Question

Can someone please help me? I am stuck on this problem of how to put it all together.

On a white board are numbers from 1 through 2012. Someone picks two numbers, erases them and writes the difference. The process continues with that person choosing at will, the two numbers to be erased continues until there is only one number left. Prove that, that number must be even?

Answer 1

If you pick 2 even numbers -> the difference is even
2 odd --> diff is even
1 even 1 odd --> diff is odd

Keep in mind that if you keep on picking 1 even and 1 odd, it will give an odd number, and eventually there will only be odd numbers
Odd - Odd = even.

Answer 2

So I have If you pick an even and odd, you write an odd. 2n-(2m+1)=2n-2m+1=2(n-m)+1=2p+1 (where p=n-m), which is odd.

If you pick an even and and even, it's even. (2n-2m=2(n-m) and n>m, n-m is an integer.

If you pick an odd and an odd, it's even. (2n+1)-(2m+1)=2n-2m=2(n-m)=2p, (where p=n-m), an even integer.

What equation would I end up with at the end to prove this? I think this is a proof by cases so i have 3 cases?

Answer 3

Hmmm, If I start with 2012 and 2011 as my first two numbers, and then keep going backwards in order creating pairs, I end up with 1 as my last remaining number which is odd.

Answer 4

The weird part is it wants you take the difference of those two numbers and put it back on the board, then pick two more numbers and repeat. Did you do that?

Answer 5

Even if I did that, I would end up with 2012 - 2011 = 1; 2010 - 1 = 2009; 2009 - 2009 = 0; 2008 - 0 = 2008; 2008 - 2007 = 1; Notice every time I subtract off an odd number, I end up with an odd number. If I kept going, I'd get to 1. Since 2 is an even number, I would end up with 2 - 0 = 2. Then 2 - 1 = 1...an ODD number

Answer 6

how much of a proof do you need? Does it need to actually be a proof, or just an explination?

Answer 7

Oh i see. Darn this problem is no fun. And i think all I need after the 3 cases I presented above is an proof, if possible and if not an explanation work.

Answer 8

Counter-examples disprove proofs

Answer 9

By the time you get to 2, you'll either have 2 - 0 = 2 or 2 - 2 = 0. Either way you still have to subtract 1 leaving you with an odd number.

Answer 10

A program can be written for this and that would be a good way to prove it.

Answer 11

Ok i follow you. And what is a program? We have not learned that yet

Answer 12

A program is a computer based tool that is used to automate tedious calculations such as these. It basically takes a "proof" and puts it into a form that is usable either for demonstration, instrumentation, or evaluation purposes. Calculators, for example are programs used for evaluating numerical equations or algebraic functions.

Answer 13

implementation* rather than instrumentation

Answer 14

Sorry my computer froze. I just realized that I am not sure we can subtract two numbers and get 0 and continue on because the difference always has to be positive and 0 isnt a positive?

Answer 15

To say we can't continue on is like treating zero as if it isn't a valid number. Rules of mathematics suggest that we can find the difference between two numbers and get zero if those two numbers are the same. If we're being asked to find the difference between a and zero, then the difference is a. Thus no rules are being violated here.

Answer 16

I was talking about the rule to the problem. Since it says the difference is always positive. And a positive number is any number greater than 0 no?

Answer 17

The question says from 1 to 2012

Answer 18

Those are just the range of numbers you start with. But then it says to take two numbers, erase them, then REPLACE them with their DIFFERENCE.

Answer 19

so if I am in a situation where 2009 - 2009 = 0, since the difference between both is zero, I have not violated any rules.

Answer 20

Ahh shoot I left out the positive part. :/ im really sorry, my teacher added that at the end of class and I didnt write it down

Answer 21

:S?

Answer 22

Then I guess the proof you come up with is going to have holes in its armor.

Answer 23

positive part?

Answer 24

The difference of the two numbers has to be a positive integer

Answer 25

So basically subtractions that yield zero are invalid

Answer 26

Yes. Sorry about that!

Answer 27

You could subtract two numbers and get 7, which means you can't choose 7 as the next number you subtract. That makes things a little less random.

Answer 28

As long as when I keep subtracting two numbers, whatever number I get, If it is listed, then I cannot choose that number randomly anyone. Are you sure that this is the case. If I subtract two numbers, and I get 701, now I can't choose 701 if it is still on the list of numbers I'm supposed to be able to choose?

Answer 29

Yes correct. You can take the 7 you got from the difference and subtract it from another number just not the other 7

Answer 30

So now, the extra part you forgot to include says that the number I subtract must be less than the first number I chose.

Answer 31

Now the proof is that much more complicated.

Answer 32

Not less it just cant be the same number. Like if I am using the numbers 1-10. And I take the difference of 10 and 3=7. The numbers I have left are 1 2 4 5 6 7 7 8 9. The second 7 was the difference of the numbers. I can subtract it from any other number than the other 7 in the set

Answer 33

You said the difference has to be positive. If b is more than a and I subtract the two, I get a - b = c, but c will be negative

Answer 34

They don't have to be in order, you subtract the bigge4st number from the samllest. n>m

Answer 35

Okay, good luck with that. I'm just disappointed that I can't use zero.

Answer 36

Me too haha