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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 (corrected 01/26/2010 7:20pm): Indeed, the book does not seem to have this statement in it proved. | + | 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 <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) = | + | |
| + | <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 | ||
| − | + | <math> (b+c) + (-(b+c)) = 0 </math> | |
| − | + | By the Theorem about the uniqueness of the inverse element, | |
| + | implies that | ||
| − | + | <math> a + c = b + c </math> | |
| − | + | 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). | ||
| − | + | ---- | |
| − | + | --[[User:Rrichmo|Rrichmo]] 16:12, 27 January 2010 (UTC) | |
| − | + | Is it given that the sum and product of two integers is again an integer? | |
| − | + | ||
| + | Dr. Alekseenko: Yes. It is given. This property is included in the definitions of | ||
| + | addition and multiplication | ||
---- | ---- | ||
Latest revision as of 17:49, 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?
Dr. Alekseenko: Yes. It is given. This property is included in the definitions of addition and multiplication
