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	<title>Comments on: SQL SERVER &#8211; Basic Calculation and PEMDAS Order of Operation</title>
	<atom:link href="http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/feed/" rel="self" type="application/rss+xml" />
	<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/</link>
	<description>Personal Notes of Pinal Dave</description>
	<lastBuildDate>Fri, 17 May 2013 15:26:57 +0000</lastBuildDate>
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	<item>
		<title>By: mathman</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-444989</link>
		<dc:creator><![CDATA[mathman]]></dc:creator>
		<pubDate>Tue, 26 Mar 2013 23:57:51 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-444989</guid>
		<description><![CDATA[I totally agree.
This equation is 6 over 2(2+1).
No one ever wrote a / bc when they meant ac / b 
That is because they are different expressions.
Distributive property cannot change the final value, 
and, when used, is 6 / (4+2).
One more thing, do the factoring for 6 + 3 = 9
Now factor 6/2 from 6 + 3 and we get:
(6/2)2 + (6/2)1 = 9
Notice how 6/2 MUST be in parentheses?
Now final step:
(6/2)(2+1)
The same simple idea applies to &quot;Foiling&quot; binomials:
x+1(x+2) = x + ((1)x + (1)2)
(x+1)(x+2) = (x+1)x + (x+1)2]]></description>
		<content:encoded><![CDATA[<p>I totally agree.<br />
This equation is 6 over 2(2+1).<br />
No one ever wrote a / bc when they meant ac / b<br />
That is because they are different expressions.<br />
Distributive property cannot change the final value,<br />
and, when used, is 6 / (4+2).<br />
One more thing, do the factoring for 6 + 3 = 9<br />
Now factor 6/2 from 6 + 3 and we get:<br />
(6/2)2 + (6/2)1 = 9<br />
Notice how 6/2 MUST be in parentheses?<br />
Now final step:<br />
(6/2)(2+1)<br />
The same simple idea applies to &#8220;Foiling&#8221; binomials:<br />
x+1(x+2) = x + ((1)x + (1)2)<br />
(x+1)(x+2) = (x+1)x + (x+1)2</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Don</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-444409</link>
		<dc:creator><![CDATA[Don]]></dc:creator>
		<pubDate>Tue, 26 Mar 2013 00:22:39 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-444409</guid>
		<description><![CDATA[The 2 in front of the parenthesis implies possession, doubling the parenthesis is the intent.  Otherwise an alternative symbol would have been used to imply multiplication. In which case 9 would be correct.
6/2*(1 + 2) = 9
6/2(1+2)= 1]]></description>
		<content:encoded><![CDATA[<p>The 2 in front of the parenthesis implies possession, doubling the parenthesis is the intent.  Otherwise an alternative symbol would have been used to imply multiplication. In which case 9 would be correct.<br />
6/2*(1 + 2) = 9<br />
6/2(1+2)= 1</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: amal george thomas</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-412166</link>
		<dc:creator><![CDATA[amal george thomas]]></dc:creator>
		<pubDate>Thu, 24 Jan 2013 05:51:03 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-412166</guid>
		<description><![CDATA[hey man....
6/2(1+2)=.?
first bracket 
6/2(3)
again bracket
so want to solve the bracket..
2(3)=3
so,6/6
ie,,6/6=1]]></description>
		<content:encoded><![CDATA[<p>hey man&#8230;.<br />
6/2(1+2)=.?<br />
first bracket<br />
6/2(3)<br />
again bracket<br />
so want to solve the bracket..<br />
2(3)=3<br />
so,6/6<br />
ie,,6/6=1</p>
]]></content:encoded>
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	<item>
		<title>By: Hannu Hoffrén</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-408461</link>
		<dc:creator><![CDATA[Hannu Hoffrén]]></dc:creator>
		<pubDate>Wed, 16 Jan 2013 10:42:23 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-408461</guid>
		<description><![CDATA[Great reply mathman! Cleared a lot for me why I got 1 for the answer. Thanks!]]></description>
		<content:encoded><![CDATA[<p>Great reply mathman! Cleared a lot for me why I got 1 for the answer. Thanks!</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: mathman</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-408105</link>
		<dc:creator><![CDATA[mathman]]></dc:creator>
		<pubDate>Tue, 15 Jan 2013 17:46:11 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-408105</guid>
		<description><![CDATA[Hi all. I researched this topic well for about 2 weeks now and have come to the following conclusions.  I will summarize what I have said in other forums 

with respect to the notations, then I will address other points. 



First,

if you want to say 0.5x, then you HAVE to write (1/2)x with parentheses or, x &quot;all over 2&quot; with a horiztonal fraction bar, or write x/2. I have never 

seen (1/2)x before I researched this equation, but since searching online, I HAVE seen fractional coefficients written this way, only because computers 

are limited to the horizontal typing space. 

Therefore:

x/2 = (1/2)x = 0.5x 

1/2n = 1/(2n) This sort of notation is used especially with pi, ln, or e. We have never had to say 1/(2pi). It was simply 1/2pi, or 1/2e^2. 

I have always used ab/cd to mean (ab)/(cd) and I topped almost all of my calculus classes since high school through university.(moot point, I know)

Just to re-iterate, to use 6/2 as a fraction, parentheses are REQUIRED. Every book will tell you this.



Now consider the Identity Law:

a = 1a = 1(a)

We know there is ALWAYS an &#039;invisible&#039; 1 as a ceofficient of a variable if no other number is there. Therefore:

a/a = 1, and if a is also 1a, then a/1a = 1. Blindly using &#039;pemdas&#039;, some folks would do this:

a/1a = a/1*a = a*a = a^2. I hope this drives home the silliness of this calculation.



Now, on to my second point:

consider: factoring, simplifying equations, and the distributive property.

Lets start with the number 6. 

6 = (4+2). There is a common factor here: 2. So let&#039;s factor it out of both terms.

(4+2) = 2(2+1). The outside 2 remains a part of of the 2 inner terms at all times. It cannot be used in an operation by itself without the rest of (4+2). 

The reverse of factoring is distribution, so, 2(2+1) = 6. This has to be true always. The argument I have seen to this is that (6/2) can be distributed. 

This is true ONLY is 6/2 is in parentheses, otherwise, the 6 and 2 are separated by a division slash, and the 2 is a factor of 2+1. 



So, let&#039;s prove the initial equation:

6/6 = 1

6/(4+2) = 1

6/2(2+1) = 1



the same can be done for other factors:

6/6 = 1

6/(3+3) = 1

6/3(1+1) = 1

Distribution is actually a part of &quot;Simplifying Equations&quot; and is not bound to the order of operations as &quot;multiplication&quot;, since it is in fact &quot;removing 

parentheses by distributing&quot;. This can be googled and several references found.

Simplifying 2(2+1) + 3(2+1) = 5(2+1). We &quot;combined like terms&quot; here, by adding, and did not perform the &quot;parntheses&quot; part of order of operations, nor did 

we multiply, which is also higher priority than adding, because we only simplified.



Lastly, I hear the argument that &quot;This is strictly numbers and you don&#039;t use algebra rules since there are no variables&quot;. That is the most asinine 

arguement I have heard yet. All axioms, laws, and properties use variables, meaning that they hold true for &quot;any number&quot;, hence the proofs with 

variables.

I welcome thoughts on this, in an intellectually formed response. I am tired of the &#039;flaming&#039; that goes on by imbciles on some other forums with 

rebuttals like &quot;it is 9. go back to grade 3 you moron&quot;, or &quot;google says it is 9&quot;, when google changes the equation to (6/2)*(2+1), and wolfram 

contradicts itself with 2n/2n = 1, and 6/2n = 3/n, but then says 6/2(2+1) is 9. wolframs &quot;terms&quot; state that any answer should be verified with common 

sense and accuracy should also be verified.]]></description>
		<content:encoded><![CDATA[<p>Hi all. I researched this topic well for about 2 weeks now and have come to the following conclusions.  I will summarize what I have said in other forums </p>
<p>with respect to the notations, then I will address other points. </p>
<p>First,</p>
<p>if you want to say 0.5x, then you HAVE to write (1/2)x with parentheses or, x &#8220;all over 2&#8243; with a horiztonal fraction bar, or write x/2. I have never </p>
<p>seen (1/2)x before I researched this equation, but since searching online, I HAVE seen fractional coefficients written this way, only because computers </p>
<p>are limited to the horizontal typing space. </p>
<p>Therefore:</p>
<p>x/2 = (1/2)x = 0.5x </p>
<p>1/2n = 1/(2n) This sort of notation is used especially with pi, ln, or e. We have never had to say 1/(2pi). It was simply 1/2pi, or 1/2e^2. </p>
<p>I have always used ab/cd to mean (ab)/(cd) and I topped almost all of my calculus classes since high school through university.(moot point, I know)</p>
<p>Just to re-iterate, to use 6/2 as a fraction, parentheses are REQUIRED. Every book will tell you this.</p>
<p>Now consider the Identity Law:</p>
<p>a = 1a = 1(a)</p>
<p>We know there is ALWAYS an &#8216;invisible&#8217; 1 as a ceofficient of a variable if no other number is there. Therefore:</p>
<p>a/a = 1, and if a is also 1a, then a/1a = 1. Blindly using &#8216;pemdas&#8217;, some folks would do this:</p>
<p>a/1a = a/1*a = a*a = a^2. I hope this drives home the silliness of this calculation.</p>
<p>Now, on to my second point:</p>
<p>consider: factoring, simplifying equations, and the distributive property.</p>
<p>Lets start with the number 6. </p>
<p>6 = (4+2). There is a common factor here: 2. So let&#8217;s factor it out of both terms.</p>
<p>(4+2) = 2(2+1). The outside 2 remains a part of of the 2 inner terms at all times. It cannot be used in an operation by itself without the rest of (4+2). </p>
<p>The reverse of factoring is distribution, so, 2(2+1) = 6. This has to be true always. The argument I have seen to this is that (6/2) can be distributed. </p>
<p>This is true ONLY is 6/2 is in parentheses, otherwise, the 6 and 2 are separated by a division slash, and the 2 is a factor of 2+1. </p>
<p>So, let&#8217;s prove the initial equation:</p>
<p>6/6 = 1</p>
<p>6/(4+2) = 1</p>
<p>6/2(2+1) = 1</p>
<p>the same can be done for other factors:</p>
<p>6/6 = 1</p>
<p>6/(3+3) = 1</p>
<p>6/3(1+1) = 1</p>
<p>Distribution is actually a part of &#8220;Simplifying Equations&#8221; and is not bound to the order of operations as &#8220;multiplication&#8221;, since it is in fact &#8220;removing </p>
<p>parentheses by distributing&#8221;. This can be googled and several references found.</p>
<p>Simplifying 2(2+1) + 3(2+1) = 5(2+1). We &#8220;combined like terms&#8221; here, by adding, and did not perform the &#8220;parntheses&#8221; part of order of operations, nor did </p>
<p>we multiply, which is also higher priority than adding, because we only simplified.</p>
<p>Lastly, I hear the argument that &#8220;This is strictly numbers and you don&#8217;t use algebra rules since there are no variables&#8221;. That is the most asinine </p>
<p>arguement I have heard yet. All axioms, laws, and properties use variables, meaning that they hold true for &#8220;any number&#8221;, hence the proofs with </p>
<p>variables.</p>
<p>I welcome thoughts on this, in an intellectually formed response. I am tired of the &#8216;flaming&#8217; that goes on by imbciles on some other forums with </p>
<p>rebuttals like &#8220;it is 9. go back to grade 3 you moron&#8221;, or &#8220;google says it is 9&#8243;, when google changes the equation to (6/2)*(2+1), and wolfram </p>
<p>contradicts itself with 2n/2n = 1, and 6/2n = 3/n, but then says 6/2(2+1) is 9. wolframs &#8220;terms&#8221; state that any answer should be verified with common </p>
<p>sense and accuracy should also be verified.</p>
]]></content:encoded>
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	<item>
		<title>By: hhoffren</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-354890</link>
		<dc:creator><![CDATA[hhoffren]]></dc:creator>
		<pubDate>Mon, 01 Oct 2012 13:15:52 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-354890</guid>
		<description><![CDATA[And to reply to myself, searched more of this and found out it is actually PE(MD)(AS) (left to right). Didn&#039;t even remember such a thing. For some reason my mind turns 2(2+1) more important than 2 * 3. Anyway, glad to see this conversation as now I don&#039;t need as much parenthesis to make things work for sure when scripting, I just need to remember the left-to-right rule of PEMDAS. :D]]></description>
		<content:encoded><![CDATA[<p>And to reply to myself, searched more of this and found out it is actually PE(MD)(AS) (left to right). Didn&#8217;t even remember such a thing. For some reason my mind turns 2(2+1) more important than 2 * 3. Anyway, glad to see this conversation as now I don&#8217;t need as much parenthesis to make things work for sure when scripting, I just need to remember the left-to-right rule of PEMDAS. :D</p>
]]></content:encoded>
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	<item>
		<title>By: Thomas</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-354863</link>
		<dc:creator><![CDATA[Thomas]]></dc:creator>
		<pubDate>Mon, 01 Oct 2012 11:36:21 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-354863</guid>
		<description><![CDATA[3+8-4+((5*9)/3)-((8/2)*1)-2+5 = 
  7  +  (45/3) -  (4*1)  -2+5 = 
  7  +    15   -    4    -2+5 = 21

http://www.google.com/search?q=3%2B8-4%2B5*9%2F3-8%2F2*1-2%2B5]]></description>
		<content:encoded><![CDATA[<p>3+8-4+((5*9)/3)-((8/2)*1)-2+5 =<br />
  7  +  (45/3) &#8211;  (4*1)  -2+5 =<br />
  7  +    15   &#8211;    4    -2+5 = 21</p>
<p><a href="http://www.google.com/search?q=3%2B8-4%2B5*9%2F3-8%2F2*1-2%2B5" rel="nofollow">http://www.google.com/search?q=3%2B8-4%2B5*9%2F3-8%2F2*1-2%2B5</a></p>
]]></content:encoded>
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	<item>
		<title>By: Hannu Hoffrén</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-354861</link>
		<dc:creator><![CDATA[Hannu Hoffrén]]></dc:creator>
		<pubDate>Mon, 01 Oct 2012 11:21:35 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-354861</guid>
		<description><![CDATA[I&#039;m slightly confused. PEMDAS: Parentheses [(2+1)], exponents [none], multiplication [2*(2+1)] = [2*3] = [6], division [6 / 2 * (2+1)] = [6 / 6], addition [none], subtraction [none]. 6 / 6 = 1.

Wouldn&#039;t this mean the difference between PEMDAS and BODMAS is the reason for different results? Thus PEMDAS can&#039;t be commonly known as BODMAS in India. Instead, BODMAS is used instead of PEMDAS in India?]]></description>
		<content:encoded><![CDATA[<p>I&#8217;m slightly confused. PEMDAS: Parentheses [(2+1)], exponents [none], multiplication [2*(2+1)] = [2*3] = [6], division [6 / 2 * (2+1)] = [6 / 6], addition [none], subtraction [none]. 6 / 6 = 1.</p>
<p>Wouldn&#8217;t this mean the difference between PEMDAS and BODMAS is the reason for different results? Thus PEMDAS can&#8217;t be commonly known as BODMAS in India. Instead, BODMAS is used instead of PEMDAS in India?</p>
]]></content:encoded>
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		<title>By: drenzul2</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-354853</link>
		<dc:creator><![CDATA[drenzul2]]></dc:creator>
		<pubDate>Mon, 01 Oct 2012 10:51:38 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-354853</guid>
		<description><![CDATA[Occam’s Razor applies here:
“Never assume malice when stupidity is equally applicable”

It is fairly basic stuff, but I’m not surprised a lot of people don’t remember it.]]></description>
		<content:encoded><![CDATA[<p>Occam’s Razor applies here:<br />
“Never assume malice when stupidity is equally applicable”</p>
<p>It is fairly basic stuff, but I’m not surprised a lot of people don’t remember it.</p>
]]></content:encoded>
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		<title>By: Jordan</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-353365</link>
		<dc:creator><![CDATA[Jordan]]></dc:creator>
		<pubDate>Wed, 26 Sep 2012 16:44:26 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-353365</guid>
		<description><![CDATA[You just got trolled hard, so hard that it consumed a lengthy blog post and much distress. Roughly 98% of the people that debate that are clearly trolling for the sake of rustling some jimmies. Congrats Pinal, you may have been their biggest catch with that bait.]]></description>
		<content:encoded><![CDATA[<p>You just got trolled hard, so hard that it consumed a lengthy blog post and much distress. Roughly 98% of the people that debate that are clearly trolling for the sake of rustling some jimmies. Congrats Pinal, you may have been their biggest catch with that bait.</p>
]]></content:encoded>
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		<title>By: pinaldave</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-353230</link>
		<dc:creator><![CDATA[pinaldave]]></dc:creator>
		<pubDate>Wed, 26 Sep 2012 09:42:11 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-353230</guid>
		<description><![CDATA[Great Point. It seems like Children know but many grown up do not. Please check the facebook link you will find more than 5000 comment with incorrect answer.]]></description>
		<content:encoded><![CDATA[<p>Great Point. It seems like Children know but many grown up do not. Please check the facebook link you will find more than 5000 comment with incorrect answer.</p>
]]></content:encoded>
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	<item>
		<title>By: Grzegorz Łyp</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-353229</link>
		<dc:creator><![CDATA[Grzegorz Łyp]]></dc:creator>
		<pubDate>Wed, 26 Sep 2012 09:29:33 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-353229</guid>
		<description><![CDATA[This SQL blog or basics of mathematics blog ??? Children that are 7 years old know all the things from this article.]]></description>
		<content:encoded><![CDATA[<p>This SQL blog or basics of mathematics blog ??? Children that are 7 years old know all the things from this article.</p>
]]></content:encoded>
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	<item>
		<title>By: Govarthan</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-353215</link>
		<dc:creator><![CDATA[Govarthan]]></dc:creator>
		<pubDate>Wed, 26 Sep 2012 08:25:56 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-353215</guid>
		<description><![CDATA[Could you please let me know how rule apply to resolve this ?
3+8-4+5*9/3-8/2*1-2+5 = ?]]></description>
		<content:encoded><![CDATA[<p>Could you please let me know how rule apply to resolve this ?<br />
3+8-4+5*9/3-8/2*1-2+5 = ?</p>
]]></content:encoded>
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	<item>
		<title>By: AbdulKuddus K</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-353174</link>
		<dc:creator><![CDATA[AbdulKuddus K]]></dc:creator>
		<pubDate>Wed, 26 Sep 2012 05:04:50 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-353174</guid>
		<description><![CDATA[Thanks pinal]]></description>
		<content:encoded><![CDATA[<p>Thanks pinal</p>
]]></content:encoded>
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	<item>
		<title>By: Kiran</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-352974</link>
		<dc:creator><![CDATA[Kiran]]></dc:creator>
		<pubDate>Tue, 25 Sep 2012 11:53:22 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-352974</guid>
		<description><![CDATA[Hi Pinal,

This is kiran i am new to SQL server and i need to list out all the tables used in stored procedure. collecting manually is very tedious job for such a long procedures so i am searching for alternative smart way. please suggest me how could i.

Thanks
Kiran]]></description>
		<content:encoded><![CDATA[<p>Hi Pinal,</p>
<p>This is kiran i am new to SQL server and i need to list out all the tables used in stored procedure. collecting manually is very tedious job for such a long procedures so i am searching for alternative smart way. please suggest me how could i.</p>
<p>Thanks<br />
Kiran</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: balaji</title>
		<link>http://blog.sqlauthority.com/2012/09/25/sql-server-basic-calculation-and-pemdas-order-of-operation/#comment-352871</link>
		<dc:creator><![CDATA[balaji]]></dc:creator>
		<pubDate>Tue, 25 Sep 2012 04:48:33 +0000</pubDate>
		<guid isPermaLink="false">http://blog.sqlauthority.com/?p=20891#comment-352871</guid>
		<description><![CDATA[thanks for reminding the basics!!!!]]></description>
		<content:encoded><![CDATA[<p>thanks for reminding the basics!!!!</p>
]]></content:encoded>
	</item>
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