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	<title>Comments on: Reading Typing Declarations</title>
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		<title>By: Blaise Pascal</title>
		<link>http://www.arsmathematica.net/2012/09/22/reading-typing-declarations/#comment-69276</link>
		<dc:creator><![CDATA[Blaise Pascal]]></dc:creator>
		<pubDate>Sat, 22 Sep 2012 17:55:27 +0000</pubDate>
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		<description><![CDATA[I have a similar problem with geometric algebra (and I suspect other Clifford Algebras as well).  The geometric product of a and b is written as ab.  It is not, generally, commutative, and so can be broken down into a commutative dot product a?b and an anti-commutative wedge product a^b (so ab = a?b + a^b, a?b = b?a, a^b=-b^a).  By convention, the geometric product has a lower precedence than the dot and wedge products.  So ab^c should be read a(b^c) and a?bc should be read (a?b)c.  

That simply looks wrong to me, ab looks more tightly bound that a^b or a?b.  It&#039;s gotten in my way of reading standard identities and understanding them.]]></description>
		<content:encoded><![CDATA[<p>I have a similar problem with geometric algebra (and I suspect other Clifford Algebras as well).  The geometric product of a and b is written as ab.  It is not, generally, commutative, and so can be broken down into a commutative dot product a?b and an anti-commutative wedge product a^b (so ab = a?b + a^b, a?b = b?a, a^b=-b^a).  By convention, the geometric product has a lower precedence than the dot and wedge products.  So ab^c should be read a(b^c) and a?bc should be read (a?b)c.  </p>
<p>That simply looks wrong to me, ab looks more tightly bound that a^b or a?b.  It&#8217;s gotten in my way of reading standard identities and understanding them.</p>
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