Some help on mathematical induction

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  • Thread starter spanker1
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  • #1
spanker1
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help guys i am really stumped on this question.
prove that

if "n" is an integer , then n^2-n+2 is even
 

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  • #2
disregardthat
Science Advisor
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Suppose n is odd. What can you say about each member of n^2-n+2 ?
Suppose n is even. What can you say about each member of n^2-n+2?
 
  • #3
Tedjn
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That is one way of thinking about the problem. If you want to do this via induction, however, what is the first step you need to take? Also, are you trying to prove that fact for n is any integer or just a positive integer?
 
  • #4
lizzie
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to prove by principle of mathemetical induction
step 1:
put n=1
1^2-1+2=1^3=1
which is false
 
  • #5
gunch
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to prove by principle of mathemetical induction
step 1:
put n=1
1^2-1+2=1^3=1
which is false

I think what is meant by n^2-n+2 is [tex]n^2 - n +2[/tex], not [tex]n^{2-n+2}[/tex] as you seem to use.
 
  • #6
CRGreathouse
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help guys i am really stumped on this question.
prove that

if "n" is an integer , then n^2-n+2 is even

A proof by induction would take the following form:

1. For n = 1, [itex]n^2-n+2[/itex] is even.
(subproof)
2. Suppose [itex]n^2-n+2[/itex] is even. Then [itex](n+1)^2-(n+1)+2[/itex] is even.
(subproof)
3. By induction (base case 1, inductive step 2), [itex]n^2-n+2[/itex] is even for all natural numbers n.

If you can fill in the subproofs you're done.

(If you really want all integers, of course, you'll have to prove them too -- maybe by induction on [itex](-n)^2-(-n)+2.[/itex])
 
  • #7
aquariss
2
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:)

help guys i am really stumped on this question.
prove that

if "n" is an integer , then n^2-n+2 is even

hmm...My suggestion :

N^2 - N + 2
= N(N-1) + 2

For any integer N, if N is even , (N-1) is odd therefore N(N-1) is divisible by 2 and is an even #, so , N(N-1) + 2 is even

For any integer N, if N is odd , (N-1) is even therefore N(N-1) is divisible by 2 and is an even #, so , N(N-1) + 2 is even

Is this logical?
 
  • #8
qspeechc
844
15
For n=1, n^2 - n +2 = 2 is even.

Now, suppose that for some integer k, that [tex]k^2 -k + 2[/tex] is even. Then for the next integer k+1

[tex](k+1)^2 -(k+1) +2 = k^2 + 2k +1 - k -1 +2 = k^2 + k +2[/tex]

Now

[tex] k^2 + k + 2 = (k^2 -k + 2) + 2k[/tex]

On the RHS, the stuff in bracket is what we started with, we assumed it was even. 2k is even for all integers k. An even number plus an even number is an even number. Therefore for the next integer, it will also be even.

Therefore for all integers n, [tex]n^2 -n +2[/tex] is even
 
  • #9
sennyk
73
0
Quick question about induction. Could you prove it true for n = 1. Then prove it for n = n + 1. Then prove it for n = n - 1. Would that be enough to prove it for all integers?
 

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