9. Prove the statement Product of two odd integers is odd
by. contradiction Positive
method
Answers
Let us therefore consider two odd integers, being ( 2n + 1 ) and ( 2n + 3 ).
In this case, the two numbers are both odd and are consecutive.
That the two numbers are consecutive is irrelevant to this question.
That they are both odd is the important fact.
Their product is ( 2n + 1 ) * ( 2n + 3 ) or
4n^2 + 6n + 2n + 3 or
4n^2 + 8n + 3 or
4n( n + 2) + 3
By definition 4n is even because 4n = 2 * 2n and both “ 2 “ and therefore “ 2n “ are even..
However “ 3 “ is odd, so [ 4n(n + 2) + 3 ] is odd.
Therefore the product of these two odd integers is odd.
Let us though take the above analysis one step further by identifying the two odd numbers as:
( 2n + 1 ) and (2n + K ) where K is an odd Integer, but not “ 1 “.
On this basis the product of the two integers is:
( 2n + 1 ) * ( 2n + K ) so
( 4n^2 + 2Kn + 2n + K ) or
( 4n^2 + 2n( 1 + K ) + K ) or
[ 2n( 2n + ( 1 + K )) + K ] ### Expression (1)
Reflecting on this expression:
2n is even
Since K is odd as I've defined it above, therefore ( 1 + K ) is even.
However, while all of Expression 1 so far is even, by definition K is not (because I have defined it as an odd integer), the Expression (1) is odd.
Therefore the product if two odd integers is ODD as required.
Step-by-step explanation:
For any Integer to be odd, it is not exactly divisible le by 2 and it can be described by the formula “ 2n + 1 “ where “ n “ is any integer.
Let us therefore consider two odd integers, being ( 2n + 1 ) and ( 2n + 3 ).
In this case, the two numbers are both odd and are consecutive.
That the two numbers are consecutive is irrelevant to this question.
That they are both odd is the important fact.
Their product is ( 2n + 1 ) * ( 2n + 3 ) or
4n^2 + 6n + 2n + 3 or
4n^2 + 8n + 3 or
4n( n + 2) + 3
By definition 4n is even because 4n = 2 * 2n and both “ 2 “ and therefore “ 2n “ are even..
However “ 3 “ is odd, so [ 4n(n + 2) + 3 ] is odd.
Therefore the product of these two odd integers is odd.
Let us though take the above analysis one step further by identifying the two odd numbers as:
( 2n + 1 ) and (2n + K ) where K is an odd Integer, but not “ 1 “.
On this basis the product of the two integers is:
( 2n + 1 ) * ( 2n + K ) so
( 4n^2 + 2Kn + 2n + K ) or
( 4n^2 + 2n( 1 + K ) + K ) or
[ 2n( 2n + ( 1 + K )) + K ] ### Expression (1)
Reflecting on this expression:
2n is even
Since K is odd as I've defined it above, therefore ( 1 + K ) is even.
However, while all of Expression 1 so far is even, by definition K is not (because I have defined it as an odd integer), the Expression (1) is odd.
Therefore the product if two odd integers is ODD as required.