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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).
--Rrichmo 16:12, 27 January 2010 (UTC)
Is it given that the sum and product of two integers is again an integer?
