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I had a question:

Let $A=[-1,4]$. Find $A^2$ and $\frac{1}{|A|}$.

Someone told me that the answers are $A^2=[0,16]$ and $\frac{1}{|A|}=[1,\infty]$.

However, if you notice in the first part we are considering a union between $[0,1]$ and $[0,16]$ while in the second part we consider an intersection between $[1,\infty]$ and $\left[\frac{1}{4}, \infty\right]$.

Is the person who told me wrong? Which one do we consider (union or intersection)? And if the person is right, why do we consider the union for one and intersection for another?

This is not me.
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1 Answers1

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If you define $A^2$ as $$A^2=\{a^2|a\in A\}$$ then $A^2=[0,16]$. This can easily be proven by showing two things:

  1. $A^2\subseteq [0,16]$
  2. $[0,16]\subseteq A^2$.

Similarly, if you define $\frac{1}{|A|}$ as $$\frac{1}{|A|}=\left\{\frac{1}{|a|}|a\in A\right\}$$

then $\frac{1}{|A|}=[\frac14,\infty)$. Again, this can easily be shown by showing that

  1. $\frac{1}{|A|}\subseteq\left[\frac14,\infty\right)$
  2. $\left[\frac14,\infty\right)\subseteq \frac{1}{|A|}$
5xum
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