Cancellation law with ba = bc vs ab = ac

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Homework Statement


if ba = bc and a is nonzero then b = c

The Attempt at a Solution


if a is nonzero I can multiply both sides by a-1 to get
ba(a-1) = bc(a-1)
b(1) = bc(a-1)
b = bc(a-1)
not sure how to proceed, is the problem statement ba = bc supposed to be ab = ac ?
 

Discussion

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i'll check it out, thanks a lot.
 
icesalmon said:

Homework Statement


if ba = bc and a is nonzero then b = c

No, that's not true at all. For example, take b=2, a=4, c=4. The equality holds, the criteria are met, yet [itex]b\neq c[/itex]

What you're looking for is

If ba = bc and b is non-zero, then a = c.

EDIT: Sorry, I missed this part:
icesalmon said:
not sure how to proceed, is the problem statement ba = bc supposed to be ab = ac ?

Yes, that's what it should be instead.
 
You do not say what algebraic structure you are working in. In a "cancellation ring" such as the set of all integers, with ordinary addition and multiplication as operations, we do NOT have "multiplicative inverses" but the cancellation law is still true: if ab= ac and a is not the additive identity, 0, then b= c.
 
I'm not sure if this is correct, but I'm thinking it's a field since they didn't specify. My book was using this as an example of how to show that axioms that hold in real arithmetic may not hold in matrix arithmetic: "for example, consider the following two laws of real arithmetic: If ab = bc and a is nonzero then b = c. [ Cancellation Law ] "
 
That so-called "cancellation law" doesn't hold for matrices. Just consider ##a=\left(\begin{array}{cc} 1&0\\0&0 \end{array}\right)## (which is nonzero), ##b=\left(\begin{array}{cc} 0&1\\1&1 \end{array}\right)## and ##c=\left(\begin{array}{cc} 0&-1\\0&1 \end{array}\right)##.