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<h2 class="hd hd-2 unit-title">Hilbert's Hotel</h2>
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<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">Hilbert’s Hotel has infinitely many rooms, one for each natural number: \(0, 1, 2, 3, \dots,\) and every room is occupied: Guest 0 is staying in room 0, Guest 1 is staying in room 1, and so forth:</span></p>
<center><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/8e26fdbc58a90a31677438a5333a77d8/asset-v1:MITx+24.118x+2T2020+type@asset+block/HilbertHotel1.png" type="saveimage" target="[object Object]" alt="" /></span></center>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">When a <em>finite</em> hotel is completely full, there is no way of accommodating additional guests. In an <em>infinite</em> hotel, however, the fact that the hotel is full is not an impediment to accommodating extra guests. </span></p>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">Suppose, for example, that Oscar shows up without a reservation. If the hotel manager wants to accommodate him, all she needs to do is make the following announcement: “Would guest <span class="math inline">\(n\)</span> kindly relocate to room <span class="math inline">\(n+1\)</span>”. Assuming everyone abides by this request, Guest 0 will end up in room 1, Guest 1 will end up in room 2, and so forth. Under the new arrangement each of the original guests has her own room, and room 0 is available for Oscar!</span></p>
<center><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/94c98289ec715380b2dd4531fd556fc8/asset-v1:MITx+24.118x+2T2020+type@asset+block/HilbertHotel2.png" type="saveimage" target="[object Object]" alt="" /></span></center>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">“But what about the guest in the <em>last</em> room”—you might wonder—“what about the guest in the room with the <em>biggest</em> number?" </span></p>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">The answer, of course, is that there is no such thing as the last room, because there is no such thing as the biggest natural number. Every room in Hilbert’s Hotel is followed by another.</span></p>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">Now suppose 5 billion new guests arrive at our completely full but infinite hotel. We can accommodate them too! This time the manager’s announcement is this: “Would guest <span class="math inline">\(n\)</span> kindly relocate to room <span class="math inline">\(n\)</span> + 5 billion”. In a similar way, we can accommodate <em>any</em> finite number of new guests! </span></p>
<p><span style="font-family: 'book antiqua', palatino; font-size: 12pt;">We can even accommodate <em>infinitely</em> many new guests. For suppose the manager announced “if you are in room <span class="math inline">\(n\)</span>, please relocate to room <span class="math inline">\(2n\)</span>”. Assuming everyone complies, this will free infinitely many rooms while accommodating all of the hotel’s original guests:</span></p>
<center><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/6207b86c2c11ae9fc9adbfa5e436fcff/asset-v1:MITx+24.118x+2T2020+type@asset+block/HilbertHotel3.png" type="saveimage" target="[object Object]" alt="" /></span></center>
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<h3 class="hd hd-2">Video Review: Hilbert's Hotel</h3>
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<h2 class="hd hd-2 unit-title">Paradox?</h2>
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<p><span style="font-family: 'book antiqua', palatino;"><span style="font-size: 12pt;">Hilbert’s Hotel might seem to give rise to paradox. </span></span></p>
<p><span style="font-family: book antiqua, palatino;">Say that the "old guests'' are the guests that were occupying the hotel prior to any new arrivals, and that the "new guests'' are the old guests plus Oscar. We seem to be trapped between conflicting considerations: </span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">On the one hand, we want to say that there there are <em>more</em> new guests than old guests. (After all: the new guests include every old guest, plus Oscar.) </span></li>
<li><span style="font-family: book antiqua, palatino;">On the other hand, we want to say that there are <em>just as many</em> new guests as there are old guests. (After all: the new guests and the old guests can be accommodated using the exact same rooms, without any multiple occupancies or empty rooms.)</span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">You'll notice that this is the same problem I introduced in <a href="/courses/course-v1:MITx+24.118x+2T2020/jump_to_id/5b1ec17c91bb47a6bb8da6a43323c0aa" target="[object Object]">my preface to the class</a>, when I talked about Galileo's treatment of infinity. Consider the squares: \(1^2, 2^2, 3^2, 4^2, \ldots\), and their roots: \(1, 2, 3, 4, \ldots\). </span></p>
<ul>
<li><span style="font-family: book antiqua, palatino;">On the one hand, we want to say that there are <em>more</em> roots than squares. (After all: the roots include every square, and more.) </span></li>
<li><span style="font-family: book antiqua, palatino;">On the other hand, we want to say that there are <em>just as many</em> roots as there are squares. (As Galileo put it: ``every square has its own root and every root has its own square, while no square has more than one root and no root has more than one square.") </span></li>
</ul>
<p><span style="font-family: book antiqua, palatino;">Here is a more abstract way of thinking about the matter. There are two principles each of which seems eminently sensible, but they turn out to be incompatible with one another in the presence of infinite sets. The first principle is this:</span></p>
<p style="padding-left: 30px;"><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"><strong>The Proper Subset Principle</strong><br />Suppose everything in <span class="math inline">\(A\)</span> is also in <span class="math inline">\(B\)</span>, but not vice-versa. Then <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span> are <em>not</em> of the same size: <span class="math inline">\(B\)</span> has more elements than <span class="math inline">\(A\)</span>.</span></p>
<p><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"> For example, suppose that <span class="math inline">\(A\)</span> is the set of kangaroos and <span class="math inline">\(B\)</span> is the set of mammals. Since every kangaroo is a mammal but not vice-versa, the Proper Subset Principle tells us that that the set of kangaroos and the set of mammals are not of the same size: there are more mammals than kangaroos.</span></p>
<p><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;">The second principle is this:</span></p>
<p style="padding-left: 30px;"><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"><strong>The Bijection Principle</strong><br />Set <span class="math inline">\(A\)</span> has the same size as set <span class="math inline">\(B\)</span> if and only if there is a <em>bijection</em> between <span class="math inline">\(A\)</span> and <span class="math inline">\(B\)</span>.</span></p>
<p><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;">What is a bijection? Suppose you have some beetles and some boxes, and that you <em>pair</em> the beetles and the boxes by placing each beetle in a different box and leaving no empty boxes. As it might be: </span><br /><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span class="math display">\[\begin{array}{ccccc} \text{Beetle} & \text{Beetle} & \text{Beetle} & \text{Beetle} & \text{Beetle} \\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ \text{Box} &\text{Box} & \text{Box} & \text{Box} & \text{Box} \end{array}\]</span></span></span></span></p>
<p><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;">A bijection is a pairing of this kind. Accordingly, </span><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"></span><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;">Hilbert's Hotel gives us a way of defining a bijection between the set of old guests and the set of new guests: </span><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span class="math display">\[\begin{array}{ccccl} \text{Guest 0} & \text{Guest 1} & \text{Guest 2} & \text{Guest 3} & \dots\\ \downarrow & \downarrow & \downarrow & \downarrow \\ \text{Oscar} &\text{Guest 0} & \text{Guest 1} & \text{Guest 2} & \dots \end{array}\] </span></span></span><span face="book antiqua, palatino" style="font-family: 'book antiqua', palatino;">(In general, a <strong>bijection</strong> from set <span class="math inline">\(A\)</span> to set <span class="math inline">\(B\)</span> is a function from <span class="math inline">\(A\)</span> to <span class="math inline">\(B\)</span> such that <span class="math inline">\((i)\)</span> each element of <span class="math inline">\(A\)</span> is assigned to a different element of <span class="math inline">\(B\)</span>, and <span class="math inline">\((ii)\)</span> no element of <span class="math inline">\(B\)</span> is left without an assignment from <span class="math inline">\(A\)</span>.)</span></p>
<p><span style="font-family: book antiqua, palatino;">The reason the Proper Subset Principle and the Bijection Principle both seem plausible is that our intuitions about size are almost exclusively developed in the context of finite sets, and both principles are true when attention is restricted to finite sets. The lesson of Hilbert's Hotel, and of Galileo's example of the perfect squares and their roots, is that the principles cannot both be true when it comes to <em>infinite sets</em>. For whereas the Bijection Principle entails that the set of old guests and the set of new guests have the same size, the Proper Subset Principle entails that they do not.</span><br /><br /><span style="font-family: book antiqua, palatino;">As it turns out, the most fruitful way of developing a theory of infinite size is to give up on the Proper Subset Principle, and keep the Bijection Principle. The crucial insight came in 1873, when the great German mathematician Georg Cantor discovered that there are infinite sets between which there can be no bijections.</span></p>
<p><span style="font-family: book antiqua, palatino;">Cantor went on to develop a rigorous theory of infinite size based on the Bijection Principle. This work made it possible, for the first time in more than two centuries, to find our way out of the paradox that Galileo had articulated so powerfully in 1638. (David Hilbert described Cantor's work on the infinite as "the finest product of mathematical genius and one of the supreme achievements of purely intellectual human activity.'')</span><br /><br /><span style="font-family: book antiqua, palatino;">Throughout this book I will follow Cantor's strategy: I will keep the Bijection Principle and discard the Proper Subset Principle. Accordingly, when I say that set \(A\) and set \(B\) "have the same size'', or when I say that \(A\) has "just as many'' members as \(B\), what I will mean is that there is a bijection from \(A\) to \(B\).</span></p>
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<h3 class="hd hd-2">Video Review: The Paradox</h3>
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<h4>Alternative Notation</h4>
<p><span style="font-family: book antiqua,palatino;">Bijections go by many different names, and I'm afraid you'll sometimes hear me use other names in the videos associated with this course.) </span></p>
<p><span style="font-family: book antiqua,palatino;">Here are some alternative ways of speaking about a bijection from \(A\) to \(B\):<br /></span></p>
<ul>
<li><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"></span></span><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;">a one-one correspondence between \(A\) and \(B\)</span></span></span></span></li>
<li><span style="font-family: book antiqua,palatino;"></span><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;">a one-one correspondence from \(A\) to \(B\)</span></span></span></span></span></span></span></span></li>
<li><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;">a one-one mapping from</span> \(A\) onto \(B\)</span></span></span></span></span></li>
<li><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;"><span style="font-family: 'book antiqua', palatino;">a one-one function from</span> \(A\) onto \(B\)</span></span></span></span></span></span></li>
</ul>
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<h2 class="hd hd-2 unit-title">Bijections Between Infinite Sets</h2>
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<p><span style="font-family: 'book antiqua', palatino;">When the Bijection Principle is in place, many different infinite sets can be shown to have the same size. Four prominent examples are the natural numbers, the integers, the rational numbers and the finite sequences of natural numbers:</span></p>
<table>
<thead>
<tr class="header"><th align="center"><span style="font-family: 'book antiqua', palatino;"><strong>Set</strong></span></th><th align="center"><span style="font-family: 'book antiqua', palatino;"><strong>Symbol</strong></span></th><th align="center"><span style="font-family: 'book antiqua', palatino;"><strong>Members</strong></span></th></tr>
</thead>
<tbody>
<tr class="odd">
<td align="center"><span style="font-family: 'book antiqua', palatino;">Natural numbers</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\mathbb{N\ }\)</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\(0, 1, 2, 3, \dots \)</span></td>
</tr>
<tr class="even">
<td align="center"><span style="font-family: 'book antiqua', palatino;">Integers</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\mathbb{Z\ }\)</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\( \dots -2, -1, 0, 1, 2, \dots \)</span></td>
</tr>
<tr class="odd">
<td align="center"><span style="font-family: 'book antiqua', palatino;">Rational numbers</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\mathbb{Q\ }\)</span></td>
<td align="center"><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(a/b\)</span> (for <span class="math inline">\(a,b \in \mathbb{Z}\)</span> and <span class="math inline">\(b \neq 0\)</span>)</span></td>
</tr>
<tr class="even">
<td align="center"><span style="font-family: 'book antiqua', palatino;">Finite sequences of natural numbers</span></td>
<td align="center"><span class="math inline" style="font-family: 'book antiqua', palatino;">\(\mathfrak{F}\)</span></td>
<td align="center"><span style="font-family: 'book antiqua', palatino;"><span class="math inline">\(\langle n_1,\dots, n_k\rangle\)</span> for <span class="math inline">\(k \geq 0\)</span> and <span class="math inline">\(n_1,\dots,n_k \in \mathbb{N}\)</span></span></td>
</tr>
</tbody>
</table>
<p><span style="font-family: 'book antiqua', palatino;">Let us verify that the set <span class="math inline">\(\mathbb{N}\)</span> of natural numbers has the same size as the set <span class="math inline">\(\mathbb{Z}\)</span> of integers. All we need to so is define a bijection <span class="math inline">\(f\)</span> from the natural numbers to the integers. One such bijection assigns the even natural numbers to the non-negative integers and the odd natural numbers to the negative integers: </span><span style="font-family: 'book antiqua', palatino;"><span class="math display">\[\begin{array}{ccccccl} 0 & 1 & 2 & 3 & 4 & 5 &\dots\\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ 0 & -1 & 1 & -2 & 2 & -3 & \dots \end{array}\]</span> </span></p>
<p><span style="font-family: 'book antiqua', palatino;">Formally: <span class="math display">\[f(n)=\begin{cases} \frac{n}{2}, \hspace{2mm} \text{if $n$ is even}\\ -\frac{(n+1)}{2}, \hspace{2mm} \text{if $n$ is odd} \end{cases}\]</span>In the next section we will verify that the set <span class="math inline">\(\mathbb{N}\)</span> of natural numbers has the same size as the set <span class="math inline">\(\mathbb{Q}\)</span> of rational numbers. After that we’ll verify that the set <span class="math inline">\(\mathbb{N}\)</span> of natural numbers has the same size as the set <span class="math inline">\(\mathfrak{F}\)</span> of finite sequences of natural numbers.</span></p>
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<h3>Reminder</h3>
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<span style="font-family: 'book antiqua', palatino;">Which of the following, if any, is a bijection from the natural numbers to the even natural numbers?</span>
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<span class="math inline">\(f(n) = 2n\)</span>
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<span class="math inline">\(f(n) = n+1\)</span>
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<span class="math inline">\(f(n) = \frac{2(2n+1)}{3}\)</span>
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<span class="math inline">\(f(n) = 2\)</span>
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<span style="font-family: 'book antiqua', palatino;">Does there exist a bijection from the natural numbers to the integers that are multiples of 7? </span>
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<span style="font-family: 'book antiqua', palatino;">(If so, can you give an example? If not, why not?)</span>
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<span style="font-family: 'book antiqua', palatino;">Which of the following, if any, are true of bijections?</span>
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<strong> Reflexivity </strong>
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<span style="font-family: 'book antiqua', palatino;"> For any set \(A\), there is a bijection from \(A\) to \(A\). </span>
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<strong> Symmetry </strong>
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<br/>
<span style="font-family: 'book antiqua', palatino;"> For any sets \(A\) and \(B\), if there is a bijection from \(A\) to \(B\), then there is a bijection from \(B\) to \(A\). </span>
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<strong> Transitivity </strong>
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<h2 class="hd hd-2 unit-title">The Rational Numbers</h2>
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<p><span style="font-family: 'book antiqua', palatino;">Recall that the set of <strong>rational numbers</strong>, <span class="math inline">\(\mathbb{Q}\)</span>, is the set of numbers equal to some fraction <span class="math inline">\(a/b\)</span>, where <span class="math inline">\(a\)</span> and <span class="math inline">\(b\)</span> are integers and <span class="math inline">\(b\)</span> is distinct from zero. (For instance, <span class="math inline">\(17/3\)</span> is a rational number, and so is <span class="math inline">\(-4=4/-1\)</span>.) The <strong>non-negative</strong> rational numbers, <span class="math inline">\(\mathbb{Q}^{\geq 0}\)</span>, are simply the rational numbers greater than or equal to 0.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">In this section we’ll prove an astonishing result: there is a bijection between the natural numbers and the non-negative rational numbers.</span></p>
<p><span style="font-family: 'book antiqua', palatino;">In other words: <em>there are just as many natural numbers as there are non-negative rational numbers</em>. (In one of the exercises below, I’ll ask you to show that there is also a bijection between the natural numbers and the full set of rational numbers, <span class="math inline">\(\mathbb{Q}\)</span>.)</span></p>
<p><span style="font-family: 'book antiqua', palatino;">The proof is arrestingly simple. Consider the matrix below, and note that every (non-negative) rational number appears somewhere on the matrix. (In fact, they all appear multiple times, under different labels. For instance, <span class="math inline">\(\frac{1}{2}\)</span> occurs under the labels <span class="math inline">\(\frac{1}{2}\)</span>, <span class="math inline">\(\frac{2}{4}\)</span>, <span class="math inline">\(\frac{3}{6}\)</span>, etc.) </span></p>
<center>
<figure><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/fe0a0b248ca6519fee8c2bd3c32ce8f5/asset-v1:MITx+24.118x+2T2020+type@asset+block/RationalNumbers1.png" type="saveimage" target="[object Object]" alt="" height="400" width="400" /></span><figcaption><span style="font-family: 'book antiqua', palatino;">The non-negative rational numbers, arranged on an infinite matrix.</span></figcaption></figure>
</center>
<p><span style="font-family: 'book antiqua', palatino;"></span></p>
<p><span style="font-family: 'book antiqua', palatino;">Now notice that the red path traverses each cell of the matrix exactly once. We can use this observation to define a bijection from natural numbers to matrix cells: to each natural number \(n\) assign the cell at the n\(th\) step of the path.</span></p>
<center>
<figure><span style="font-family: 'book antiqua', palatino;"><img src="/assets/courseware/v1/9983ed518dda1a22b2ae11fb6e860f11/asset-v1:MITx+24.118x+2T2020+type@asset+block/RationalNumbers2.png" type="saveimage" target="[object Object]" alt="" height="400" width="400" /></span><figcaption><span style="font-family: 'book antiqua', palatino;">A path that goes through each cell of the matrix exactly once.</span></figcaption></figure>
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<p><span style="font-family: 'book antiqua', palatino;"></span></p>
<p><span style="font-family: 'book antiqua', palatino;">If each rational number appeared only once on the matrix, this would immediately yield a bijection from natural numbers to rational numbers. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">But we have seen that each rational number appears multiple times, under different labels. Fortunately, it is easy to get around the problem. We can follow the same route as before, except that this time we skip any cells corresponding to rational numbers that had already been counted. The resulting assignment is a bijection from the natural numbers to the (non-negative) rational numbers. </span></p>
<p><span style="font-family: 'book antiqua', palatino;">(I was so excited when I first learned about this result that I decided that I wanted to spend the rest of my life thinking about this sort of thing!)</span></p>
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<span style="font-family: 'book antiqua', palatino;">We've seen that there is a bijection between the natural numbers and the non-negative rational numbers.</span>
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<span style="font-family: 'book antiqua', palatino;">Is there also a bijection between the natural numbers and the full set of (negative and non-negative) rational numbers?</span>
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<span style="font-family: 'book antiqua', palatino;">(If so, can you given an example? If not, why not?)</span>
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Problem 2
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<span style="font-family: 'book antiqua', palatino;">Is there a bijection between the natural numbers and as many copies of the natural numbers as there are natural numbers?</span>
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<span style="font-family: 'book antiqua', palatino;">(If so, can you given an example? If not, why not?)</span>
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Problem 3
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<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><p>
<span style="font-family: 'book antiqua', palatino;">Is there a bijection between the natural numbers and some subset of the rational numbers that includes only rational numbers that are larger than 0 and smaller than or equal to 1? </span>
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<span style="font-family: 'book antiqua', palatino;">(If so, can you given an example? If not, why not?)</span>
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