Home / IBDP Maths AA: Topic 1.6 :Simple deductive proof, numerical and algebraic HL Paper 2

IBDP Maths AA: Topic 1.6 :Simple deductive proof, numerical and algebraic HL Paper 2

Question

Consider two consecutive positive integers, n and n + 1 .
Show that the difference of their squares is equal to the sum of the two integers.

▶️Answer/Explanation

Ans:

diff. of squares =\(\left(n+1\right)^2-n^2\)

=\(\left(n+1+n\right)\left(n+1-n\right)\)

= \(\left(n+1+n\right)\left(n+1-n\right)\)

=2n+1

=n+1+n

=sum of 2 integers

Question

[Maximum mark: 4] [without GDC]
Prove the following statements
(a) There exist no integers \(a\) and \(b\) for which \(6a + 2b = 35 \)                             [2]
(b) There exist no integers \(a\) and \(b\) for which \(6a + 9b = 35  \)                            [2]

▶️Answer/Explanation

Ans.

(a) Suppose they exist integers \(a\) and \(b\) for which \(6a + 2b = 35\).
But LHS is a multiple of 2 while RHS is not. Contradiction.
(b) Suppose they exist integers \(a\) and \(b\) for which \(6a + 9b = 35\) .
But LHS is a multiple of 3 while RHS is not. Contradiction.

Question

[Maximum mark: 8] [without GDC]
Consider the equation of integers
\(a + b + c = d\)
(a) Use a deductive proof to prove the statements
(i) “if \(a,b, c \) are all even then \(d\) is also even;
(ii) “if \(a,b, c\) are all odd then \(d\) is also odd”                                                                         [3]
(b) Use a counterexample to disprove the statement
“if \(d\) is odd then \(a,b, c\) are all odd”.                                                                                       [1]
(c) State whether the following statement is true or false and prove your claim.
“if \(d\) is even then \(a,b, c\) are all even”.                                                                                   [2]
(d) Use contradiction to prove the statement
“if \(d\) is even then at least one of \(a,b, c\) is even”.                                                                 [2]

▶️Answer/Explanation

Ans.

(a)     (i) Let \(a = 2n,b = 2m, c = 2k\) , where \(n,m, k ∈ Z\) .
Then \(d = a + b + c = 2n + 2m + 2k = 2(m + n + k)\) even.
(ii) Let \(a = 2n + 1,b = 2m +1, c = 2k +1\) , where \(n,m, k ∈ Z\) .
Then \(d = a + b + c = 2n +1+ 2m + 1+ 2k +1= 2 (m + n + k +1) +1\) odd.

(b) Let \(a = 2,b = 4, c = 5\) . Then \(d =11\) odd but \(a,b, c\) are not all odd

(c) It is false. We use again a counter example:
Let \(a = 2,b = 3, c = 5\) . Then \(d =10\) even but \(a,b, c\) are not all even.
(d) Suppose that the conclusion is false, that is \(a,b, c\) are all odd. Then by (a)(ii), \(d\) is odd
Contradiction. [in fact, this statement is the contrapositive of the statement (a) (ii)]

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