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Algebra Question

  1. Oct 6, 2013 #1

    I wanted to confirm that I am understanding something correctly. I have been doing this all the time, and have never had any problems with it. For some reason now I am wanting to confirm it.

    So, if you have something such as: x3-6+y2-√z it would be equivalent to write it as: y2+x3-√z-6

    Can you always rearrange terms like this, no matter what they are? What I do is write it as an addition problem so I can change the order, since you can change the order in addition.

    For example if you have: log(x)-x3 you could write it as log(x)+(-x3) then as -x3 + log(x).

    Just as -11+5-3+5=-4 would be the same as 5-11-3+5=-4 or -11-3+5+5=-4

    Does my reasoning seem correct?

    Last edited: Oct 6, 2013
  2. jcsd
  3. Oct 6, 2013 #2
    Yes that is correct. I believe that falls under the commutative property which essentially states

    x + y = y + x
  4. Oct 6, 2013 #3
    And this would also go for multiplication, but not division correct?
  5. Oct 6, 2013 #4

    Example using decimal notation:

    2*4 = 8 and 4*2 = 8

    8 = 8 TRUE

    2/4 = .5 and 4/2 = 2

    .5 = 2 FALSE

    And here's a tip, never be afraid to go back to the most basic of examples to solidify your thoughts if you start question them. Forget the log(x) and y^3 stuff and just use a scratch piece of paper and walk through simple things like 1+2 = 2+1 to convince yourself that the order of addition does not matter.
  6. Oct 7, 2013 #5


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    Hi, ThomasMagnus!
    As the others have said, you are correct.
    The two operations multiplication and addition are COMMUTATIVE, i.e, you can change the order in which they appear.
    Division and subtraction are NOT commutative, we can call them ANTI-commutative if you like.

    But, as you yourself noted, can't we in 5-4 rewrite this as 5+(-4)=(-4)+5?

    Yes, we can, by using negative numbers in addition, we can think of the operation of subtraction as "redundant", i.e, the info is contained in the use of adding with negative numbers instead.
    That makes for easier calculations, because you then can go about commuting as much as you want!
    In 5-(-4), for example, we just write 5-(-4)=5+(-(-4))=(-(-4))+5 and so on.

    Does something similar exist with division relative to multiplication?

    Yes, because division by a non-zero number "a" can always be thought of as multiplication with the reciprocal of "a", i.e, 1/a.

    Thus, 2/4=2*(1/4)=(1/4)*2.

    By thinking in terms of "negative numbers" and "reciprocals", you really have only two basic operations to worry about, addition and multiplication, rather than four operations.

    (And, if you think of multiplication as repeated addition, there's only one fundamental operation to worry about..:smile:)
  7. Oct 7, 2013 #6


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    Yes, and note that it is largely because (well, perhaps even more that addition and multiplication are "associative" while subtraction and division are not) that we do not consider subtraction and division as separate "operations" but just the inverses of additon and multiplication. That is, that "a subtract b" is just "a plus the additive inverse of b" and "a divided by b" is "a multiplied by the multiplicative inverse of b".
  8. Oct 7, 2013 #7


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    To elaborate on HallsofIvy's reply:
    When we think of ONE operation to be performed, commutativity is the only property that comes into the picture.
    However, when we have two or more operations to be performed, we must also take into account what we call the associativity of the operations, that is, what relevance it has which operation we perform first.

    We have, for example, for numbers "a", "b" and "c" that:
    (a+b)+c=a+(b+c), that is, it doesn't matter WHICH addition we perform first. Addition is associative.

    But, is:

    You easily see that subtraction is NOT associative at all;

    We have that (a-b)-c=a-(b+c), whereas a-(b-c)=(a-b)+c
  9. Oct 7, 2013 #8
    Thank you for the info! So the next time I see something in an answer key as something along the lines of x-b+c and I have -b+c+x I can assume they are the exact same, correct? Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6 ?

  10. Oct 7, 2013 #9


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    "Also, does this go for all terms, even something such as ln|x-1|-ln|x|+6=-ln|x|+ln|x-1|+6"

    That's right! :smile:
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