If you take the reciprocal in an inequality, would it change the &gt; / &lt; signs?Example: − 16 &lt; 1 x − 1 4 &lt; 16In the example above, if you take the reciprocal of 1 x − 1 4 = x 1 − 4 1 would that flip the &lt; to &gt; or not?In another words, if you take the reciprocal of − 16 &lt; 1 x − 1 4 &lt; 16would it be like this: 1 − 16 &gt; x 1 − 4 1 &gt; 1 16

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If you take the reciprocal in an inequality, would it change the   &amp;gt;      /    &amp;lt; signs?Example:  −  16  &amp;lt;      1    x    −      1    4    &amp;lt;  16In the example above, if you take the reciprocal of      1    x    −      1    4    =      x    1    −      4    1  would that flip the   &amp;lt; to   &amp;gt; or not?In another words, if you take the reciprocal of  −  16  &amp;lt;      1    x    −      1    4    &amp;lt;  16would it be like this:      1          −      16        &amp;gt;      x    1    −      4    1    &amp;gt;      1    16

timmeraared · Accepted Answer

If a and b have the same nonzero sign, then  a  &amp;lt;  b    ⟺        1    a    &amp;gt;      1    b  (i.e., taking reciprocals reverses the inequality).If a and b have opposite (nonzero) signs, then  a  &amp;lt;  b    ⟺        1    a    &amp;lt;      1    b  (i.e., taking reciprocals preserves the inequality).These follow from the fact that the function   f  (  x  )  =  1      /    x defined on the nonzero reals is strictly decreasing and sign-preserving on each component   (  −  ∞  ,  0  ) and   (  0  ,  ∞  ) of its domain.If either of a or b is zero, then you can&#039;t take reciprocals.Finally, compound inequalities like   a  &amp;lt;  b  &amp;lt;  c should be separated into &quot;  a  &amp;lt;  b and   b  &amp;lt;  c&quot; and each component considered separately.I also remark that inverting a sum is not the same as inverting the addends separately.

migongoniwt · Accepted Answer

It depends if   x and   y are the same sign.Case 1:   0  &amp;lt;  x  &amp;lt;  y then   0  &amp;lt;  x  (  1      /    y  )  &amp;lt;  y  (  1      /    y  ) and   0  &amp;lt;  x      /    y  &amp;lt;  1 and   0  &amp;lt;  x      /    y  (  1      /    x  )  &amp;lt;  1  (  1      /    x  ) so   0  &amp;lt;  1      /    y  &amp;lt;  1      /    xIf both positive, flip.Case 2:   x  &amp;lt;  0  &amp;lt;  y then   x      /    y  &amp;lt;  0  &amp;lt;  1. Then as   x  &amp;lt;  0 we flip when we do   x      /    y  ∗  (  1      /    x  )  &amp;gt;  0  &amp;gt;  1  ∗  (  1      /    x  ) so   1      /    y  &amp;gt;  0  &amp;gt;  1      /    x so   1      /    x  &amp;lt;  0  &amp;lt;  1      /    y. Don&#039;t flip.Case 3:   x  &amp;lt;  y  &amp;lt;  0 then   x      /    y  &amp;gt;  1  &amp;gt;  0 and   1      /    y  &amp;lt;  1      /    x  &amp;lt;  0. Flip if they are the same sign.But FOR THE LOVE OF GOD!!!!!!! the reciprical of   1      /    x  −  1      /    4 is !!!!!!!NOT!!!!!!   x      /    1  −  4      /    1!!!!!!!!It is       1          1              /            x      −      1              /            4        =            4      x              4      −      x

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