smc-5116-20-Contrapositive

Proof, SPEC2 2023 VCAA 1 MC

Consider the following statement.

'If my football team plays badly, then they are not training enough.'

Which one of the following statements is the contrapositive of the statement above?

If they are not training enough, then my football team plays badly.
If my football team plays badly, then they need more training.
If they are training enough, then my football team does not play badly.
If my football team doesn't play badly, then they are training enough.
If they are training enough, then my football team will most likely win.

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\( C\)

Show Worked Solution

\(\text{Statement is conditional:}\ X \Rightarrow Y \)

\(\text{Logically equivalent contrapositive statement:}\ \neg Y \Rightarrow \neg X\)

\(\Rightarrow C\)

Proof, EXT2 P1 2022 HSC 13a

Prove that for all integers `n` with `n >= 3`, if `2^(n)-1` is prime, then `n` cannot be even. (3 marks)

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`text{Proof (See Worked Solutions)}`

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`text{Contrapositive statement:}`

`text{If}\ n\ text{is even}, 2^n-1\ text{is NOT prime.}`

`text{Let}\ \ n=2k,\ \ (kinZZ and k>=2)`

`2^n-1`	`=2^(2k)-1`
	`=(2^k)^2-1`
	`=(2^k-1)(2^k+1)`

`text{S}text{ince}\ \ k>=2\ \ =>\ \ 2^k-1>=3 and 2^k+1>=5`

`:.2^n-1\ \ text{is not prime if}\ n\ text{is even, as it has two non-trivial integer factors.}`

`:.\ text{By contrapositive statement, if}\ 2^(n)-1\ text{is prime}, n\ text{cannot be even.}`

Proof, EXT2 P1 2022 HSC 3 MC

Let `A, B, P` be three points in three-dimensional space with `A \ne B`.

Consider the following statement.

If `P` is on the line `A B`, then there exists a real number `\lambda` such that `vec{A P}=\lambda \vec{A B}`.

Which of the following is the contrapositive of this statement?

If for all real numbers `\lambda, \vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
If for all real numbers `\lambda, \vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.
If there exists a real number `\lambda` such that `\vec{A P}=\lambda \vec{A B}`, then `P` is on the line `A B`.
If there exists a real number `\lambda` such that `\vec{A P} \ne \lambda \vec{A B}`, then `P` is not on the line `A B`.

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`B`

Show Worked Solution

`text{Statement:}`

`text{If}\ P\ text{is on}\ AB\ \ =>\ \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}`

`text{Contrapositive statement:}`

`text{If}\ not \ EE lambda\ \ text{such that}\ \ vec{A P}=\lambda \vec{A B}\ \ =>\ \ P\ text{is not on}\ AB`

`text{In other words …}`

`text{If for all real numbers}\ lambda, \vec{A P} \ne \lambda \vec{A B}, text{then}\ P\ text{is not on the line}\ A B.`

`=>B`

Proof, EXT2 P1 2021 HSC 9 MC

Four cards have either RED or BLACK on one side and either WIN or TRY AGAIN on the other side.

Sam places the four cards on the table as shown below.

A statement is made: ‘If a card is RED, then it has WIN written on the other side’.

Sam wants to check if the statement is true by turning over the minimum number of cards.

Which cards should Sam turn over?

1 and 4
3 and 4
1. 2 and 4
1, 3 and 4

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`B`

Show Worked Solution

`text{L} text{ogically equivalent statements are:}`

`text{Red}`	`=> \ text{Win} \ …\ (1)`
`¬ \ text{Win}`	`=> \ ¬ \ text{Red} \ …\ (2)`

`text{To confirm statement is true}`

♦ Mean mark 48%.

`text{Card 1 – no need to turn}`

`text{Card 2 – no need to turn}`

`text{Card 3 – turn to confirm (2)}`

`text{Card 4 – turn to confirm (1)}`

`=> B`

Proof, EXT2 P1 2021 HSC 4 MC

Consider the statement:

‘For all integers `n`, if `n` is a multiple of 6, then `n` is a multiple of 2’.

Which of the following is the contrapositive of the statement?

There exists an integer `n` such that `n` is a multiple of 6 and not a multiple of 2.
There exists an integer `n` such that `n` is a multiple of 2 and not a multiple of 6.
For all integers `n`, if `n` is not a multiple of 2, then `n` is not a multiple of 6.
For all integers `n`, if `n` is not a multiple of 6, then `n` is not a multiple of 2.

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`C`

Show Worked Solution

`text{L} text{ogically equivalent statements:}`

`text{If} \ n \ text{is multiple of 6} \ => \ n \ text{is a multiple of 2}`

`¬\ n\ text{is a multiple of 2}\ => \ ¬\ n \ text{is a multiple of 6}`

`text{i.e. if} \ n \ text{is not multiple of 2} \ => \ n \ text{is not a multiple of 6}`

`=>\ C`

Proof, EXT2 P1 2020 HSC 15a

In the set of integers, let `P` be the proposition:

'If `k + 1` is divisible by 3, then `k^3 + 1` divisible by 3.'

Prove that the proposition `P` is true. (2 marks)

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Write down the contrapositive of the proposition `P`. (1 mark)

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Write down the converse of the proposition `P` and state, with reasons, whether this converse is true or false. (3 marks)

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`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

i. `text{Let} \ \ k + 1 = 3N, \ N∈ Z`

`=> k = 3N – 1`

`k^3 + 1`	`= (3N -1)^3 + 1`
	`= (3N)^3 + 3(3N)^2 (-1) + 3(3N)(-1)^2 + (-1)^3 + 1`
	`= 27N^3 – 27N^2 + 9N – 1 + 1`
	`= 3 (9N^3 – 9N^2 + 3N)`
	`= 3Q \ , \ Q ∈ Z`

`therefore \ text{If} \ \ k+ 1 \ \ text{is divisible by 3}, text{then} \ \ k^3 + 1 \ \ text{is divisible by 3.}`

ii. `text{Contrapositive}`

`text{If} \ \ k^3 + 1 \ \ text{is not divisible by 3, then}\ \ k + 1\ \ text{is not divisible by 3.}`

♦♦ Mean mark part (iii) 36%.

iii. `text{Converse:}`

`text{If} \ \ k^3 + 1\ \ text{is divisible by 3, then}\ \ k + 1\ \ text{is divisible by 3.}`

`text(Contrapositive of converse:)`

`text{If}\ \ k + 1\ \ text{is not divisible by 3, then}\ \ k^3 + 1\ \ text{is not divisible by 3.}`

`text(i.e.)\ \ k + 1 \ \ text{is not divisible by 3 when}\ \ k + 1 = 3Q + 1\ \ text{or}\ \ k + 1 = 3Q + 2, text{where}\ Q ∈ Z`

`text{If} \ \ k + 1`	`= 3Q + 1\ \ => \ k=3Q`
`k^3 + 1`	`= (3Q)^3 + 1`
	`= 27Q^3 + 1`
	`= 3(9Q^3) + 1`
	`= 3M + 1 \ \ (text{not divisible by 3,}\ M ∈ Z)`

`text{If} \ \ k + 1`	`= 3Q + 2\ \ => \ k=3Q+1`
`k^3 + 1`	`= (3Q + 1)^3 + 1`
	`= (3Q)^3 + 3(3Q)^2 + 3(3Q) + 1 + 1`
	`= 27Q^3 + 27Q^2 + 9Q + 2`
	`= 3(9Q^3 + 9Q^2 + 3Q) + 2`
	`= 3M + 2 \ (text{not divisible by 3,}\ M ∈ Z) `

`therefore \ text{By contrapositive, if}\ \ k^3 + 1\ \ text {is divisible by 3, k + 1 is divisible by 3.}`

Proof, EXT2 P1 2020 HSC 7 MC

Consider the proposition:

'If `2^n - 1` is not prime, then `n` is not prime'.

Given that each of the following statements is true, which statement disproves the proposition?

`2^5 - 1` is prime
`2^6 - 1` is divisible by 9
`2^7 - 1` is prime
`2^11 - 1` is divisible by 23

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`D`

Show Worked Solution

`text(Strategy 1 – Contradiction)`

`text(Consider option)\ D,`

`text(S)text(ince)\ \ 2^11 -1\ \ text(is divisible by 23, it is NOT prime.)`

`text(The proposition states that 11 is not prime which is false.)`

`:. 2^11 – 1\ \ text(is divisible by 23, disproves the proposition.)`

`text(Strategy 2 – Contrapositive)`

`text{The proposition is conditional}`

`X => Y`

`text{L}text{ogically equivalent contrapositive statement}`

`not \ Y => not \ X`

`text{i.e. If} \ n \ text{is prime} \ => \ 2^n – 1 \ text{is prime.}`

`text{Consider D:}`

`n = 11 \ text{(prime)}`

`2^11 – 1 \ text{is divisible by 23 (not prime)}`

`therefore \ text{Contrapositive statement is false and disproves the proposition.}`

`=> \ D`

Proof, EXT2 P1 SM-Bank 13

If `(n - 3)^2` is an even integer, prove by contrapositive that `n` is odd. (2 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Statement)`

`text(If) \ \ (n – 3)^2 \ \ text(is even) => n \ text(is odd)`

`text(Contrapositive)`

`text(If) \ \ n \ not \ text(odd) => (n – 3)^2 \ not \ text(even)`

`text{(i.e.)}\ \ \ n \ text(even) => (n – 3)^2 \ text(is odd)`

`text(If)\ \ n \ \ text(even), \ ∃ \ k, \ k ∈ Ζ \ \ text(where)\ \ n = 2k`

`(n – 3)^2`	`= (2k – 3)^2`
	`= 4k^2 – 12k + 9`
	`= 4(k^2 – 12k + 2) + 1`

`=> (n – 3)^2 \ \ text(is odd)`

`:. \ text(If) \ \ n\ \ text(is even), (n – 3)^2 \ \ text(is odd)`

`:. \ text(By contrapositive, statement is true.)`

Proof, EXT2 P1 SM-Bank 12

If `ab` is divisible by 3, prove by contrapositive that `a` or `b` is divisible by 3. (3 marks)

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`text{Proof (See Worked Solutions)}`

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`text(Statement)`

`ab \ text(is divisible by 3)\ => a \ text(or) \ b \ text(is divisible by) \ 3`

`text(Contrapositive)`

`a \ text(or) \ b not \ text(divisible by 3)\ => ab not \ text(divisible by 3)`

`text(Let) \ \ a`	`= 3x + p, \ text(where) \ \ x ∈ Ζ \ \ text(and) \ \ p= 1 \ text(or) \ \ 2`
`b`	`= 3y + q, \ text(where) \ \ y ∈ Ζ \ \ text(and) \ \ q= 1 \ text(or) \ \ 2`

`ab`	`= (3x + p)(3y + q)`
	`= 9xy + 3qx + 3py + pq`
	`= 3(3xy + qx + py) + pq`

`text(Possible values of) \ \ pq = 1, 2, 4`

`=> \ pq \ text(is not divisible by 3)`

`:. \ ab \ text(is not divisible by 3)`

`:. \ text(By contrapositive, statement is true.)`

Proof, EXT2 P1 SM-Bank 11

If `a^2-4a + 3` is even, `a ∈ Ζ`,

prove by contrapositive that `a` is odd. (3 marks)

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`text{Proof (See Worked Solutions)}`

Show Worked Solution

`text(Proof by contrapositive)`

`a not \ text(odd)`	`\ =>\ \ a^2-4x + 3 not \ text(even)`
`text{(i.e) If} \ \ a \ text(is even) `	`\ => \ a^2-4x + 3 \ \ text(is odd)`

`text(If) \ \ a \ \ text(is even) , ∃ \ k , k ∈ Ζ , text(such that) \ \ a = 2k`

`text(Substitute) \ \ 2k \ \ text(into) \ \ a^2-4x + 3`

`(2k)^2-4(2k) + 3`	`= 4k^2-8k + 3`
	`= 2(2k^2-4k + 1) + 1`

`:. \ text(By contrapositive, if) \ \ a^2-4x + 3 \ \ text(is even) \ => \ a \ \ text(is odd.)`

Proof, EXT2 P1 SM-Bank 5 MC

Four cards are placed on a table with a letter on one face and a shape on the other.

You are given the rule: "if N is on a card then a circle is on the other side."

Which cards need to be turned over to check if this rule holds?

N and G
G and triangle
circle and N
N and triangle

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`=> \ D`

Show Worked Solution

`text(Solution 1)`

`text(L)text(ogically equivalent statements are:)`

`N`	`=> \ text(circle)`
`not\ text(circle)`	`=> not N`

`text(To confirm rule is not broken,)`

`N \ text(must be turned)`

`text{Triangle (not circle) must be turned – only other shape not a circle.}`

`=> D`

`text(Solution 2)`

`text(Consider the flip side of each card.)`

`text(If circle has) \ N \ text(on the other side (or not)) – text(tells us nothing.)`

`text(If) \ G \ text(has a circle on the other side (or not)) – text(tells us nothing.)`

`text(If) \ N \ text(doesn’t have a circle on other side) – text(rule broken.)`

`text(If triangle has an) \ N \ text(on other side) – text(rule broken.)`

`:. \ text(Need to turn) \ N \ text(and triangle)`

Proof, EXT2 P1 SM-Bank 1 MC

Consider the following statement.

"If you have no treasure, I have no kingdom."

Which of the following is logically equivalent to this statement?

If I have no kingdom then you have no treasure.
If you have treasure then I have a kingdom.
If you have no kingdom then I have no treasure.
If I have a kingdom then you have treasure.

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`D`

Show Worked Solution

`text(The statement is conditional.)`

`text(If)\ X\ text(then)\ Y\ \ text(or)\ \ X=>Y`

`text(The contrapositive statement is logically equivalent.)`

`text{(i.e.)}\ \ ¬Y => ¬X`

`text(If I have a kingdom then you have treasure.)`

`=> D`