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My math textbook says that "y is directly proportional to x if there is a positive constant k such that y = kx"

But of course Wolfram Alpha says that "Two quantities y and x are said to be directly proportional, proportional, or "in direct proportion" if y is given by a constant multiple of x, i.e., y=cx for c a constant." where the condition on c being positive is not given.

But my teacher said two quantities are directly proportional if one quantity increases as the other increases or one quantity decreases the other quantity decreases i.e. implying k is positive, following the 1st definition. But I don't agree with this.

Freddie
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    Your teacher is wrong, or rather, they are using language loosely and assuming $k>0.$ – Adam Rubinson Aug 25 '21 at 10:16
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    I think it's better to take $k \in \mathbb R$ rather than $k>0$, in order to be more general. But this also implies that both your mathbook and teacher are wrong. That happens ... – Matti P. Aug 25 '21 at 10:17
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    Your teacher is very wrong; consider $y=-\dfrac{1}{x}$. As $x$ increases, $1/x$ decreases and so $y$ increases. But are they directly proportional? No. – ultralegend5385 Aug 25 '21 at 10:48
  • Both definitions are wrong: $k$ just needs to be nonzero. – ryang Jun 27 '23 at 04:01

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