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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. | ||
| − | Dr. Alekseenko: | + | Dr. Alekseenko (corrected 01/26/2010 7:20pm): Indeed, the book does not seem to have this statement in it proved. |
| − | <math> ( | + | First of all, let us establish that <math> -(b+c)=(-b)+(-c)</math>. Indeed, |
| + | consider: | ||
| + | <math> -(b+c) = \mathrm{axiom\ of\ zero\ ele.}\ } -(b+c)+ 0 + 0 = </math> | ||
| + | Use the axiom of additive inverse: | ||
| + | <math> = -(b+c) + b + (-b) + c + (-c) = </math> | ||
| + | Use commutative law on terms #3 and #4: | ||
| + | <math> = -(b+c) + b + c+ (-b) + (-c) = </math> | ||
| + | Use associative law: | ||
| + | <math> = -(b+c) + (b + c)+ (-b) + (-c) = </math> | ||
| + | Finally, use axiom of inverse (A4) on terms #1 and #2: | ||
| + | <math> = 0 + (-b) + (-c) = </math> | ||
| + | and the axiom of zero: | ||
| + | <math> = (-b) + (-c) = </math> | ||
| − | + | Now consider | |
| + | <math> (a+c) + (-(b+c)) = </math> | ||
| + | Use the above identity: | ||
| + | <math> = a + c + ((-b) + (-c)) =</math> | ||
| + | and the associative property | ||
| + | <math> = a + c + (- b) + (-c) =</math> | ||
| + | and the commutative property on terms #2 and #3: | ||
| + | <math> = a + (- b) + c + (-c) =</math> | ||
| + | Use the additive inverse property (A4): | ||
| + | <math> = a + (- b) + 0 =</math> | ||
| + | and the zero element property: | ||
| + | <math> = a + (- b)</math> | ||
| + | |||
| − | <math> a + ( | + | Now if <math> a=b </math> then <math> a + (-b) = b + (- b)=0 </math> |
| − | |||
| − | |||
next associative law, again, | next associative law, again, | ||
| − | <math> a | + | <math> a - b + ((-c) + c) =</math> |
finally use axiom of zero, i.e., <math> (-c)+c =0 </math> we obtain | finally use axiom of zero, i.e., <math> (-c)+c =0 </math> we obtain | ||
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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) = \mathrm{axiom\ of\ zero\ ele.}\ } -(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 $
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).
