Difference between revisions of "MA341 HW discussion Section 2 1" - Rhea

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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).