smc-5116-10-Conjectures

Proof, EXT2 P1 2023 HSC 4 MC

Consider the following statement about real numbers.

"Whichever positive number \(r\) you pick, it is possible to find a number \(x\) greater than 1 such that

\(\dfrac{\ln x}{x^3}<r\). "

When this statement is written in the formal language of proof, which of the following is obtained?

\(\forall x>1 \quad \exists r>0 \quad \dfrac{\ln x}{x^3}<r\)
\(\exists x>1 \quad \forall r>0 \quad \dfrac{\ln x}{x^3}<r\)
\(\forall r>0 \quad \exists x>1 \quad \dfrac{\ln x}{x^3}<r\)
\(\exists r>0 \quad \forall x>1 \quad \dfrac{\ln x}{x^3}<r\)

Show Answers Only

\(C\)

Show Worked Solution

\(\text{Whichever positive number}\ r\ \text{you pick …}\ \forall r>0 \)

\(\text{It is possible to find a number}\ x\ \text{greater then 1 …}\ \exists x>1 \)

\(\text{Such that}\ \ \dfrac{\ln x}{x^3}<r\)

\(\Rightarrow C\)

Proof, EXT2 P1 2020 HSC 8 MC

Consider the statement:

'If `n` is even, then if `n` is a multiple of 3, then `n` is a multiple of 6'.

Which of the following is the negation of this statement?

`n` is odd and `n` is not a multiple of 3 or 6.
`n` is even and `n` is a multiple of 3 but not a multiple of 6.
If `n` is even, then `n` is not a multiple of 3 and `n` is not a multiple of 6.
If `n` is odd, then if `n` is not a multiple of 3 then `n` is not a multiple of 6.

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`B`

Show Worked Solution

`text{Proposition: If} \ \ X =>Y`

♦♦ Mean mark part 33%.

`X \ text{is a compound statement}`

`text{“If} \ n \ text{is even and a multiple of 3.”}`

`Y \ text{states “} n \ text{is a multiple of 6.”}`

`text{Negation if} \ X \ text{but} \ ¬ \ Y.`

`=> \ B`

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 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.

Show Answers Only

`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`