prove that $n >2$ and $n$ is a prime then $n$ is odd

Question

$n > 2$ and $n$ is a prime number then $n$ is odd.

Prove by contradiction
assume $n$ is even then there is some $k\in\Bbb N, n = 2k$ then $n >2, 2k > 2 , k > 1$

Is this a sufficient proof?

What is your contradiction? $k>1$? You have the right idea but remember what prime means. — Carlos Eugenio Thompson Pinzón, Nov 06 '13 at 19:21
@CarlosEugenioThompsonPinzón Im trying to say n is not a prime since it is divisible by 2 — user105696, Nov 06 '13 at 19:23
How do you define a prime number and why is 2 a prime number? — Carlos Eugenio Thompson Pinzón, Nov 06 '13 at 19:25
prime number has only two factors, itself and 1 2= (1)(2) I understand the concept but I dont know how to prove it — user105696, Nov 06 '13 at 19:27
And your other even prime candidate $n=2k$, which factors does it have? — Carlos Eugenio Thompson Pinzón, Nov 06 '13 at 19:29
@user105696 That is a terrible definition of prime. Note that $-1|2, -2|2, 1|2, \ \mathrm{and}\ 2|2$, so by your definition $nothing$ is prime. Moreover, what would you say is a prime number in the Gaussian integers?
EDIT: sorry, not trying to be mean, but there are definitely more elegant and general ways of defining a prime number — Tim Ratigan, Nov 06 '13 at 19:30
Relax this proof is for an introductory course... so this definition is sufficient and n > 2 so your point is invalid — user105696, Nov 06 '13 at 19:32
But you haven't arrived at (or explicitly stated) a contradiction! — amWhy, Nov 06 '13 at 19:33
@amWhy exactly that is what I am trying to figure out with your help.... — user105696, Nov 06 '13 at 19:34
@Tim.Ratigan: divisibility by only $1$ and itself seems to me a good enough definition as longer as we are sure we are on $\Bbb N$ (and itself$\ne1$). Negative divisors are excluded from $\Bbb N$. — Carlos Eugenio Thompson Pinzón, Nov 06 '13 at 19:45
@CarlosEugenioThompsonPinzón How would I actually show this by using what I have so far?? — user105696, Nov 06 '13 at 19:48

score 2 · Answer 1 · edited Nov 06 '13 at 20:09

2

To write a proof by contradiction, you must assume the premise and the negation of the conclusion:

Assume there exists an even prime $n > 2$. So $n = 2k$ where $k>1$ is an integer. (Here you can use your work showing why $k$ must be greater than 1).

Hence $n$ has at least two factors greater than or equal to $2$: $2,$ and $k \geq 2$. Hence...

edited Nov 06 '13 at 20:09

Matthew Conroy

11,234

answered Nov 06 '13 at 19:31

amWhy

209,954

Hence n is even but this contradicts the fact that n>2 and n is a prime number since n =2(k) and k>=2 – user105696 Nov 06 '13 at 19:40
The fact that $n$ has two or more factors greater than or equal to $2$ contradicts the premise that $n$ is prime. – amWhy Nov 06 '13 at 19:49

score 0 · Answer 2 · answered Nov 06 '13 at 19:32

0

In your proof,because $n>2$ then $k\geq 2$ and thus $n$ has at least two factors grater than 2

answered Nov 06 '13 at 19:32

Haha

5,648

prove that $n >2$ and $n$ is a prime then $n$ is odd

2 Answers2

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