Solve 2 x &lt; 3 over the reals.I was using the method of intervals to solve this Question. I seemed to miss 0 on the number line while using the method of intervals. Why is 0 included here in the method of intervals?

Question

Solve       2    x    &amp;lt;  3 over the reals.I was using the method of intervals to solve this Question. I seemed to miss   0 on the number line while using the method of intervals. Why is   0 included here in the method of intervals?

Madden Huber · Accepted Answer

2    x    −  3  &amp;lt;  0or            2      −      3      x        x    &amp;lt;  0  ,which gives the answer:   x  &amp;gt;      2    3   or   x  &amp;lt;  0

demitereur · Accepted Answer

First suppose   x  &amp;gt;  0. Then      2    x    &amp;lt;  3    ⟺    3  x  &amp;gt;  2    ⟺    x  &amp;gt;  2      /    3.If   x  &amp;gt;  0 and   x  &amp;gt;  2      /    3, then   x  &amp;gt;  2      /    3. Now suppose that   x  &amp;lt;  0. Now when we multiply by   x the inequality sign flips. Hence      2    x    &amp;lt;  3    ⟺    3  x  &amp;lt;  2    ⟺    x  &amp;lt;  2      /    3.If   x  &amp;lt;  0 and   x  &amp;lt;  2      /    3, then   x  &amp;lt;  0. Thus the solution set is   (  −  ∞  ,  0  )  ∪  (  2      /    3  ,  ∞  )

Solve 2/x <3 over the reals. I was using the method of intervals to solve this Question. I seemed to miss 0 on the number line while using the method of intervals. Why is 0 included here in the method of intervals?

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