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Can 2+2=5?

  1. Jun 14, 2004 #1
    Could someone help me with this. I'm not very good with math and the only thing I know about this is that in the division, you can't divide by zero (somebody said that's why the formula is wrong). :grumpy:

    How do you know a + b - c is zero? :confused:

    I don't even know what a, b, and c equal. :cry: What does a, b, and c equal? :shy:

    Let a + b = c where a, b, c are real numbers

    5a – 2a – 2a +5b – 2b – 2b = 5c – 2c – 2c

    5a + 5b – 5c = 2a + 2b – 2c + 2a + 2b – 2c

    5(a + b – c) = 2(a + b – c) + 2(a + b – c)

    Divide through by a + b – c:
    5 = 2 + 2

    Therefore, 2 + 2 = 5

    sorry for the stupid question but I'd like to know and this forum looked like the only place that might tell me. :redface:

    thank you.
  2. jcsd
  3. Jun 14, 2004 #2
    oh, and when a number is beside a letter like that (5b) does that mean multiply 5 x b?

    Could someone rewrite that formula and replace the letters with numbers for me? So I can see how to put it all together?

    I'm sorry if I'm being a pest. :shy:

    thank you. :smile:
    Last edited: Jun 14, 2004
  4. Jun 14, 2004 #3
    Then a + b - c = c - c, which simplifies into a + b - c = 0.

  5. Jun 14, 2004 #4
    thank you :)

    but what did a, b, and c equal to start with? Like what numbers do I put in place of those letters?
  6. Jun 14, 2004 #5
    It's irrelevant exactly what the numbers are, as long as they satisfy a + b = c. You can put any real numbers you want, as long as they satisfy that relation. a = 1, b = 1, c = 2 or a = b = pi, c = 2pi for example. Of course, as soon as you choose the numbers, the fallacy becomes (even more) clear...
    Last edited: Jun 14, 2004
  7. Jun 14, 2004 #6
    oh! I see! :biggrin:

    Now it makes sense. When my dad said to research it on the net (I don't think he likes math questions at 5am :grumpy: ) or goto physicsforums.com and to find the answer he wasn't kidding!!! ha! :smile:

    Now when I go to school today, I won't look so dumb! :approve:

    Thanks! :shy:

    PS - I love math, I'm just not very good at it yet. :(
  8. Jun 14, 2004 #7


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    Jen, Even I don't like math problems at 5 A.M.!!!
  9. Jun 14, 2004 #8
    hehe... good thing Muzza didn't mind it at 5am! :biggrin:

    I don't usually get up that early but I was at my aunts birthday party Sunday evening and never got my homework done so I got up early to do it. :zzz:

    This wasn't part of our homework. Someone brought it to class and the teacher thought some of us might find it fun to figure out even though it was like way out of our league. At least I got it right with some help... it was fun to find out how it works! :rolleyes:

    You guys could probably do this stuff in your sleep though ha! :bugeye:
  10. Jun 14, 2004 #9
    I'm going to post because I just missed one of the actual Math help questions I would've been able to solve. Poor me and my Grade 10 Math level knowledge. Just incase you need further clarification:

    How do you know a + b - c is zero? Rearrange and:
    a + b = c
    Therefore each of the sides total values equal each other.
    Move the C over it becomes:
    a + b - c = 0
    You know a+b=c so a+b-c (which equals a+b) = 0

    This is just incase you weren't sure with the substitution of numbers. I used to use number substitution for equations because I didn't bother to learn how they all worked. Eventually calculators make the problems more confusing than actually learning the theory behind them.
  11. Jun 15, 2004 #10
    How did you get from "a+b=c" to "5a – 2a – 2a +5b – 2b – 2b = 5c – 2c – 2c" ? I see they are equal, I am curious as to the methodology used to get to one point to another.

    I was under the impression 2+2=4, not 2+2=5 unless of course the second equation is for greater values of 2? What does your conclusion indicate exactly?
  12. Jun 15, 2004 #11


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    It seems that people have explained to you how a + b = c implies that a + b - c = 0. Now, you know that 2 + 2 is not 5, it is 4, so where's the problem? It's when you said "Divide through by a + b - c," because you essentially said "Divide through by 0" (because a + b - c = 0). But you should know that you cannot divide by zero.
  13. Jun 15, 2004 #12
    all you did was prove that your initial assumption was wrong, except in the case a+b = c, in which case you are dividing by zero, and your simplification fails anyway.
  14. Jun 15, 2004 #13
    Yeah, but it's noon here when it's 5 am where you live.

    The probably worked backwards from something like 5 * 0 = 4 * 0 <=> 5(a + b - c) = 4(a + b - c) etc, to arrive at that equation.

    As has been mentioned, they divide by zero which renders the entire argument faulty. See Dr Math's FAQ section for several explanations why division by zero isn't defined/"allowed".
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