# Practice with deductions

• Aug 13th 2014, 11:56 AM
michele
Practice with deductions

(a) Below are two attempts at deductions. In each case, suppose that we know that the first two propositions are true, and wish to deduce that the third is true. In each of parts (i) and (ii), state whether the deduction is valid. If it is valid, give a deduction combining proof by contradiction and Modus Ponens. If it is not valid, explain why not.

1. (i) We know that:
If I have my car keys, then I can open my car door.I cannot open my car door.
We conclude that:
I do not have my car keys.

2. (ii) We know that:
If I have my car keys, then I can open my car door.I can open my car door.
We conclude that:
I have my car keys.

(b) (i)Show that for any positive integer n, n^4 −n^2 = n × (n−1)n(n+1)

(ii) Deduce that n^4 − n^2 + 1 always has remainder 1 upon division by 3, for any positive integer n.

• Aug 13th 2014, 12:08 PM
HallsofIvy
Re: Practice with deductions
1 and 2. "If A then B" does NOT imply "if B then A".
in both A is "if I have my keys". That says nothing about "if I can open the door".

bi. An obvious way to prove that is to do the multiplications indicated on the right.

bii. You want to prove that n^4- n^2+ 1= 3k+ 1 for some number k. Subtracting 1 form both sides, n^4- n^2= 3k. Now look again at bi.
• Aug 13th 2014, 02:09 PM
michele
Re: Practice with deductions
Thanks Hallsofivy but I'm still got some doubts in how to work my way out in part b(ii)
• Aug 13th 2014, 02:33 PM
Cartesius24
Re: Practice with deductions
Proove first n(n-1)(n+1) is always divisible by 3 .It will allow you to use bi
• Aug 13th 2014, 02:55 PM
HallsofIvy
Re: Practice with deductions
What Cartesius24 said! Think about the three numbers n-1, n, and n+ 1. ONE of those MUST be divisible by 3. Do you see why?
• Aug 14th 2014, 04:10 AM
michele
Re: Practice with deductions
Not really, I'm totally confused in how to see how this numbers can be divided by 3.
• Aug 14th 2014, 05:58 AM
HallsofIvy
Re: Practice with deductions
Seriously? Don't you see that if you have three consecutive numbers one of them must be a multiple of 3?

For a formal proof, use the fact that any integer is of the form 3k or 3k+ 1 or 3k+ 2 for some integer k.

If n itself is not divisible by 3 then it must be of the form 3k+ 1 or 3k+ 2.

If n= 3k+ 1 then n- 1= 3k is divisible by 3. If n= 3k+ 2, then n+1= 3k+ 3= 3(k+ 1) is divisible by 3.