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Direct and indirect proportion

  1. Apr 22, 2012 #1
    I read that y=-2x is a direct proportion graph. But how is this so? Since as one variable increases the other variable decreases proportionately so how can this be considered to be a direct proportion case? So does it mean that k>0? Since when it is less than 0 it then becomes a case which does not fit into any of the scenarios? Or if we were to write it then since it decreases proportionately then the graph would be written as y=2/x. So on the whole, will k always be positive?

    Thanks for the help guys! :smile:
  2. jcsd
  3. Apr 22, 2012 #2
    Hi, sgstudent,
    the notions of "direct" and "indirect" proportion are taught at an early age, where possibly they don't want to complicate the picture with negative numbers. The important fact is the power of [itex]x[/itex] ([itex]1[/itex] or [itex]-1[/itex]), rather than whether the result increases or decreases (the two notions, of course, coincide for positive quantities).

    The terms come from the idea of a "proportion" or "ratio":[tex]
    \frac y a = \frac k x
    [/tex]or, "[itex]y[/itex] is to [itex]a[/itex] as [itex]k[/itex] is to [itex]x[/itex]". You can always write this proportion so that [itex]y[/itex] is "above" (in the numerator) of the left-hand side (flip the whole thing otherwise). Then you have only two cases for [itex]x[/itex]: either it is "up" or it is "down". One you call it "direct", the other "inverse". There's nothing more to it. It's a notion as old as the ancient Greeks, as it already shows itself in book V of Euclid's "Elements"; he would talk in the context of distances or measures, though, which were always non-negative.
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