Question about a cancellation law

[icesalmon]

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 ?

 

[DrOnline]

I am not familiar with this, but "similar threads" perhaps has some info for you:

ex: https://www.physicsforums.com/showthread.php?t=590003

 

[icesalmon]

i'll check it out, thanks a lot.

 

[Mentallic]

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 $b\neq c$

What you're looking for is

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

EDIT: Sorry, I missed this part:

not sure how to proceed, is the problem statement ba = bc supposed to be ab = ac ?

Yes, that's what it should be instead.

 

[HallsofIvy]

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.

 

[icesalmon]

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 ] "

 

[Michael Redei]

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 $c=\left(\begin{array}{cc} 0&-1\\0&1 \end{array}\right)$.