IBDP Maths AA:Topic : SL 1.1: Operations with numbers: IB style Questions HL Paper 1

Question 📵 Hard❓

Question:

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.

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