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This thread is just what the title says it is. It is all about Math. You can post math jokes(the jokes don't have to be good), math questions, what you like about Math, or even why you hate Math.
My math joke: Resistance is not futile. It is voltage divided by amps.
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A [0,1] range would be if you never messed up. Key the word should.
So you are saying I did not account for all of my mistakes? Because I'm pretty sure I did...
Remember, you can't think of this as a sequence of partial sums. Its one big sum that results in the number of ducks my friend has. Which is 57.
as long as the +57s are equal to the -57s, right?
= 57 + (-57 + 57) + (-57 +57)+ (-57 +57)....
Since by the properties of numbers, a+b+c = a+(b+c)
=57 + 0 + 0 + 0 + 0.....
= 57.
I think that the way your grouped was wrong. What you show right now is an extra 57 meaning there would be 1 more 57 then -57.
So you are saying I did not account for all of my mistakes?
Your mistake being that you gave him 57 ducks and then took 57 ducks instead of giving 1 duck and taking 1 duck.
What you show right now is an extra 57 meaning there would be 1 more 57 then -57.
But, I have infinitely many (-57)s and every (-57) is paired off with a positive 57. Every time I give 57 ducks, I immediately follow it by taking 57 ducks. So, how could I possibly have an "extra" 57 ducks? Its not like there is a (-57) at the "end" of my sum that isn't included in the bracketing and should be paired with the first 57, because there is no "end" of the sum. It just goes on forever. So, there are just as many (+57)s as there are (-57)s, giving a total of 57 ducks.
If I were to group it like this:
57-57+57-57+57-57....=
(57-57)+(57-57)+(57-57)...
=0+0+0+0.....
You would have no problem, right?
But instead I grouped it like this:
57 + (-57 + 57) + (-57 +57)+ (-57 +57)....
=57+0+0+0+0....
=57
But since the "tails" of these series are equal to each other, and are infinitely long (meaning that no matter how I group it, there will be no term "left out" of a grouping at the end), it shouldn't matter how I group it, right?
I mean,
a +b +c + d+...
=
(a+b)+(c+d)+....
=a+(b+c)+(d+e)+....
So why do you only have a problem with the second grouping?
using 2, 6, 3, 11, 9 how can you make the answer equal 13?
Unbelievably easy for me, I got it in about 3 seconds.
2+6=8, 8+3=11, 11+11=22, 22-9= 13
Maybe I'm just having a good day.
anyways, I'm in geometry and it's way better then algebra! XD
Really? I like Algebra better, although I think I might be better at geometry.
As for the duck problem, I can't really understand it. If you give and take away a duck leaving every one accounted for eventually, then how does he have 57 ducks?
Really? I like Algebra better, although I think I might be better at geometry.
Same here. Algebra is more fun for me.
If you give and take away a duck leaving every one accounted for eventually, then how does he have 57 ducks?
That is what aknerd and I are talking about. I think that the answer is 0. He thinks that it is 57.
That is what aknerd and I are talking about. I think that the answer is 0. He thinks that it is 57.
Well, now that I think about it, maybe it shouldn't be 57. Because, if I were to "split up" every 58th term, and sprinkle through the previous 57 terms (for both the positive and negative 57s), I would get something like:
57+1 -57-1 +57+1 -57-1.... +0-0+ 57+1 - 57-1....
Where the first zero is where the 58th +57 normally would be, and the second zero is where the 58th -57 would be. So, I didn't really change anything, because I added/subtracting everything back into the sum that I took out.
But this sum is just:
58-58+58-58+58.....
Which is just
58- (58-58) - (58-58).... (distribute the negative sign, if you want to verify this)
which is just
58 -0 -0 -0 -0 -0....
Which is just 58.
So, actually, there should be 58 ducks, not 57. My bad. Actually, wait. If I were to distribute the 59th terms ...
I mean,
a +b +c + d+...
=
(a+b)+(c+d)+....
=a+(b+c)+(d+e)+....
Wouldn't this only work if a>0, b>0, c>0 etc.? If one or more are actually negative, you'd get two different results (obviously, else you wouldn't have that 0 or 57 thing). It does matter where you put your brackets then.
Wouldn't this only work if a>0, b>0, c>0 etc.? If one or more are actually negative, you'd get two different results (obviously, else you wouldn't have that 0 or 57 thing). It does matter where you put your brackets then.
You might be on to something there.
Here is a Wikipedia link about the associative property that aknerd was using. Look under Non-associativity.
Wouldn't this only work if a>0, b>0, c>0 etc.? If one or more are actually negative, you'd get two different results (obviously, else you wouldn't have that 0 or 57 thing). It does matter where you put your brackets then.
Ehh... I've always thought of the non associative subtraction thing as just bad math. It has nothing to do with whether or not a, b, c are positive. If you look at subtraction as adding a negative, then it is associative.
Observe:
1-2+3-4+5 = (1-2)+(3-4)+5 = 1+(-2+3)+(-4+5)
Which is how I used association in my sum. I said that
57-57+57-57...= 57+(-57+57)+(-57+57)+... and
57-(57-57)-(57-57)-....
Which is very different from
57 -(57+57)-(57+57)-....
The real reason things go wrong is addressed in Dalek's link: infinite sums are non associative. They even used my original example!
So, really, I did not make a mistake. The series 1-1+1-1+1-1+1.... can be rearranged to equal any integer.
The series 1-1+1-1+1-1+1.... can be rearranged to equal any integer.
Does that mean I was pretty much right saying it could be from [0,57] or can it only be 0 and 57?
Does that mean I was pretty much right saying it could be from [0,57] or can it only be 0 and 57?
No, it can be literally any integer, without bounds. An easier way to rearrange this would be in the form:
X-1+1-1+1-1+1-1+1....
Where X is some arbitrary integer, and you group the 1's so that they all cancel. You can have X be any integer without unbalancing the sum, because you will still have the same amount of positive and negative 1s. Think about it: if I were to pull 100 ones out of the original sum and put them all at the beginning, I would still be able to find another 100 ones to put in the places where I took the original ones from, and so on.
For instance, if I had
1-1+1-1+1-1....
I could make it
1+1-1-1+1-1+1...
By switching the numbers in bold. Then, I could just keep switching numbers until I got:
2+(-1+1)+(-1+1)+...
=2
Because there will always be somewhere to put the one. I can do this how many time I want, with positive or negative numbers.
I also showed how you can make
57-57+57-57+....
=58 (or 59, or 60, or...)
By "splitting up" every 58th terms.
No, it can be literally any integer, without bounds.
Ahhhhh.. OK. That is the problem with infinity. There is never really a practical use for it. The only use for it is theoretical.
Okay, I thought of another one:
using 2, 6, 3, 11, 9 how can you make the answer equal 13?
6/(2*3)+11=13 in base 9. Bam (the 13 is in base nine, the rest is in base 10). Can anyone think of a solution that doesn't involve any addition or subtraction (without repeating any numbers)?
11, being a prime number, makes it pretty tricky. It might not be possible...
Ahhhhh.. OK. That is the problem with infinity. There is never really a practical use for it. The only use for it is theoretical.
ummmmmmmm I wouldn't say that. I mean, infinity is used in many mathematical concepts, for instance limits. And limits are used to define derivatives, and derivatives are one of the most practical mathematical concepts.
I just think this example is very amusing, because there are infinitely many solutions to the sum.
ummmmmmmm I wouldn't say that. I mean, infinity is used in many mathematical concepts, for instance limits. And limits are used to define derivatives, and derivatives are one of the most practical mathematical concepts.
Since I have not been to Calculus yet I have not learned derivatives. I have learned a bit of limits though but not much.
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