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Subtract or multiply by a number

  1. Jan 9, 2005 #1
    I have a quick basic question , perhaps Arildno could could help me as he was very good at explaining before.....



    2 + root 9 = 5

    If I add or subtract or multiply by a number , the equality holds as long as its done to both sides but if I square both sides it means I multiply one side by

    (2 + root 9 ) while I multiply the other side by 5.

    But nevertheless, the equality clearly remains valid.

    Why is this ?


    Thanks


    Roger
     
  2. jcsd
  3. Jan 9, 2005 #2

    Curious3141

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    I don't see the mystery. If the original equality holds true (as it does in this case), then squaring both sides will give a valid equality. Muliplying the LHS by the LHS and the RHS by the RHS will give a correct result. In essence, it's the same thing as multiplying both sides by either the LHS or the RHS (since they are equal anyway). However, taking the new equality and taking positive and negative roots of either side will give a wrong result if you're careless about which roots to take (e.g. "nonsense" results like [itex] 2 + \sqrt{9} = -5[/itex]. This is because the functional mapping [itex]f:x \rightarrow x^2[/itex] is not injective over the reals.
     
    Last edited: Jan 9, 2005
  4. Jan 9, 2005 #3
    Well remember that 2+root 9 and 5 are equal so multiplying a number by 2+root 9 is the same thing as multiplying a number by 5. It doesn't really matter that the two sides are written differently; you're still multiplying both sides by the same number.
     
  5. Jan 9, 2005 #4

    Please could you explain this in detail (the highlighted part).


    Roger
     
  6. Jan 9, 2005 #5

    Curious3141

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    A function is something that maps elements in one set called the domain onto elements in another set called the range. Both of these should be defined when you're defining a function. When you say [itex]f:x \rightarrow y[/itex] you mean "f maps domain x onto range y". Another way of saying the same thing is [itex] f(x) = y[/itex] and I'm sure you've seen that before.

    An injective function is one where every element in the range is mapped to by AT MOST one element in the domain. That means no two elements in the domain will get mapped onto one particular element of the range.

    In the functional mapping [itex]f:x \rightarrow x^2[/itex], where the domain is all real numbers and the range is nonnegative real numbers, you find that a particular value of the range (say 4) can be mapped to by two possible values in the domain (-2 and + 2). This is because (-2)^2 = (2)^2 = 4. Hence [itex]f:x \rightarrow x^2[/itex] is not an injective function.

    There are other terms relating to functions, like surjective and bijective. This link will make things a lot clearer : http://en.wikipedia.org/wiki/Injective_function

    Hope this helps. :smile:
     
    Last edited: Jan 9, 2005
  7. Jan 9, 2005 #6
    Injective means one-to-one. One input (and no more than one) to one output.
    It is not injective because , for example, f(-1)=f(1). The output 1 has two inputs, 1 and -1. In fact, f(-x)=f(x) for all x.

    Facts:
    1. If A=B then f(A)=f(B). For example, 1=2-1 and 1^2=(2-1)^2.

    2. If f(A)=f(B), then A=+ or - B. For example, 9=3^2=(-3)^2=9.
     
  8. Jan 10, 2005 #7

    HallsofIvy

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    However, asking about "functional mappings" in order to see why, if a= b, doing anything to both sides results in an equality that is still valid is overkill.

    The only mathematical point is that "a= b" means "a and b are names for the same thing". If I start with the same thing and then do exactly the same thing, it doesn't matter if I call then "Bill" or "Annie", the result is the same!

    (And writing "root 9" instead of "3" has nothing whatever to do with it!)
     
  9. Jan 10, 2005 #8

    Curious3141

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    Agreed. I wanted to emphasise that squaring was OK but taking square roots was not, and only expanded when Roger asked for a detailed clarification.
     
  10. Jan 10, 2005 #9

    HallsofIvy

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    For this direction, taking the square root is perfectly good (as long as the square root exists). It's only when you try to reverse a function that is not one-to-one that you run into trouble.
     
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