# An elementary doubt on Inequalities

Given the following expression:

$$(7y-1)(y-7) \le 0$$

To me this inequality implies $y \le 7$ and $y \le \frac{1}{7}$ but the correct expression (from my module) happens to be $\frac{1}{7} \le y \le 7$

Where exactly I am wrong?

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$- \times - = +$. – Aryabhata Dec 7 '10 at 20:07
@Moron:if suppose $(7y-1)(y-7) \ge 0$ ..does then $y \ge 7 \text { and } y \ge \frac{1}{7}$ would be correct? – Quixotic Dec 7 '10 at 20:13
@Tretwick Marian: It depends on $y$ ;) – AD. Dec 7 '10 at 20:21
I am not getting ... can any body post the rules? Or may be some links might be helpful. – Quixotic Dec 7 '10 at 20:22
You see, if $a\cdot b\le0$ then either $a\le0$ and $b\ge0$, or $a\ge0$ and $b\le0$. – AD. Dec 7 '10 at 20:28

Hint: If the product of two numbers is nonpostive, one of them must be nonpositive and one of them must be nonnegative.

Edit: $(7y-1)(y-7)\leq 0$. So either, $7y-1 \leq 0$ and $y-7 \geq 0$ or $7y-1 \geq 0$ and $y-7 \leq 0$. Now, can you eliminate one of these two cases?

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Now I am confused with your nonpositive and nonnegative 's.Can I read them as negative and positive respectively? – Quixotic Dec 7 '10 at 20:11
@Tretwick: Nonpositive means either negative or zero. Your inequality is not strict. – Timothy Wagner Dec 7 '10 at 20:22
@Tretwick Marian: Almost, non-negative means it is $0$ or positive, and non-positive means it is $0$ or negative - that is it might be zero which is neither positive or negative. – AD. Dec 7 '10 at 20:24
Hm... thanks for the update :) – Quixotic Dec 7 '10 at 20:24
@Timothy Wagner: reload problem :) – AD. Dec 7 '10 at 20:24

Another way is to analyze as a quadratic function $f(y)=7y^2-50y+7$.

Note that you want solve : $f(y)=7y^2-50y+7 \leq 0$, whose graph is:

The intersections with the x-axis are : $x=\frac{1}{7}$ and $x=7$, this graph we say that $f(y) \leq 0$ in $[\frac{1}{7},7]$, this wanted to see.

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The Zero-Product Property ($ab=0\implies a=0\text{ or }b=0$) is the mechanics behind going from $(7y-1)(y-7)=0$ to $y=\frac{1}{7}$ or $y=7$. There is no directly analogous property for inequalities.

The method that I'd suggest for examining inequalities that compare an expression to 0 is the boundary algorithm—find the values that make the expression equal to 0, which are the boundary points of a set of intervals on the number line, then test each interval to see if it satisfies the original equation.

In your specific example, the boundary points are $y=\frac{1}{7}$ and $y=7$, so test some value of $y$ below $y=\frac{1}{7}$ (for example, $y=0$), some value of $y$ between $y=\frac{1}{7}$ and $y=7$ (for example, $y=\frac{1}{2}$), and some value of $y$ above $y=7$ (for example, $y=10$). The original inequality is false below $y=\frac{1}{7}$ and above $y=7$ and true between $y=\frac{1}{7}$ and $y=7$. Since the original inequality included = (it was ≤), the solution includes the boundary points that solved the corresponding equation, so the solution is $\frac{1}{7}\le y\le 7$.

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HINT $\$ If $\rm\ a < b\$ then $\rm\ (x-a)\ (x-b) < 0\ \iff\ a < x < b$

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