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Maybe a stupid question but should we prove if a=b then (a+c)=(b+c) if we want to use this property? Can't find where it is given in the book.
 
Maybe a stupid question but should we prove if a=b then (a+c)=(b+c) if we want to use this property? Can't find where it is given in the book.
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Dr. Alekseenko: Apparently, this equation a consequence of
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<math> (a+c)-(b+c) = </math>
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use associative law
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<math> a + (c - b) + c =</math>
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use commutative law
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<math> a + (b + (-c)) + c =</math>
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next associative law, again,
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<math> a + b + ((-c) + c) =</math>
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finally use axiom of zero, i.e.,  <math> (-c)+c =0 </math> we obtain
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<math> a + b + 0 </math>
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then, again by the zero axiom
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<math> a + b </math>
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Since we have just established it, you do not need to prove it in your homework.
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(but you have to put a reference to this proof).
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Revision as of 05:11, 26 January 2010

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Maybe a stupid question but should we prove if a=b then (a+c)=(b+c) if we want to use this property? Can't find where it is given in the book.

Dr. Alekseenko: Apparently, this equation a consequence of

$ (a+c)-(b+c) = $

use associative law

$ a + (c - b) + c = $

use commutative law

$ a + (b + (-c)) + c = $

next associative law, again,

$ a + b + ((-c) + c) = $

finally use axiom of zero, i.e., $ (-c)+c =0 $ we obtain

$ a + b + 0 $

then, again by the zero axiom

$ a + b $

Since we have just established it, you do not need to prove it in your homework.

(but you have to put a reference to this proof).

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett