# Field: If A=B; Prove A+C = B+C

• Apr 9th 2009, 07:10 AM
aman_cc
Field: If A=B; Prove A+C = B+C
This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
Seems like we have to rely of definition rather than a formal prove.

One approach:
Let us for the moment assume
A=B => B-A = 0 Let is call it (1)

Then
A+C = A + C + B - A = B + C

But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

Thanks
• Apr 9th 2009, 11:15 AM
clic-clac
Hi

Isn't this just a property of equality?
• Apr 9th 2009, 11:33 AM
ThePerfectHacker
Quote:

Originally Posted by aman_cc
This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
Seems like we have to rely of definition rather than a formal prove.

One approach:
Let us for the moment assume
A=B => B-A = 0 Let is call it (1)

Then
A+C = A + C + B - A = B + C

But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

Thanks

There is nothing to prove here.
Addition, is a binary operation $+:F\times F\to F$.
Therefore, $+(A,C) = +(B,C)$ since $A=B$.
• Apr 9th 2009, 07:33 PM
aman_cc
Quote:

Originally Posted by clic-clac
Hi

Isn't this just a property of equality?

Thanks. I am new to Algebra. If you could please elaborate. Sorry if this is too trivial.
• Apr 9th 2009, 07:41 PM
K A D C Dilshan
Quote:

Originally Posted by aman_cc
This is trivial. But I could not work out a formal prove for this (using the axioms of addition).
Seems like we have to rely of definition rather than a formal prove.

One approach:
Let us for the moment assume
A=B => B-A = 0 Let is call it (1)

Then
A+C = A + C + B - A = B + C

But to prove (1) above we somewhere have to rely on A=B => A+C = B+C

Thanks

Lets assume ,
A=B => A+C != B+C (!= stands for not equal)
=> A+C > B+C or A+C < B+C
=> A > B or A < B (by definitions of inequalities)

What do you think ?
• Apr 9th 2009, 08:00 PM
aman_cc
Quote:

Originally Posted by K A D C Dilshan
Lets assume ,
A=B => A+C != B+C (!= stands for not equal)
=> A+C > B+C or A+C < B+C
=> A > B or A < B (by definitions of inequalities)

What do you think ?

Yes - Makes sense. Though I had to prove to myself the following result
"=> A+C > B+C or A+C < B+C
=> A > B or A < B (by definitions of inequalities)"

You can also prove the original from the definition of mapping.
If x=y then f(x) = f(y), as others suggested.

Thanks
• Apr 9th 2009, 11:26 PM
clic-clac
Quote:

Lets assume ,
A=B => A+C != B+C (!= stands for not equal)
=> A+C > B+C or A+C < B+C
=> A > B or A < B (by definitions of inequalities)
Doing that, you assume your field is totally ordered, which is often wrong.

The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").
• Apr 9th 2009, 11:46 PM
aman_cc
Quote:

Originally Posted by clic-clac
Doing that, you assume your field is totally ordered, which is often wrong.

The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").

Yes clic-clac, that makes lot more sense as the result will hold even in fields which are not ordered. Like you said it can be proved/argued based on the logic of equality and definition of mapping ("+"). Atleast I;m convinced with the argument now ! Thanks very much
• Apr 10th 2009, 03:21 AM
K A D C Dilshan
Quote:

Originally Posted by clic-clac
Doing that, you assume your field is totally ordered, which is often wrong.

The answer is just the fact that if A=B, then any sentence in which you replace A by B or B by A will keep its meaning ("a property of equality").

But I think I have used only definitions ...