# Thread: proof

1. ## proof

if A union B = A union C, then B=C.

proof: Assume that A union B = A union C. Suppose x is in B. Then x is in A U B. By the assumption x is also in A U C. But we did not assume x was in A so x must be in C. Thus,we have shown B is a subset of C and by a parallel argument C is a subset of B

is this right

2. Originally Posted by leinadwerdna
if A union B = A union C, then B=C.

proof: Assume that A union B = A union C. Suppose x is in B. Then x is in A U B. By the assumption x is also in A U C. But we did not assume x was in A so x must be in C. Thus,we have shown B is a subset of C and by a parallel argument C is a subset of B

is this right
Not quite. Let $\displaystyle A=\{1, 2, 3\}$, $\displaystyle B=\{3,4\}$. Can you perhaps see the problem in your proof by looking at these examples?

However, you can quite easily tweak your proof to make it valid, taking into account the problem I've hinted at.

3. Originally Posted by leinadwerdna
if A union B = A union C, then B=C
Originally Posted by Swlabr
However, you can quite easily tweak your proof to make it valid, taking into account the problem I've hinted at.
No! No matter how one tweaks it, we can’t prove something that is false to be true.
$\displaystyle \begin{gathered} A = \left\{ {1,2,3} \right\},\,B = \left\{ {1,4} \right\}\;\& \,C = \left\{ {3,4} \right\} \hfill \\ \left\{ {1,2,3,4} \right\} = A \cup B = A \cup C\text{ but }B \ne C \hfill \\ \end{gathered}$

4. thank you