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I am trying to help my student understand why it is true that

$$0 < \frac{x}{x+y} < 1 \impliedby 0 < x \text{ , } 0 < y$$

I so far say that intuitively, $\frac{x}{x}=1$ and since $0<y$ we know $$\frac{x}{x+y} \lt 1$$

when $y$ is close to $0$.

But when $y$ is much bigger

$$0<\frac{x}{x+y}$$.

She really has trouble with "intuition" but she does very well with rigorous algebraic proofs.

Is there an "algebraic" approach to this situation?

A would appreciate your help.

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    $\begingroup$ You can multiply the whole inequality by $x+y$(it's not zero and it's positive) then you get $0<x<x+y$ which is the same as $0<x$ and $x<x+y$ and this is same as $0<y$ $\endgroup$
    – kingW3
    Jul 2, 2019 at 22:34
  • $\begingroup$ If $y > 0$ then $x + y > x$. So $1 = \frac {x+y}{x+y} > \frac {x}{x+y}$. Likewise $0 < \frac {x}{x+y} < 1$ if and only if $\frac{x+y}{x} > 1$. And $\frac {x+y}x = \frac xx + \frac yx = 1 + \frac yx > 1$. $\endgroup$
    – fleablood
    Jul 2, 2019 at 23:27

4 Answers 4

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Since $x$ and $y$ are greater than zero, we can divide by $x + y$ without messing up inequalities.

So if $0 < x < x + y$, which is clear, then $\frac{0}{x + y} < \frac{x}{x+y} < \frac{x+y}{x + y}$, which just states $0 < \frac{x}{x+y} < 1$

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    $\begingroup$ Oh, my, THIS is exactly the type of answer that I was looking for! Thank you very much!! $\endgroup$
    – hyg17
    Jul 2, 2019 at 22:25
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    $\begingroup$ @hyg17 You're welcome. I hope it helps your student. $\endgroup$
    – Ruben
    Jul 2, 2019 at 22:28
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    $\begingroup$ It has. The biggest challenge is that she is 7 and her language skills are not developed but she understands math at pre-calc level. Her mind works differently and she unexpectedly gets stuck at concepts that I would assume that she understands but her verbal skills and math skills are for some reason independent from each other. $\endgroup$
    – hyg17
    Jul 2, 2019 at 22:32
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    $\begingroup$ @hyg17 seven years old, that seems like quite the challenge. I wish you the best of luck. $\endgroup$
    – Ruben
    Jul 2, 2019 at 23:11
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If: $$x,y>0$$ then: $$x+y>x$$ $$x+y>y$$ and so: $$\frac x{x+y}<1$$ $$\frac y{x+y}<1$$ also since all terms are greater than $0$, you have a positive number divided by a positive number, which must be positive, and so: $$\frac x{x+y}>0$$ $$\frac y{x+y}>0$$

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Similar to one of the answers, but succinctly: since $x, y > 0,$

$$0 = \frac{0}{x + y} < \frac{x}{x + y} < \frac{x}{x} = 1$$

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We can explain by the following simple rules:

  • Rule 1. Sum of two positive numbers is a positive number.

  • Rule 2. Quotient of two positive numbers is a positive number.

  • Rule 3. Subtracting a positive number we get a smaller result.

Since $x>0,y>0$, $x+y>0$ by Rule 1, and $\displaystyle\frac{x}{x+y}>0$ and $\displaystyle\frac{y}{x+y}>0$ by Rule 2. Finally, by Rule 3, we have

$$ \frac{x}{x+y}=\frac{(x+y)-y}{x+y}=1-\frac{y}{x+y}<1. $$

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