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$$A:= \{ x^2 | x \in (2,3\sqrt{2}] , x \in \mathbb{R} \textrm{ and } x^2 \in \mathbb{Z}\}$$

$$B:= \{ x^3 | x \in (-2\sqrt[3]{3},3] , x \in \mathbb{R} \textrm{ and } x^3 \in \mathbb{Z}\}$$


s(A) + s(B) = ?

s(A) is number of elements of A set. For example if $\ A = \{a,b,c\} $

s(A) = 3
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I fixed the TeX for you. –  Willie Wong Mar 13 '11 at 14:04
    
Thank you Willie :) I hope question is clear, now . –  Eray Mar 13 '11 at 14:05
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1 Answer 1

up vote 4 down vote accepted

Given $x \in (2, 3\sqrt{2}]$, what can $x^2$ be? It must be

$$ 2^2 = 4 < x^2 \leq 18 = (3\sqrt{2})^2 $$

So $A$ comprises of all integers ($x^2 \in \mathbb{Z}$) bigger than 4 and less-than-or-equal-to 18. There are 14 of them.


Given $x \in (-2\sqrt[3]{3}, 3]$, you similarly find that

$$ -24 < x^3 \leq 27 $$

which tells you that there are $27 - (-24) = 51$ elements in $B$


Add them together you get $s(A) + s(B) = 65$.

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+1. It is as easy as that. Meanwhile $s(A \cup B) = 51$ and $s(A \cap B) = 14$. I don't really see why someone would want to calculate this sort of thing. –  Henry Mar 13 '11 at 14:22
    
but x must be a $\mathbb{R} $ , so x can be $\ 2\sqrt{3} $ for example. And $\ (2\sqrt{3})^2 = 12 $ and $\ 12 \in \mathbb{Z} $ –  Eray Mar 13 '11 at 14:40
    
@Henry: This is surely an exercise in understanding and manipulating basic notation in how things are defined: using $\{... \| ... \}$, implicit definitions $x^2$, and using close and open bounds to ranges. –  Mitch Mar 13 '11 at 14:47
    
@Mitch: Perhaps you are right. I would have been happier if it asked for $s(A)$ and $s(B)$ separately, but it was asking for the sum which made me dubious. –  Henry Mar 13 '11 at 15:01
    
@Eray Alakese: Willie Wong has counted 12 as an element of $A$ (and, incidentally, also of $B$). –  Henry Mar 13 '11 at 15:01
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