Question

prove that the product of any two odd numbers is odd

Answer 1

product not sum

Answer 2

product is multiplication**

Answer 3

7x7=49

Answer 4

oh Product, sorry.
1 x 1 = 1.
3 x 3 = 9.
All odd number will be odd with multiplication in place.

Answer 5

*Numbers

Answer 6

11x11=121

Answer 7

Say if you multiply 3 x 3
You get 9, that is an odd number.
If you multiply 9 x 9,
you get 81, that is an odd number.

Answer 8

thats is showing, not proving.

Answer 9

Any odd number can be expressed in the form of 2n+1
where n is a whole number
Let other odd number be 2n+3
Let's take their product
\[(2n+1)(2n+3)\]
we get
\[4n^2+8n+3\]
here the first two numbers are even and last is odd. So product of any two numbers is odd

Answer 10

every odd numbers,multiplied by it self the result is odd

Answer 11

(2n+1) and (2k+1) are odd numbers
(2n+1)(2k+1) = 4nk +2n +2k +1
= 2(2nk +n+k) +1
and this is also odd

Answer 12

hoblos' proof is correct, ash's one needs some further work.

Answer 13

@hoblos is k and even number

Answer 14

*an

Answer 15

n&k are any integers

Answer 16

Actually, any odd number can be expressed in the form of 2n+1 where \( n\in \mathbb{Z} \)

This can be proved by induction.

Answer 17

\[4(n^2+2n)+3\]
sum of even and an odd number is odd

Answer 18

so, from \[ (2n+1)(2k+1) = 4nk +2n +2k +1
= 2(2nk +n+k) +1 \]

Set, \( (2nk +n+k) = p \in \mathbb{z} \)

Answer 19

Ash, for a complete proof you have to show that sum of even and an odd number is odd ...

Answer 20

But if you just take out the factor 2, you are done.

Answer 21

Okay:)
\[2(2n^2+4n)+2+1\]
\[2(2n^2+2n+1)+1\]
this is an odd number

Answer 22

it is often difficult to prove the obvious :)

Answer 23

usually, i firm slap to the back of the head is proof enough for the obvious ;)