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In the set of integers, let P be the proposition: 'If k + 1 is divisible by 3, then k^2 + 1 is divisible by 3' - HSC - SSCE Mathematics Extension 2 - Question 15 - 2020 - Paper 1

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In the set of integers, let P be the proposition: 'If k + 1 is divisible by 3, then k^2 + 1 is divisible by 3'. (i) Prove that the proposition P is true. (ii) Wri... show full transcript

Worked Solution & Example Answer:In the set of integers, let P be the proposition: 'If k + 1 is divisible by 3, then k^2 + 1 is divisible by 3' - HSC - SSCE Mathematics Extension 2 - Question 15 - 2020 - Paper 1

Step 1

(i) Prove that the proposition P is true.

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Answer

To prove that the proposition P is true, we will assume that k+1k + 1 is divisible by 3. This implies that there exists an integer jj such that:

k+1=3jk + 1 = 3j

From here, we can derive:

k=3j1k = 3j - 1

Now, we will calculate k2+1k^2 + 1:

k2+1=(3j1)2+1k^2 + 1 = (3j - 1)^2 + 1 =9j26j+1+1= 9j^2 - 6j + 1 + 1 =9j26j+2= 9j^2 - 6j + 2

Notice that we can factor this as follows:

=3(3j22j)+2= 3(3j^2 - 2j) + 2

Now, since the only case when k+1k + 1 is divisible by 3 is when jj is an integer, we find that k2+1k^2 + 1 is also divisible by 3, confirming that proposition P is true.

Step 2

(ii) Write down the contrapositive of the proposition P.

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Answer

The contrapositive of the proposition P is:

'If k2+1k^2 + 1 is not divisible by 3, then k+1k + 1 is not divisible by 3.'

Step 3

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

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Answer

The converse of the proposition P is:

'If k2+1k^2 + 1 is divisible by 3, then k+1k + 1 is divisible by 3.'

To determine if this converse is true, consider the case when k2+1k^2 + 1 is divisible by 3. This does not necessarily mean that k+1k + 1 must also be divisible by 3, as shown through a counter-example:

  • Let k=1k = 1:
    • Then k2+1=12+1=2k^2 + 1 = 1^2 + 1 = 2, which is not divisible by 3.
    • However, if we consider k=2k = 2, then k2+1=4+1=5k^2 + 1 = 4 + 1 = 5, and k+1=3k + 1 = 3, which is divisible by 3.
  • Upon checking multiple integers, we find scenarios of contradicting cases. Thus, the converse of proposition P is false.

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