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We know $-1\le a \le 1$ and $-1\le b \le 1$ . Now find the interval of $ab$

In general , I want to know if we have intervals of $a$ and $b$ , what we can say about $ab$ interval .

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    @Clayton I think the right answer is $-1 \le ab \le 1$ – 2017-02-25

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Hint: $-1 \le a \le 1 \;\land\; -1 \le a \le 1 \iff |a| \le 1 \;\land\; |b| \le 1 \implies |ab| = |a| \cdot |b| \le1\,$ therefore $-1 \le ab \le 1\,$. Since the bounds are attained for $a=1, b=\pm1$ it follows that the interval is $[-1,1]$.

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    Okay , what we can say about general intervals ? – 2017-02-25
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    @S.H.W The above proves that the range of $ab$ is the full $[-1,1]$ interval. Not sure what you mean by `general intervals`. – 2017-02-25
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    My mean is what we can say about interval of $ab$ if $a_0\le a\le a_1$ and $b_0\le b\le b_1$ ? – 2017-02-25
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    @S.H.W For intervals symmetric about $0$ (i.e. $a_0=-a_1, b_0=-b_1$) you could use a similar argument. For arbitrary intervals, you'll need to break it down by cases, see Hagen von Eitzen's answer. – 2017-02-25
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    So that formula is right and trustable ? Can you provide a proof for it ? – 2017-02-25
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    @S.H.W Yes, that answer is correct. You could prove it by cases, for example if $a_0 \ge 0$ and $b_0 \ge 0$ then $a \ge a_0 \,\land\, b \ge b_0 \implies ab \ge a_0b_0\,$, then you'd have to consider the other sign combinations. – 2017-02-26
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    Okay , thank you a lot. – 2017-02-26
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Here, $-1\le ab\le 1$.

If $a_0\le a\le a_1$ and $b_0\le b\le b_1$, then several cases have to be considered depending on the signs of the bounds. However, it turns out that we simply have $$\min\{a_0b_0,a_1b_0,a_0b_1,a_1,b_1\} \le ab\le\max\{a_0b_0,a_1b_0,a_0b_1,a_1,b_1\}$$

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    Can you provide a proof for this formula ? – 2017-02-25
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    I think he is considering all the possible combinations of products of the minimum and maximum of $a$ and $b$. The maximum of these products should be the upperbound for $ab$, and the minmimum of the products should be the lowerbound for $ab$ – 2017-02-25
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    @mrnovice So you think this formula is right and trustable in all situations ? – 2017-02-25
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    No, but in this case, it seems pretty clear that the formula should hold. – 2017-02-25