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Proving a = 2a

  1. Jul 7, 2004 #1
    i went into physics class the other day and saw on the board

    prove a = 2a
    a = b
    a^2 = ab
    a^2 - b^2 = ab - b^2
    (a+b)(a-b) = b(a-b)
    hence a+a = a

    it makes scence to me, but wen i showed my friend sometime latr
    he told me it couldnt be
    but he couldnt explain it.
    could somone tell me why 2a = a
    from the prove provided

    thanx heaps
  2. jcsd
  3. Jul 7, 2004 #2


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    Note, that first of all, a = b, therefore a - b = 0.

    Now, you have:

    (a+b)(a-b) = b(a-b)
    (a+b) = b

    But to go from one line to the one after, you had to divide by (a - b), which means you divided by 0, but that's not allowed, and there's your problem. Division by zero is not defined.
  4. Jul 8, 2004 #3
    ok, thanx heaps
    so just simply substituting a-b=0 from the first line.

    thanx again
  5. Jul 8, 2004 #4
    This is a common example of how a rule that we think is ticky-tack can create a completely nonsensible result. It pays to be careful.
  6. Jan 2, 2009 #5

    There is a small clause which no one remembers:
    the formula (a^2 - b^2)=(a+b)(a-b) if and only if (a=/b)
  7. Jan 2, 2009 #6
    No... a^2 - b^2 is always (a+b)(a-b), however if you have a-b on both sides, you can only "cancel" i.e. divide both sides by (a-b) if a <> b i.e. so you don't divide by 0.
  8. Jan 2, 2009 #7


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    Thank goodness no one remembers it because it is not true! if a= b= 2 then 4- 4= 4(0) is certainly true.

    What you are thinking of, I suspect, is that
    [tex]\frac{a^2- b^2}{a- b}= a+ b[/itex]
    only if [itex]a\ne b[/itex].

    That provision is given in every pre-calculus book so it is not true that "no one remembers". It is only the students that don't remember it!
  9. Jan 2, 2009 #8
    Those students are my best friends! They are willing to do things like give me $2 for my $1!
  10. Jan 2, 2009 #9
    If I have no cars. I can multiply by anything and I still have no cars.
    Fine. No loss to me there.


    If I have five cars
    and I multiply the number of cars I have by nothing

    Why don't I still have my five cars?

    English and math together, sometimes cause confusions.
  11. Jan 2, 2009 #10


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    You still have five cars! No arithmetic operation is going to change the number of cars you have!

  12. Jan 2, 2009 #11
    5 x 0 = 5

    Dang. I must of got a math teacher with a book that had an error in it.
  13. Jan 2, 2009 #12
    Or you just need to define what "multiply" means in English. To me it means how many times you take something and add it up i.e. take 0 cars and multiply by anything i.e. add 0 to itself any number of times and you get 0. If you have 5 cars and multiply by 0 i.e. add that number to itself zero number of times, you have 0 since you are not doing anything.

    Maybe both your Math and English teachers failed you, x is a letter. :)
  14. Jan 2, 2009 #13
    hehehe quite possibly.
    x is just a lazy way to say multiply the starting number by the following number.

    Commutative Property of multiplication.
    I accept it, ( because it works ) but it just never sat well when I was told about it the first time

    If I start with nothing or zero, it seems correct that I can't do anything to it.
    but, It's the idea that I start with something and then do nothing to it ...poof it's gone.
    It still sticks in my head that if I divide something by nothing I haven't actually done anything and the answer should still be what I started with, instead of the 'undefined' thing.

    hehehe as I have said before, I'm not a big fan of 'zero' as a concept.
  15. Jan 2, 2009 #14
    As far as division by zero, think of division as inverse of multiplication.

    You said you accept that 0*anything = 0 so basically let's fix two values for "anything" i.e. 1 and 2. You would agree that 0*1 = 0 and 0*2 = 0, if division by zero was allowed then 1 = 0/0 and 2 = 0/0 so 1 = 2?
  16. Jan 2, 2009 #15


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  17. Jan 2, 2009 #16
    no problem with
    0*anything = 0

    no problem with
    anything * 0 = anything

    the order of operation seems to be a problem for me.
    the commutative property of multiplication only works for a second term > 0

    I seem to remember that this shows up when trying to multiply matrices also.

    If you guys want to spend some time trying to teach me another way to look at it ( the zero concept ) I would be thankful, but I was only responding to the OP in that zero can mess up ideas.
    In the OP case it is a hidden division by zero error that makes it look like 2a=a
    For multiplication it looks like the five cars I start with vanish if I try to multiply them by zero.
    Last edited by a moderator: Jan 2, 2009
  18. Jan 2, 2009 #17
    But if anything*0 = anything, then when you divide both sides by "anything" you get 0 = 1
  19. Jan 3, 2009 #18
    I'm probably wrong but with
    [tex]\frac{a^2- b^2}{a- b}= a+ b[/itex]
    only if [itex]a\ne b[/itex]

    Why isn't it possible for you to just manipulate this by multiplying both sides by the denominator?

    [tex]a^2- b^2}= (a+ b)(a - b)[/itex]
    only if [itex]a\ne b[/itex]
  20. Jan 3, 2009 #19
    That is fine. Halls was just illustrating the reverse process, i.e., dividing both sides by (a-b) if it is non-zero.
  21. Jan 3, 2009 #20


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    IF a is not equal to b, saying
    [tex]\frac{a^2- b^2}{a-b}= a+ b[/tex]
    lets you say [itex]a^2- b^2= (a- b)(a+ b)[/itex] with the proof you state. If a= b, then that proof does not work.

    But saying a particular proof does not work doesn't mean the statement itself is not true!

    Whether a= b or not 0a= 0b is always true. If a= b, [itex]a^2- b^2= (a- b)(a+ b)[/itex] because both sides are equal to 0.
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