
#1
Aug2813, 10:12 PM

P: 246

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. 



#2
Aug2813, 10:54 PM

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P: 4,499

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




#4
Aug2813, 11:22 PM

P: 135

Does this small odd and even proof works?
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.




#5
Aug2813, 11:51 PM

P: 246





#6
Aug2913, 12:40 AM

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P: 4,499





#7
Aug2913, 01:31 AM

P: 246

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 ##2c1 = 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) 



#8
Aug2913, 09:07 AM

Mentor
P: 4,499

That's a very nice proof



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