This is taken from Peter J. Eccles, Introduction to Mathematical Reasoning, page 17. This is not a homework because it is an example in the text. Prove that 101 is an odd number. The text has given a way of proving it and I just want to do it with my own approach. Prove that 101 is an odd number. Assume 101 is even. There is a number ##b## such that ##101=2b##. Adds 1 to both side. ##102=2b+1## The right side shouldn't be divisible by 2 but the left side can be divided by 2. A contradiction? Hence 101 is an odd number.
Given the level of the question it seems like you should prove that 102 can be divided by 2 (by stating what the multiplication is). At that point it's just as easy to say 101 = 50*2+1 so is odd but them's the shakes
Your proof also uses the fact that if a number isn't even, then it's odd. Again considering the level of the question, this fact may not have been proven yet.
Then you should probably think about proving that if a number is of the form 2b+1 then it's odd (in particular you have to prove it's not even)
Prove that ##2b+1## is odd. Suppose ##2b+1## is even, then it exists an integer ##c##, where ##2b+1=2c## ##2c+1=2b+2## and ##2c-1 = 2b##, hence ##2b<2c<2b+2##. Further ##b<c<b+1## But there is no integer which is larger and smaller than the next consecutive integer. So ##2b+1## must be odd. I think this opens a new bag of cats, but at least I found this myself from scratch! (The book standard proof implicitly assumes this I believe)