Line 14: Line 14:
 
First of all, let us establish that <math> -(b+c)=(-b)+(-c)</math>. Indeed,  
 
First of all, let us establish that <math> -(b+c)=(-b)+(-c)</math>. Indeed,  
 
consider:
 
consider:
<math> -(b+c) = \mathrm{axiom\ of\ zero\ ele.}\  -(b+c)+ 0 + 0 = </math>
+
 
 +
<math> -(b+c) = </math>
 +
 
 +
Use the axiom of zero element
 +
 
 +
<math> -(b+c)+ 0 + 0 = </math>
 +
 
 
Use the axiom of additive inverse:
 
Use the axiom of additive inverse:
 +
 
<math> = -(b+c) + b + (-b) + c + (-c)  = </math>
 
<math> = -(b+c) + b + (-b) + c + (-c)  = </math>
 +
 
Use commutative law on terms #3 and #4:
 
Use commutative law on terms #3 and #4:
 +
 
<math> = -(b+c) + b + c+ (-b) + (-c)  = </math>
 
<math> = -(b+c) + b + c+ (-b) + (-c)  = </math>
 +
 
Use associative law:
 
Use associative law:
 +
 
<math> = -(b+c) + (b + c)+ (-b) + (-c)  = </math>
 
<math> = -(b+c) + (b + c)+ (-b) + (-c)  = </math>
 +
 
Finally, use axiom of inverse (A4) on terms #1 and #2:
 
Finally, use axiom of inverse (A4) on terms #1 and #2:
 +
 
<math> =  0 + (-b) + (-c)  = </math>
 
<math> =  0 + (-b) + (-c)  = </math>
 +
 
and the axiom of zero:
 
and the axiom of zero:
 +
 
<math> =  (-b) + (-c)  = </math>
 
<math> =  (-b) + (-c)  = </math>
 +
 +
  
 
Now consider  
 
Now consider  
 +
 
<math> (a+c) + (-(b+c)) = </math>  
 
<math> (a+c) + (-(b+c)) = </math>  
 +
 
Use the above identity:  
 
Use the above identity:  
 +
 
<math> = a + c + ((-b) + (-c)) =</math>
 
<math> = a + c + ((-b) + (-c)) =</math>
 +
 
and the associative property
 
and the associative property
 +
 
<math> = a + c + (- b) + (-c) =</math>
 
<math> = a + c + (- b) + (-c) =</math>
 +
 
and the commutative property on terms #2 and #3:
 
and the commutative property on terms #2 and #3:
 +
 
<math> = a + (- b) + c + (-c) =</math>
 
<math> = a + (- b) + c + (-c) =</math>
 +
 
Use the additive inverse property (A4):
 
Use the additive inverse property (A4):
 +
 
<math> = a + (- b) + 0 =</math>
 
<math> = a + (- b) + 0 =</math>
 +
 
and the zero element property:
 
and the zero element property:
 +
 
<math> = a + (- b)</math>
 
<math> = a + (- b)</math>
 
  
 
Now if <math> a=b </math> then <math> a + (-b) = b + (- b)=0 </math>
 
Now if <math> a=b </math> then <math> a + (-b) = b + (- b)=0 </math>
 +
Thus, according to our derivation,
  
 +
<math> (a+c) + (-(b+c)) = 0 </math>
  
 +
which together with
  
next associative law, again,
+
<math> (b+c) + (-(b+c)) = 0 </math>  
 
+
<math> a - b + ((-c) + c) =</math>
+
 
+
finally use axiom of zero, i.e.,  <math> (-c)+c =0 </math> we obtain
+
 
+
<math> a + b + 0 </math>
+
  
then, again by the zero axiom
+
By the Theorem about the uniqueness of the inverse element,
 +
implies that
  
<math> a + b </math>
+
<math> a + c = b + c  </math>  
  
 
Since we have just established it, you do not need to prove it in your homework.  
 
Since we have just established it, you do not need to prove it in your homework.  

Revision as of 14:38, 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 (corrected 01/26/2010 7:20pm): Indeed, the book does not seem to have this statement in it proved.

First of all, let us establish that $ -(b+c)=(-b)+(-c) $. Indeed, consider:

$ -(b+c) = $

Use the axiom of zero element

$ -(b+c)+ 0 + 0 = $

Use the axiom of additive inverse:

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

Use commutative law on terms #3 and #4:

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

Use associative law:

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

Finally, use axiom of inverse (A4) on terms #1 and #2:

$ = 0 + (-b) + (-c) = $

and the axiom of zero:

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


Now consider

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

Use the above identity:

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

and the associative property

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

and the commutative property on terms #2 and #3:

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

Use the additive inverse property (A4):

$ = a + (- b) + 0 = $

and the zero element property:

$ = a + (- b) $

Now if $ a=b $ then $ a + (-b) = b + (- b)=0 $ Thus, according to our derivation,

$ (a+c) + (-(b+c)) = 0 $

which together with

$ (b+c) + (-(b+c)) = 0 $

By the Theorem about the uniqueness of the inverse element, implies that

$ a + c = b + c $

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).

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