Question

Just need some clarification. I follow the even and odd part that in order to be even its 2k and odd is 2k+1. but where do they get the difference of 3 and 4? Each integer n>11 can be written as the sum of two composite numbers. Proof. Case 1: (n is even) Since n is even, n=2kfor some k≥6. Hence, n=2(k−3)+6 Case 2: (n is odd) Since n is odd, n=2k+1for some k≥5. n=2(k−4)+9 (The sum of two composites)

Answer 1

not sure what you are asking exactly, but \(2(k-3)+6=2k\) is all the first one says.

Answer 2

they are rewriting \(n=2k=2(k-3)+6\)

Answer 3

Thanks for helping out to start off. I know that they are rewriting for a new k but why did they chose the number three to subtract k from? is it just a random number or is there a purpose to it?

Answer 4

they picked 3 because \(n>11\) and so 3 is will work

Answer 5

ok so the 6 is because its a composite, and the number they are subtracting from k is just any number that will at the end give us a number >1?

Answer 6

if \(n<11\) you could not say this. they are picking the smallest number they can because the assertion is for \(n>11\)

Answer 7

yes, what you said

Answer 8

alright cool! so to prove this I can use those two examples?

Answer 9

looks like it works
algebra i mean

Answer 10

second one basically says any odd number greater than 11 can be written as some even number plus 9. seem rather obvious

Answer 11

subtract 9 from an odd number, get an even number.

Answer 12

haha i know. The teacher I have skips a lot so I am still figuring things out. but thank you sooo much!

Answer 13

yw