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

Revision as of 11:12, 27 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).

--Rrichmo 16:12, 27 January 2010 (UTC)

Is it given that the sum and product of two integers is again an integer?