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Let $A = \{x^2 : x \in \mathbb{N} \text{ and } 0 \leq x^2 \leq 90\}$.

Define a 1-1 onto function with domain $A$ onto a set of the form $\{1, 2, \ldots, n\}$ to show the cardinality of $A$ is $n$.


I understand that if a bijection exists between two sets, then the cardinality of the sets must be equal.

What I do not understand, is that it seems the domain $A$ has a fixed cardinality of $9$. This is the size of the set of the squares of the natural numbers that fit $0 \leq x^2 \leq 90$.

However, if we pick a value of $n \neq 9$, how can a bijection exist?

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It can't. The size of the set is $9$ (or if $0$ is a natural number, then $10$). –  mixedmath Mar 3 '13 at 23:22
    
You are just asked to find a bijection $f : A \to \{1, 2, \dots , 9\}$. Also, depending on whether you consider $0$ a natural number, your set may have $10$ elements. –  JavaMan Mar 3 '13 at 23:23
    
@JavaMan Thank you for editing my markup and for the answer. I interpreted the problem statement 'onto a set of the form $\{1, 2, ..., n\}$' to mean that for any value of n we pick, we can find a bijection. –  Zach Mar 3 '13 at 23:25

1 Answer 1

up vote 1 down vote accepted

$A = \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81\}$

In this case, you need $n = 10:$ there are 10 values $x$ including $0$ such that $0\leq x^2 \leq 90$.

So the range is $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

Try $f(a) = \sqrt{a} + 1$, where $a = x^2 \in A$.

Now just show that $f: A \to B$ is one-to-one and onto.

Also note, you can show that for any $n > 10$, any function $g: A \to B_n$ would fail to be onto $B_n$, where $B_n$ is the set $\{1, 2, 3, ... , n\}, n>10$. And show that any function $h: A \to B_n$ would fail to be injective if $B_n$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 10$.


If $A$ cannot contain $0^2 = 0$, then use the following:

Simply use $A = \{1, 4, 9, 16, 25, 36, 49, 64, 81\}$,

$B = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$, (so $n = 9$),

$f(x) = \sqrt a, \;a\in A$, for your one-to-one function $f:A\to B$,

Then show that $f$ is one-to-one and onto. In this case, you can show why, for any $n > 9,\,$ any function $\,g: A \to B_n\,$ would fail to be onto $\,B_n$, where $\,B_n\,$ is the set $\{1, 2, 3, ... , n\}, n>9$. And finally, show why any function $\,h: A \to B_n\,$ would fail to be injective if $\,B_n\,$ is the set $\{1, 2, ... n\},$ $1 \leq n\lt 9$.

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I didn't realize I had a handful of "unupvoted" answers out there! Thank you, amzoti! –  amWhy Apr 30 '13 at 1:01

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