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mr_coffee
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Prove that for all integers n, n^2-n+3 is odd. stuck on algebra part :(
The question in the book is the following bolded statement.
Prove that for all integers n, n^2-n+3 is odd.
I rewrote it as this, is that right?
For all integers n, there exists an integer such that n^2-n+3 is odd.
I'm having problems proving this, the professor said the algebra isn't supose to be hard, but i can't get that. Doing the form of the proof isn't what I'm stuck on though.
Heres my start:
Proof:
Let n be an integer.
We must show n^2-n+3 is an odd integer.
Since n is an integer, then n(n-1)+3.
I don't think I'm doing this the right way, this chapter goes over the quotient remainder theorem. Which is Given any integer n and postie integer d, there exists unique integers q and r such that n = dp + r and 0 =< r < d.
From this you can see, that integers can be written in the form,
n = 4q or n = 4q + 1 or n = 4q + 2 or n = 4q + 3, for some integer q.
Where n = 4q is even, and 4q + 2 is even, and 4q + 1 is odd as well as 4q + 3.
This sounds like the right place, becuase the 4q+3 looks really similar to the n(n-1)+3. but I'm not sure what I'm doing becuase I think i can do the following:
Since n is an integer, and they say: We must show n^2-n+3 is an odd integer. can i do somthing like this?
n^2-n+3 = 4q+3?, where q is an integer.
becuase 4q + 3 is an odd integer.
Then i come out with:
n^2-n = 4q. #but now it seems it lost the odd integer form, and now looks even!
n(n-1)=4q;
q = n(n-1)/4
But now I'm lost again...any help or suggestions? Thanks!
The question in the book is the following bolded statement.
Prove that for all integers n, n^2-n+3 is odd.
I rewrote it as this, is that right?
For all integers n, there exists an integer such that n^2-n+3 is odd.
I'm having problems proving this, the professor said the algebra isn't supose to be hard, but i can't get that. Doing the form of the proof isn't what I'm stuck on though.
Heres my start:
Proof:
Let n be an integer.
We must show n^2-n+3 is an odd integer.
Since n is an integer, then n(n-1)+3.
I don't think I'm doing this the right way, this chapter goes over the quotient remainder theorem. Which is Given any integer n and postie integer d, there exists unique integers q and r such that n = dp + r and 0 =< r < d.
From this you can see, that integers can be written in the form,
n = 4q or n = 4q + 1 or n = 4q + 2 or n = 4q + 3, for some integer q.
Where n = 4q is even, and 4q + 2 is even, and 4q + 1 is odd as well as 4q + 3.
This sounds like the right place, becuase the 4q+3 looks really similar to the n(n-1)+3. but I'm not sure what I'm doing becuase I think i can do the following:
Since n is an integer, and they say: We must show n^2-n+3 is an odd integer. can i do somthing like this?
n^2-n+3 = 4q+3?, where q is an integer.
becuase 4q + 3 is an odd integer.
Then i come out with:
n^2-n = 4q. #but now it seems it lost the odd integer form, and now looks even!
n(n-1)=4q;
q = n(n-1)/4
But now I'm lost again...any help or suggestions? Thanks!
Last edited: