+1 to the person who can . . .

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    Re: +1 to the person who can . . .

    Post by bunnywink on Sat Dec 22, 2012 4:46 pm

    Wait, I thought you were gonna post another one? LOL :shock:


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    Re: +1 to the person who can . . .

    Post by reim0027 on Sat Dec 22, 2012 5:13 pm

    I can. But, I thought it would be better if the person who guessed correctly would post a new one (and give the +1 rep to the next person). Keeps it fresh.


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 5:15 pm

    But wasn't Sloth the one who got the right answer first?


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    Re: +1 to the person who can . . .

    Post by reim0027 on Sat Dec 22, 2012 5:17 pm

    Good point. Sloth - you're up.


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 5:18 pm

    Ughh, don't look at me... the only riddles I know can be solved through google :|



    Last edited by Sloth9230 on Sun Dec 23, 2012 5:43 pm; edited 1 time in total


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    Re: +1 to the person who can . . .

    Post by reim0027 on Sat Dec 22, 2012 5:25 pm

    Ok. I'll give a math problem. Please, don't go Google the answer. Where is the error?



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    Re: +1 to the person who can . . .

    Post by Emergence on Sat Dec 22, 2012 5:27 pm

    Did I miss something?


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 5:27 pm

    In the picture not loading silly
    I'm really hoping it's not my browser, then I'd feel dumb lol


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 5:31 pm

    There's an error with the picture tongue


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 5:34 pm

    Uparkaam wrote:There's an error with the picture tongue

    If you open it in a new tab, you can see it.




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    Re: +1 to the person who can . . .

    Post by Emergence on Sat Dec 22, 2012 5:37 pm

    In who's post?


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 5:38 pm

    Mozilla doesn't show anything there that could be opened in another tab. I'm gonna try Explorer.


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 5:44 pm

    I'm pretty sure that regardless of the numbers used, since a=b, the answer would always end up as zero. So how they got 2=1 is... confusing, or it's just wrong.

    Edit: Ah I get it, it's saying that two (a - b)'s is the same as one (a-b), which technically it is. Though that's not really math.


    Last edited by Sloth9230 on Sat Dec 22, 2012 5:47 pm; edited 3 times in total


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 5:45 pm

    My guess:

    2=1 should be 2=2


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    Re: +1 to the person who can . . .

    Post by reim0027 on Sat Dec 22, 2012 5:48 pm

    Sloth is close. But, I'm looking for the specific error (and in what step it is).


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 5:53 pm

    It's in the second to last step, I'm not sure what it's called though. But if they do 2(a - b) = 1(a - b) then it's (a - b)+(a - b) or (2a - 2b) = (a - b) which is (0 + 0) or (0 - 0) = 0 not 2=1

    the problem is that they cancelled the (a - b) parts out and just left it as 2=1, which you can't do because the number in the parentheses is 0.

    I'm only in elemntary algebra dang it, this is too hard! :evil:


    Last edited by Sloth9230 on Sat Dec 22, 2012 6:14 pm; edited 4 times in total


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 6:01 pm

    Do I have it now?


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    Re: +1 to the person who can . . .

    Post by Emergence on Sat Dec 22, 2012 6:10 pm

    Where is the problem lol?


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 6:11 pm

    It's hidden. You have to find it winking


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 6:12 pm



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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 6:13 pm

    If the second last part was 2(a-b)=1(a-b) it could be solved like this (I guess):

    2(a-b)=1(a-b) if I divide all of this with (a-b) it would result in this:

    2=1

    Which means that the last part is actually true.
    The error is in the second last part. The right hand side should be 1(a-b) instead of the a-b.


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 6:15 pm

    Also, E, are you using Mozilla?


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 6:16 pm

    Uparkaam wrote:If the second last part was 2(a-b)=1(a-b) it could be solved like this (I guess):

    2(a-b)=1(a-b) if I divide all of this with (a-b) it would result in this:

    2=1

    Which means that the last part is actually true.
    The error is in the second last part. The right hand side should be 1(a-b) instead of the a-b.

    But by divideing by (a - b) you're dividing by zero which you can't do. Booyah!


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    Re: +1 to the person who can . . .

    Post by Uparkaam on Sat Dec 22, 2012 6:19 pm

    But I'm not treating (a-b) as zero at this phase. I'm treating it as (a-b).
    (I don't know if that made any sense lol)


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    Re: +1 to the person who can . . .

    Post by Sloth9230 on Sat Dec 22, 2012 6:22 pm

    I know what you mean, but the instant you start dividing by zero, the numbers themselves become meaningless since it's "undifined"

    therefore you could get something like 6=1, 6=10, 100=37

    or in this case 2=1

    that's the mistake they made, they divided by zero


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