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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.'

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

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

  3. 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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  1. `text{See Worked Solutions}`
  2. `text{See Worked Solutions}`
  3. `text{See Worked Solutions}`
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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 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.)`