarrow 8c^{3}=2b^{3}\Leftrightarrow 4c^{3}=b^{3}\). Thus b3 is even and thus \(b\) is even.
Contradiction since \(a,b\) have no common factors
Therefore, \(\sqrt[3]{2}\) is irrational.
Question:
a) Prove that the number \(14 641\) is the fourth power of an integer in any base greater than \(6\).
▶️Answer/Explanation
Ans: \(14641\) (base \(a > 6\) ) \( = {a^4} + 4{a^3} + 6{a^2} + 4a + 1\) ,
\( = {(a + 1)^4}\) this is the fourth power of an integer
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
Ans: (i) \(aRa\) since \(\frac{a}{a} = 1 = {2^0}\) , hence \(R\) is reflexive A1
\(aRb \Rightarrow \frac{a}{b} = {2^k} \Rightarrow \frac{b}{a} = {2^{ – k}} \Rightarrow bRa\)
so R is symmetric A1
\(aRb\) and \(bRc \Rightarrow \frac{a}{b} = {2^m}\), \(m \in \mathbb{Z}\) and \(bRc \Rightarrow \frac{b}{c} = {2^n}\) , \(n \in \mathbb{Z}\) M1
\( \Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{a}{c} = {2^{m + n}}\) , \(m + n \in \mathbb{Z}\) A1
\( \Rightarrow aRc\) so transitive R1
hence \(R\) is an equivalence relation AG
(ii) equivalence classes are {1, 2, 4, 8} , {3, 6} , {5, 10} , {7} , {9} A3
Note: Award A2 if one class missing, A1 if two classes missing, A0 if three or more classes missing.
Question:
Find the positive square root of the base 7 number \({(551662)_7}\), giving your answer as a base 7 number.
▶️Answer/Explanation
Ans: converting to base 10
\({(551662)_7} = 2 + 6 \times 7 + 6 \times {7^2} + 1 \times {7^3} + 5 \times {7^4} + 5 \times {7^5}\) (M1)
\( = 96721\) (A1)
\(\sqrt {96721} = 311\) A1
converting back to base 7
\(7)\underline {311} \) (M1)
\()\underline {44} (3\)
\()\underline 6 (2\) (A1)
it follows that \(\sqrt {{{(551662)}_7}} = {(623)_7}\) A1
Note: Accept \(623\).
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
Ans: Suppose that the result is not true. That is \(n = 2k + 1\) is odd. Then
\(n^{2}+2n+5=(2k+1)^{2}+2(2k+1)+5=4k^{2}+4k+1+4k+2+7=4k^{2}+8k+10\)
which is even. Contradiction.