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prove that

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

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- #1

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prove that

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

- #2

disregardthat

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Suppose n is even. What can you say about each member of n^2-n+2?

- #3

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- #4

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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

step 1:

put n=1

1^2-1+2=1^3=1

which is false

- #5

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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.to prove by principle of mathemetical induction

step 1:

put n=1

1^2-1+2=1^3=1

which is false

- #6

CRGreathouse

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A proof by induction would take the following form:

prove that

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

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

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hmm...My suggestion :

prove that

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

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

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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

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