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<channel>
	<title>Godly Musings &#187; Glee!</title>
	<atom:link href="http://memethief.com/category/glee/feed/" rel="self" type="application/rss+xml" />
	<link>http://memethief.com</link>
	<description>Being the thoughts, musings, aggravations and sarcastic whinings of the local god</description>
	<lastBuildDate>Tue, 22 Dec 2009 05:43:55 +0000</lastBuildDate>
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			<item>
		<title>Godless gets some press</title>
		<link>http://memethief.com/2009/08/13/godless-gets-some-press/</link>
		<comments>http://memethief.com/2009/08/13/godless-gets-some-press/#comments</comments>
		<pubDate>Thu, 13 Aug 2009 17:06:26 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Announcement]]></category>
		<category><![CDATA[Atheism]]></category>
		<category><![CDATA[Cool Linkage]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/?p=489</guid>
		<description><![CDATA[As noted by, among others, past Godless Radio guest Larry Moran, the U of T eBulletin has done a writeup on Godless. Hurrah!
]]></description>
			<content:encoded><![CDATA[<p>As noted by, among others, past <a href="http://godlessradio.ca">Godless Radio</a> guest <a href="http://sandwalk.blogspot.com/2009/08/atheist-radio-at-university-of-toronto.html">Larry Moran</a>, the U of T eBulletin has done a <a href="https://webapps.utsc.utoronto.ca/ose/story.php?id=1682&#038;sectid=1">writeup on Godless</a>. Hurrah!</p>
]]></content:encoded>
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		<title>Why Number Theory is Cool</title>
		<link>http://memethief.com/2009/01/11/why-number-theory-is-cool/</link>
		<comments>http://memethief.com/2009/01/11/why-number-theory-is-cool/#comments</comments>
		<pubDate>Sun, 11 Jan 2009 05:03:09 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Glee!]]></category>
		<category><![CDATA[Math]]></category>

		<guid isPermaLink="false">http://memethief.com/?p=390</guid>
		<description><![CDATA[I&#8217;ve started doing an intro to number theory, and one of the cool things about it is that it examines some simple questions with complicated answers. Here&#8217;s a quick example:

Show that  is even, for any given integer n.

And here&#8217;s the proof: we can easily see that . Since either  or  must be [...]]]></description>
			<content:encoded><![CDATA[<p>I&#8217;ve started doing an intro to number theory, and one of the cool things about it is that it examines some simple questions with complicated answers. Here&#8217;s a quick example:</p>
<blockquote><p>
Show that <img src="http://memethief.com/wp-content/cache/tex_b759c74cbace3266576305032324ed0b.png" align="middle" class="tex" alt="n^2 - n" /> is even, for any given integer <i>n</i>.
</p></blockquote>
<p>And here&#8217;s the proof: we can easily see that <img src="http://memethief.com/wp-content/cache/tex_ac5e0ccf173b8c953d0a360bdf8047d9.png" align="middle" class="tex" alt="n^2 - n = n(n-1)" />. Since either <img src="http://memethief.com/wp-content/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.png" align="middle" class="tex" alt="n" /> or <img src="http://memethief.com/wp-content/cache/tex_a438673491daae8148eae77373b6a467.png" align="middle" class="tex" alt="n-1" /> must be even, and the product of an even number with any other number is even, their product must also be even.</p>
<p>This sort of problem has a property similar to many geometry problems, which is that the original statement is easily comprehensible to people without a mathematical background. In this case, the solution also is widely accessible. There are some more complex proofs, though, and I&#8217;ve put one below the fold if you&#8217;re interested.<br />
<span id="more-390"></span></p>
<blockquote><p>It is impossible to pick two integers a and b greater than two such that <img src="http://memethief.com/wp-content/cache/tex_ac8a3b4b80db2731f61364ace982f1e7.png" align="middle" class="tex" alt="2^b - 1" /> divides evenly into <img src="http://memethief.com/wp-content/cache/tex_dbe6b338d550355f611a245e7fa80d33.png" align="middle" class="tex" alt="2^a + 1" />.</p></blockquote>
<p>This is what I mean: I think this is a statement that is easily understood by anyone who remembers any 8th-grade math. Finding a proof of this, though, is trickier. We use an iterative approach, manipulating the expression until we get results.</p>
<p>First of all, for the sake of brevity, I&#8217;m going to use a bit of shorthand. When I say <img src="http://memethief.com/wp-content/cache/tex_8ee0942e8c69bc44598c3c6dbb1bd409.png" align="middle" class="tex" alt="x|y" /> what I mean is &#8220;x divides evenly into y&#8221;. This is the same as saying <img src="http://memethief.com/wp-content/cache/tex_d2f4bc2dac8484d205afd4eafdff6741.png" align="middle" class="tex" alt="y = px" /> for some integer value of p. The opposite of this is <img src="http://memethief.com/wp-content/cache/tex_712b7c04496659e6f5390f8be7de0f79.png" align="middle" class="tex" alt="x\not|y" />. Now the proof itself.</p>
<p><em>Proof.</em> Let a and b be any (not necessarily different) integers greater than 2.</p>
<p>Suppose for the time being that <img src="http://memethief.com/wp-content/cache/tex_c069c00c60591e85e2b9bcf3bcd0cf5a.png" align="middle" class="tex" alt="2^b - 1|2^a + 1" />. We are going to accept this assumption and show that it leads to a contradiction. This will prove that the assumption is false.</p>
<p>Now,<br />
<center><img src="http://memethief.com/wp-content/cache/tex_d65d6d973e05337fd3692817cbd7e845.png" align="middle" class="tex" alt="\begin{align}2^a+1 &#038; = (2^{a-b} \times 2^{b}) + 1 \\ &#038;= (2^{a-b} \times 2^{b} - 2^{a-b} + 2^{a-b}) + 1 \\ &#038;= 2^{a-b}(2^{b} - 1 + 1) + 1 \\ &#038;= 2^{a-b}(2^{b} - 1) + (2^{a-b} + 1)\end{align}" /></center></p>
<p>Since we&#8217;ve assumed that <img src="http://memethief.com/wp-content/cache/tex_ac8a3b4b80db2731f61364ace982f1e7.png" align="middle" class="tex" alt="2^b - 1" /> divides evenly into <img src="http://memethief.com/wp-content/cache/tex_dbe6b338d550355f611a245e7fa80d33.png" align="middle" class="tex" alt="2^a + 1" />, we must be able to subtract any multiple of the first from the second, and the result will still be divisible by the first (if this is not obvious, try it with some integers and you&#8217;ll see why it works). So we can subtract <img src="http://memethief.com/wp-content/cache/tex_a789a40d670f81515bde4b057c8ea03f.png" align="middle" class="tex" alt="2^{a-b}(2^{b} - 1)" /> from the above expression, and it should still be divisible by <img src="http://memethief.com/wp-content/cache/tex_ac8a3b4b80db2731f61364ace982f1e7.png" align="middle" class="tex" alt="2^b - 1" />.</p>
<p>Here, we&#8217;ve shown that if <img src="http://memethief.com/wp-content/cache/tex_c069c00c60591e85e2b9bcf3bcd0cf5a.png" align="middle" class="tex" alt="2^b - 1|2^a + 1" /> then <img src="http://memethief.com/wp-content/cache/tex_c9aaa0ab7aa925b51cf70350a55b69a6.png" align="middle" class="tex" alt="2^b - 1|2^{a-b} + 1" />.</p>
<p>Now, we have three possible cases:</p>
<ol>
<li><img src="http://memethief.com/wp-content/cache/tex_45f46d6a35751f1f0868e7d5321578ad.png" align="middle" class="tex" alt="a-b=b" /><br />
In this case, we have <img src="http://memethief.com/wp-content/cache/tex_1923a1dccbbc6354b3121a7a225563e3.png" align="middle" class="tex" alt="2^b - 1|2^{a-b} + 1 = 2^b + 1" />. We can do the same thing as above, where we subtract a multiple of the first number, to get:<br />
<center><img src="http://memethief.com/wp-content/cache/tex_44134e1e833e565f2e9e6e252130f682.png" align="middle" class="tex" alt="2^b - 1|(2^b + 1) - (2^b - 1)" /></center><br />
<center><img src="http://memethief.com/wp-content/cache/tex_dd8c3f942163e77a0af7ec50cdfb04b4.png" align="middle" class="tex" alt="2^b - 1|2" /></center><br />
This means that <img src="http://memethief.com/wp-content/cache/tex_ac8a3b4b80db2731f61364ace982f1e7.png" align="middle" class="tex" alt="2^b - 1" /> is a factor of 2, namely either 1 or 2. Thus <img src="http://memethief.com/wp-content/cache/tex_5c7f956d705fab11713e6e4090970315.png" align="middle" class="tex" alt="2^b = 2" /> or <img src="http://memethief.com/wp-content/cache/tex_4c7cafdc5d167f21cde14699b5525b37.png" align="middle" class="tex" alt="2^b = 3" />. 3 is impossible if <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" /> is an integer, and 2 is only possible if <img src="http://memethief.com/wp-content/cache/tex_3c94d884933477acdc14fc70da4b987a.png" align="middle" class="tex" alt="b=1" />, but we know that <img src="http://memethief.com/wp-content/cache/tex_87810e708b0402312962bb334f1e7c82.png" align="middle" class="tex" alt=" b \ge 3 " /> so this is impossible.
</li>
<li><img src="http://memethief.com/wp-content/cache/tex_45a74df50b6dfe185b7fcea88a48e21a.png" align="middle" class="tex" alt="a-b < b" /><br />
Since <img src="http://memethief.com/wp-content/cache/tex_c9aaa0ab7aa925b51cf70350a55b69a6.png" align="middle" class="tex" alt="2^b - 1|2^{a-b} + 1" />, the second term must be equal to or greater than the first. We can&#8217;t have <img src="http://memethief.com/wp-content/cache/tex_8ee0942e8c69bc44598c3c6dbb1bd409.png" align="middle" class="tex" alt="x|y" /> if <img src="http://memethief.com/wp-content/cache/tex_0868cac5772a0bddd80fcea379a6271a.png" align="middle" class="tex" alt="y < x" />. So this gives us:<br />
<center><img src="http://memethief.com/wp-content/cache/tex_43d533f96106415227fcd22b674a6bd9.png" align="middle" class="tex" alt="2^b - 1 \le 2^{a-b} + 1" /></center><br />
<center><img src="http://memethief.com/wp-content/cache/tex_61ef314036391b7e408a69486356f91b.png" align="middle" class="tex" alt="2^b - 2^{a-b} \le 2" /></center><br />
Now since <img src="http://memethief.com/wp-content/cache/tex_5078125398a9403f7fbec82a62fb99d2.png" align="middle" class="tex" alt="2^b \ge 8" />, this means <img src="http://memethief.com/wp-content/cache/tex_ce5be109612bc1b874b25f904f44956d.png" align="middle" class="tex" alt="2^{a-b} \ge 6" /> and therefore <img src="http://memethief.com/wp-content/cache/tex_79a854abc854f822fb43a42e5eeeaca1.png" align="middle" class="tex" alt="a-b &gt; 2" />. But we divide the last line by two to get:<br />
<center><img src="http://memethief.com/wp-content/cache/tex_4216c2d165fedb3ae546d77b8eead69a.png" align="middle" class="tex" alt="2^{b-1} - 2^{a-b-1} \le 1" /></center><br />
Note that this difference cannot be zero, since the exponents are not equal. If both <img src="http://memethief.com/wp-content/cache/tex_b1d10db2016c2f83c13b25fcb170cdeb.png" align="middle" class="tex" alt="b-1" /> and <img src="http://memethief.com/wp-content/cache/tex_82b6bc614c33addcc09aff5d7fd379a1.png" align="middle" class="tex" alt="a-b-1" /> are greater than 0 (and not equal), then their difference will be a multiple of 2. But it clearly is not. So one of the two must be less than or equal to 0. Which means either <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" /> or <img src="http://memethief.com/wp-content/cache/tex_8ca2ed590cf2ea2404f2e67641bcdf50.png" align="middle" class="tex" alt="a-b" /> is less than or equal to 1. It can&#8217;t be <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" />, because we know it&#8217;s greater than 2. And it can&#8217;t be <img src="http://memethief.com/wp-content/cache/tex_8ca2ed590cf2ea2404f2e67641bcdf50.png" align="middle" class="tex" alt="a-b" /> because we already determined that it&#8217;s greater than 2. So we have another impossibility.
</li>
<li><img src="http://memethief.com/wp-content/cache/tex_8adac32a2ae6bca0b913101ed344976e.png" align="middle" class="tex" alt="a-b &gt; b" /><br />
This is the only possibility left, therefore it must be true (if our initial assumption is true).
</li>
</ol>
<p>Now we have <img src="http://memethief.com/wp-content/cache/tex_c9aaa0ab7aa925b51cf70350a55b69a6.png" align="middle" class="tex" alt="2^b - 1|2^{a-b} + 1" />, and we know that <img src="http://memethief.com/wp-content/cache/tex_79a854abc854f822fb43a42e5eeeaca1.png" align="middle" class="tex" alt="a-b &gt; 2" />, since it&#8217;s greater than <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" />. Thus, we can plug <img src="http://memethief.com/wp-content/cache/tex_8ca2ed590cf2ea2404f2e67641bcdf50.png" align="middle" class="tex" alt="a-b" /> into this whole process, in place of <img src="http://memethief.com/wp-content/cache/tex_0cc175b9c0f1b6a831c399e269772661.png" align="middle" class="tex" alt="a" />.</p>
<p>Note that <img src="http://memethief.com/wp-content/cache/tex_8ca2ed590cf2ea2404f2e67641bcdf50.png" align="middle" class="tex" alt="a-b" /> is smaller than <img src="http://memethief.com/wp-content/cache/tex_0cc175b9c0f1b6a831c399e269772661.png" align="middle" class="tex" alt="a" /> since <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" /> is positive. So when we wind up at this same step, with a result of <img src="http://memethief.com/wp-content/cache/tex_9be583d6e683444f65ae016a941f876c.png" align="middle" class="tex" alt="(a-b) - b = a-2b &gt; b" />, we&#8217;re dealing with an even smaller number. We keep going, and eventually we get a number <img src="http://memethief.com/wp-content/cache/tex_74a3f0545b0dd1be9291283b33cdcd00.png" align="middle" class="tex" alt="a-nb" /> that is smaller than or equal to <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" /> (I&#8217;m not going to prove that this is inevitable, but you should be able to convince yourself of this). This leads us to possibility 1 or 2, above, which is a contradiction.</p>
<p>Now the contradiction cascades back through the proof. Since a contradiction is generated in one of those three options, the conclusion that <img src="http://memethief.com/wp-content/cache/tex_83855b3d9cd8029b9fea8a7caa440c1c.png" align="middle" class="tex" alt="a-nb \le b" /> must be false, therefore it cannot be the case that <img src="http://memethief.com/wp-content/cache/tex_bb184a42cbd072098bbef744802f2d69.png" align="middle" class="tex" alt="2^b - 1 | 2^{a-nb} + 1" />. Thus it cannot be true that <img src="http://memethief.com/wp-content/cache/tex_5ec0e8e7edbc70f249b131c4de3e6d5c.png" align="middle" class="tex" alt="2^b - 1 | 2^{a-(n-1)b} + 1" /> and so on, back through the iterations of the proof. Thus it cannot be true that <img src="http://memethief.com/wp-content/cache/tex_c76253f5dc5080b3cef3c0eebe62c6f0.png" align="middle" class="tex" alt="2^b - 1 | 2^{a-b} + 1" />, therefore it cannot be true that <img src="http://memethief.com/wp-content/cache/tex_e3be2745299ef58f88c91841f6767b64.png" align="middle" class="tex" alt="2^b - 1 | 2^a + 1" />. This was our initial assumption, therefore we have proved the opposite.</p>
<p>Now, the final step is to generalize the whole thing. Sure, this may work for some choice of <img src="http://memethief.com/wp-content/cache/tex_0cc175b9c0f1b6a831c399e269772661.png" align="middle" class="tex" alt="a" /> and <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" />, but is this always true? Well, the proof does not depend on the values of the variables, except for the restriction that they both be greater than 2. So regardless of what values you put in, the original statement (in blockquote) holds. <img src="http://memethief.com/wp-content/cache/tex_77e61462d1adabba326f954b07b72122.png" align="middle" class="tex" alt="\blacksquare" /></p>
<p>One final thing: why do we require that <img src="http://memethief.com/wp-content/cache/tex_0cc175b9c0f1b6a831c399e269772661.png" align="middle" class="tex" alt="a" /> and <img src="http://memethief.com/wp-content/cache/tex_92eb5ffee6ae2fec3ad71c777531578f.png" align="middle" class="tex" alt="b" /> be greater than 2? Because you can find counterexamples otherwise. For instance, if <img src="http://memethief.com/wp-content/cache/tex_d2efd682355d6172acea54879367bcdb.png" align="middle" class="tex" alt=" a = 3, b = 2" />, then the statement is false. And if <img src="http://memethief.com/wp-content/cache/tex_d8a35c47c0f29d1f868a2769cd3003c5.png" align="middle" class="tex" alt="b = 1" /> then the statement is always false.</p>
<p>Anyway, I hope I haven&#8217;t bored you too much. I find this sort of work very interesting, and this is just the basic groundwork of the field of Number Theory. Some of the more complicated theorems to do with, for instance, prime numbers, are very deep. Hopefully I&#8217;ll succeed in getting to work on this as a research project over the summer, and I can write more about it.</p>
]]></content:encoded>
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		<title>William Gibson weighs in</title>
		<link>http://memethief.com/2008/10/29/william-gibson-weighs-in/</link>
		<comments>http://memethief.com/2008/10/29/william-gibson-weighs-in/#comments</comments>
		<pubDate>Wed, 29 Oct 2008 13:32:00 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Cool Linkage]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/?p=380</guid>
		<description><![CDATA[Over on William Gibson&#8217;s blag, a discussion of undecided voters, and an apt analogy. If you think WG is as awesome as I do, go over and check it out.
]]></description>
			<content:encoded><![CDATA[<p>Over on <a href="http://www.williamgibsonbooks.com/blog/">William Gibson&#8217;s blag</a>, a <a href="http://www.williamgibsonbooks.com/blog/2008_10_01_archive.asp#5668695234493185636">discussion of undecided voters</a>, and an apt analogy. If you think WG is as awesome as I do, go over and check it out.</p>
]]></content:encoded>
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		<item>
		<title>My birthday&#8217;s comin up!</title>
		<link>http://memethief.com/2008/10/15/my-birthdays-comin-up/</link>
		<comments>http://memethief.com/2008/10/15/my-birthdays-comin-up/#comments</comments>
		<pubDate>Wed, 15 Oct 2008 13:44:39 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Announcement]]></category>
		<category><![CDATA[Glee!]]></category>
		<category><![CDATA[Oh, the Angst!]]></category>
		<category><![CDATA[birthday]]></category>
		<category><![CDATA[stuff]]></category>

		<guid isPermaLink="false">http://memethief.com/?p=374</guid>
		<description><![CDATA[&#8230; hint, hint.
Also, I&#8217;m running out of trustworthiness. So if you need to rely on me for something you have about a month left to do it.
]]></description>
			<content:encoded><![CDATA[<p>&#8230; <a href="http://www.questionablecontent.net/merch.php">hint</a>, <a href="http://www.thinkgeek.com/">hint</a>.</p>
<p>Also, I&#8217;m <a href="http://www.bartleby.com/73/1828.html">running out of trustworthiness</a>. So if you need to rely on me for something you have about a month left to do it.</p>
]]></content:encoded>
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		<item>
		<title>PHP-Generated Comic Test</title>
		<link>http://memethief.com/2008/04/25/php-generated-comic-test/</link>
		<comments>http://memethief.com/2008/04/25/php-generated-comic-test/#comments</comments>
		<pubDate>Fri, 25 Apr 2008 01:01:12 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Comics]]></category>
		<category><![CDATA[Glee!]]></category>
		<category><![CDATA[Ooh... pretty...]]></category>

		<guid isPermaLink="false">http://memethief.com/2008/04/25/php-generated-comic-test/</guid>
		<description><![CDATA[I&#8217;ve been toying with this idea for a while, and hopefully this summer I&#8217;ll get to work on it more. The idea is that I create a database-driven comic strip. All the characters, plots and so on are stored in a MySQL database, and get interpreted by PHP and turned into SVG.
The amusing thing about [...]]]></description>
			<content:encoded><![CDATA[<p>I&#8217;ve been toying with <a href="http://fsckme.com/comic">this idea</a> for a while, and hopefully this summer I&#8217;ll get to work on it more. The idea is that I create a database-driven comic strip. All the characters, plots and so on are stored in a MySQL database, and get interpreted by PHP and turned into <a href="http://www.w3.org/TR/SVG11/" title="Scalable Vector Graphics">SVG</a>.</p>
<p>The amusing thing about this is that, as I work on it and tweak it, the art will retroactively get better. For instance, if a particular character&#8217;s template is badly drawn, and I update it as I create new strips, the old strips will automagically get an art update as well :)</p>
<p>Anyway, I&#8217;ve been having fun so far, working on this for an hour at a time, once every couple of months over the past year. Now I&#8217;ve got a framework that looks like it&#8217;ll speed up development immensely. So long as I create a database schema that conveys all the information I need it to.</p>
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		<title>Gained in Translation</title>
		<link>http://memethief.com/2008/03/16/gained-in-translation/</link>
		<comments>http://memethief.com/2008/03/16/gained-in-translation/#comments</comments>
		<pubDate>Sun, 16 Mar 2008 02:16:57 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Glee!]]></category>
		<category><![CDATA[Hmmm]]></category>

		<guid isPermaLink="false">http://memethief.com/2008/03/16/gained-in-translation/</guid>
		<description><![CDATA[Last night I was given a fortune cookie. The English side was a little harsh:
&#8220;Enjoy your own company. If you don&#8217;t, who will?&#8221;
But the French side was sexay:
&#8220;Il faut savoir se plaire &#224; soi-m&#234;me. Sinon, qui d&#8217;autre peut le faire?&#8221;
]]></description>
			<content:encoded><![CDATA[<p>Last night I was given a fortune cookie. The English side was a little harsh:</p>
<blockquote><p>&#8220;Enjoy your own company. If you don&#8217;t, who will?&#8221;</p></blockquote>
<p>But the French side was sexay:</p>
<blockquote><p>&#8220;Il faut savoir se plaire &agrave; soi-m&ecirc;me. Sinon, qui d&#8217;autre peut le faire?&#8221;</p></blockquote>
]]></content:encoded>
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		<title>Nerdgasm!</title>
		<link>http://memethief.com/2008/02/12/nerdgasm/</link>
		<comments>http://memethief.com/2008/02/12/nerdgasm/#comments</comments>
		<pubDate>Tue, 12 Feb 2008 04:02:57 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Cool Linkage]]></category>
		<category><![CDATA[Gaming]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/2008/02/12/nerdgasm/</guid>
		<description><![CDATA[OMG, this sounds awesome: Shadowrun mashed up with Don&#8217;t Rest Your Head. Fred Hicks calls it Don&#8217;t Sleep on the Run. drool.
]]></description>
			<content:encoded><![CDATA[<p>OMG, this sounds awesome: <a href="http://canonpuncture.wordpress.com/2008/02/10/dont-rest-your-head-meets-shadowrun/">Shadowrun mashed up with Don&#8217;t Rest Your Head</a>. <a href="http://drivingblind.livejournal.com">Fred Hicks</a> calls it <a href="http://drivingblind.livejournal.com/329222.html">Don&#8217;t Sleep on the Run</a>. drool.</p>
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		<title>New Horizons</title>
		<link>http://memethief.com/2007/11/25/new-horizons/</link>
		<comments>http://memethief.com/2007/11/25/new-horizons/#comments</comments>
		<pubDate>Sun, 25 Nov 2007 14:48:25 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Announcement]]></category>
		<category><![CDATA[Cool Linkage]]></category>
		<category><![CDATA[Feminism]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/2007/11/25/new-horizons/</guid>
		<description><![CDATA[Hey, is anyone as excited about New Horizons as I am? It&#8217;s a supplement for Spirit of the Century, and aims &#8220;to show some of the real failings of the pulp era when it comes to fairness and justice in order to provide rich and vibrant new possibilities for adventure roleplaying in a bygone era.&#8221; [...]]]></description>
			<content:encoded><![CDATA[<p>Hey, is anyone as excited about <a href="http://bruceb.livejournal.com/240455.html" title="New Horizons: a game about gender, race, sexuality and politics in the 1920s pulps">New Horizons</a> as I am? It&#8217;s a supplement for Spirit of the Century, and aims &#8220;to show some of the real failings of the pulp era when it comes to fairness and justice in order to provide rich and vibrant new possibilities for adventure roleplaying in a bygone era.&#8221; </p>
<p>Until recently it seemed like a cool project that wasn&#8217;t really going anywhere, but the author has been writing a bunch of production notes on <a href="http://bruceb.livejournal.com/tag/new+horizons" >his blag</a>, and the Evil Hat folks are on board too. This look like it&#8217;s going to not only be a real, live product but also be the sort of professional, good-looking product we have learned to expect from Evil Hat Productions.</p>
<p>Check this summary out:</p>
<blockquote><p>
She is the strongest human being alive, her muscles super-charged by her own scientific processes. She’s fought dinosaurs barehanded and lived to tell the tale. But she can’t join any professional society for engineers, or even hold the patents for her inventions in her own name.</p>
<p>He is a supernaturally good poet. He can smell truth and lies from across the street. He’s saved the life of one president, two prime ministers, and a future pope. But if he goes out for a gourmet meal with friends, managers will insist he go in through the servants’ entrance.</p>
<p>Two men share a mystical union, pooling their health, knowledge, and magical essence. They bind demons and champion the falsely accused in courts on three continents. But if they ever once acknowledge the love they share along with their power, they’ll be disbarred and shunned by decent people everywhere.</p>
<p>The band of five fought in two wars for liberty, first against invading armies and then against tyrants at home. They free serfs, fight the architects of murder, and have twice stopped mad schemes of genocide. But they’re communists, and can&#8217;t even get visas to visit other heroes and scholars in the US.</p>
<p>Brother and sister are heirs to a millennia-old family tradition of serving justice and knowledge. Their ancestors commanded armies, delved into ancient tombs to lay ghosts—and worse things—to rest, taught the founders of new schools of philosophy and military strategy. But in the New World, he&#8217;s barely tolerated as a ditch digger—and she&#8217;ll be deported if she teaches English to other immigrants.</p>
<p>These are the other heroes, the ones who must fight for their dignity and liberty just as fiercely as they take on the challenges all pulp heroes face.</p>
<p>New Horizons is a new supplement for Spirit of the Century. Each chapter addresses a marginalized group from the pulps, kept outside by their sex, their race, their lifestyle, or their beliefs. In New Horizons you’ll find information about real-life heroic individuals and teams, the challenges they face and some of the solutions they find to the problems of dealing with 1920s society. You’ll also find heroes and villains ready for use, plot hooks, and ties to the mysteries around the Century Club. The life of heroes outside the mainstream may seem as strange as the secret language of Atlantis, but can be as exciting and powerful in play as a zeppelin armada.
</p></blockquote>
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		<title>OMG Joss Whedon is Back!</title>
		<link>http://memethief.com/2007/11/05/omg-joss-whedon-is-back/</link>
		<comments>http://memethief.com/2007/11/05/omg-joss-whedon-is-back/#comments</comments>
		<pubDate>Mon, 05 Nov 2007 00:10:49 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Announcement]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/2007/11/05/omg-joss-whedon-is-back/</guid>
		<description><![CDATA[Via Slashdot, it seems that Joss Whedon is slated to do another TV show! It stars Eliza Dushku, and is called &#8220;Dollhouse&#8220;. It sounds pretty awesome. Fucken sw33t!
]]></description>
			<content:encoded><![CDATA[<p>Via <a href="http://slashdot.org/article.pl?sid=07/11/02/216218">Slashdot</a>, it seems that Joss Whedon is slated to do another TV show! It stars <acronym title="Faith">Eliza Dushku</acronym>, and is called &#8220;<a href="http://www.variety.com/article/VR1117975136.html?categoryid=14&#038;cs=1">Dollhouse</a>&#8220;. It sounds pretty awesome. Fucken sw33t!</p>
]]></content:encoded>
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		<title>JJ Abrams + TNG = ???</title>
		<link>http://memethief.com/2007/10/27/jj-abrams-tng/</link>
		<comments>http://memethief.com/2007/10/27/jj-abrams-tng/#comments</comments>
		<pubDate>Sat, 27 Oct 2007 13:56:21 +0000</pubDate>
		<dc:creator>god</dc:creator>
				<category><![CDATA[Cool Linkage]]></category>
		<category><![CDATA[Glee!]]></category>

		<guid isPermaLink="false">http://memethief.com/2007/10/27/jj-abrams-tng/</guid>
		<description><![CDATA[I&#8217;m a little behind the curve here, but via Ella comes a frackin hilarious TNG parody, starring our very own Hundred-Watt Warlock. See this, even if it means delaying putting out that fire in your kitchen for a few minutes.
]]></description>
			<content:encoded><![CDATA[<p>I&#8217;m a little behind the curve here, but via <a href="http://cherriescabaret.com/wordpress/2007/10/17/epic-lulz-the-next-generation/">Ella</a> comes a frackin hilarious <a href="http://www.galacticast.com/2007/10/15/starfleet-academy/">TNG parody</a>, starring our very own <a href="http://inoveryourhead.net/">Hundred-Watt Warlock</a>. See this, even if it means delaying putting out that fire in your kitchen for a few minutes.</p>
]]></content:encoded>
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