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	<title>Comments on: Once more on Devlin (but not on teaching)</title>
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	<link>http://jd2718.wordpress.com/2008/08/19/once-more-on-devlin-but-not-on-teaching/</link>
	<description>Education, Math, Teaching, New York, Bronx, Union, Language, Travel</description>
	<lastBuildDate>Mon, 16 Nov 2009 16:03:43 +0000</lastBuildDate>
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		<title>By: Joshua Fisher</title>
		<link>http://jd2718.wordpress.com/2008/08/19/once-more-on-devlin-but-not-on-teaching/#comment-39858</link>
		<dc:creator>Joshua Fisher</dc:creator>
		<pubDate>Mon, 08 Dec 2008 05:36:18 +0000</pubDate>
		<guid isPermaLink="false">http://jd2718.wordpress.com/?p=1190#comment-39858</guid>
		<description>Nope.

Still wrong.

Merry Christmas!</description>
		<content:encoded><![CDATA[<p>Nope.</p>
<p>Still wrong.</p>
<p>Merry Christmas!</p>
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	<item>
		<title>By: Joe Niederberger</title>
		<link>http://jd2718.wordpress.com/2008/08/19/once-more-on-devlin-but-not-on-teaching/#comment-39146</link>
		<dc:creator>Joe Niederberger</dc:creator>
		<pubDate>Thu, 28 Aug 2008 19:58:26 +0000</pubDate>
		<guid isPermaLink="false">http://jd2718.wordpress.com/?p=1190#comment-39146</guid>
		<description>If you don&#039;t question his math, its because you haven&#039;t tried to pin him down and get his (yes, very stubborn and) absurd replies.

When asked point blank whether, restricted to integers only, and under the condition that one can say that function F *is* G iff. for every integer x, F(x) = G(x), and we give F as the usual (if informal) &quot;repeated-addition&quot; definition, and &quot;G&quot; is his ethereally defined &quot;multiplication&quot; - well, under those conditions, can one not say that &quot;multiplication&quot; *is* &quot;repeated-addition&quot;. Well, guess what. Devlin says - no, absolutely false.

Absurd.</description>
		<content:encoded><![CDATA[<p>If you don&#8217;t question his math, its because you haven&#8217;t tried to pin him down and get his (yes, very stubborn and) absurd replies.</p>
<p>When asked point blank whether, restricted to integers only, and under the condition that one can say that function F *is* G iff. for every integer x, F(x) = G(x), and we give F as the usual (if informal) &#8220;repeated-addition&#8221; definition, and &#8220;G&#8221; is his ethereally defined &#8220;multiplication&#8221; &#8211; well, under those conditions, can one not say that &#8220;multiplication&#8221; *is* &#8220;repeated-addition&#8221;. Well, guess what. Devlin says &#8211; no, absolutely false.</p>
<p>Absurd.</p>
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