Question

How can i prove an number is odd

Answer 1

try to factor it

Answer 2

proof by induction.

Answer 3

i dont really like induction, it takes to long

Answer 4

"an number"? Don't you mean a number?

Answer 5

yes that is what i mean

Answer 6

How can i prove a number is odd

Answer 7

"i"? Don't you mean I?

Answer 8

You got me kid, I have no idea.

Answer 9

Do you watch the show Breaking Bad ever?

Answer 10

yeah that show is awesome

Answer 11

Yeah it is!

Answer 12

Divide it by 2 and if you are getting a remainder as 1 then it is odd..

Ha ha ha...

Answer 13

I can't think of any other way other than induction.

Answer 14

For any integer n, the number \(n^2+n+1\) is odd.\[\text{---------}\]
Proof:
\[\text{If \(n\) is an integer, then it is either zero, odd, or even.}\]

Case I; \(n\) is zero, \(n=0\)\[\left(0\right)^2+(0)+1=1\]\[\text{One is an odd number}\]\[\text{Hence, when \(n\) is zero, \(n^2+n+1\) is odd.}\]


Case II; \(n\) is even, \(n=2m,\qquad m\in \mathbb Z\)\[\left(2m\right)^2+(2m)+1=4m^2+2m+1\]\[\qquad\qquad\qquad\qquad\qquad=2(2m^2+m)+1\]\[\text{Any number that has a factor of two is even}\]\[\text{Any number that is a sum of an even and odd number is odd.}\]\[\text{Hence when \(n\) is even, \(n^2+n+1\) is odd.}\]


Case III; \(n\) is odd, \(n=2m+1,\qquad m\in \mathbb Z\)
\[\left(2m+1\right)^2+(2m+1)+1=\left(4m^2+4m+1\right)+(2m+1)+1\]\[\qquad\qquad\qquad=4m^2+6m+3\]\[\qquad\qquad\qquad\qquad=2(2m^2+3m)+3\]\[\text{Three is an odd number}\]\[\text{Hence when \(n\) is even, \(n^2+n+1\) is odd.}\]


\[\text{All cases of integer (zero, even or odd), imply that \(n^2+n+1\) is odd.}\]
This proves For any integer \(n\), the number \(n^2+n+1\) is odd.\[\square\]

Answer 15

there must be a simpler way

Answer 16

Isn't any number that's 2n+1 odd?

Answer 17

yes

Answer 18

Baracka flocka flame one hood retrice nigguh!

Answer 19

Okay I need to prove it.

Answer 20

I'm the head of the muhflutterin state nigguh, I brought you change nigguh, what the fuhck you thinkin nigguh

Answer 21

aadsadsf, your not helping

Answer 22

I've got a mayne feather, i've gotta mistress, a couple daughters, I'm so hood rich

Answer 23

I'm sorry, I thought we were done here

Answer 24

Proof by Induction:

Let 2n+1 be odd

1) Test for n=1

2(1)+1 = 5

This is odd

2) Assume this is true for n=k

2k+1 = odd (Induction hypothesis)

3) Prove statement for n=k+1

2(k+1)+1 = 2k+1

2k+3 = 2k+1

Substituting any number into k yields an odd number on both sides of the equality.

therefore, by induction, this statement holds true for all odd numbers.

Answer 25

Or:

3) Prove statement for n=k+1

2(k+1)+1 = 2k+1

2k+3 = 2k+1

In step two 2k+1= odd

2k+3=odd

Answer 26

We can repeat this infinitely many times for this expression and therefore any number in this format is always odd.

Answer 27

Or we could use proof by condradiction.

Answer 28

is there anything wrong with my proof

Answer 29

No. It's just long.

Answer 30

yeah it is way to long

Answer 31

We could also use proof by condradiction.

Answer 32

Assume 2n+1 = 2q is even for a interger 2q

2n-2q=-1

n-q= 1/2 which is supposed to be even but it's odd.

So therefore 2n+1 must be odd.

Answer 33

I can't think of anything else.