Proving "If A U C = B U C then B = C" Without Drawing a Venn Diagram

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Homework Help Overview

The discussion revolves around the statement "If A U C = B U C then B = C" in the context of set theory. Participants explore the validity of this statement and consider methods to prove or disprove it without the use of Venn diagrams.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the idea of proving the statement false by providing counter-examples, with some suggesting the use of simple sets to illustrate the point. Others question how to approach the proof without assigning specific elements to the sets.

Discussion Status

The conversation is ongoing, with various participants contributing thoughts on counter-examples and the nature of the original statement. Some guidance has been offered regarding the structure of negation in proofs, and the discussion reflects a mix of interpretations and approaches to the problem.

Contextual Notes

Participants note the importance of understanding the implications of the statement and the conditions under which it may hold true or false. There is also mention of potential typos in the original question that could affect interpretation.

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I am trying to prove this as false. Let A, B, C be any three sets.

If A U C = B U C then B = C. I can draw a Venn Diagram to prove this and I can assign values to the sets to prove it, but how can I prove without doing this? Also is the counter value A U C = B U C then B not equal to C? Can you prove using this? I am a bit confused by this problem. Maybe I am thinking too hard about it.
 
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To prove something is not true, the nicest way of doing so is coming up with a counter-example.

For example if someone said that any irrational number times any irrational number is yet another irrational number then you could say something like:

\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1

This really is the nicest way to prove something is not true.


As for your 2nd part, if is not true at least once, you can not then say from that it is always not true. If you want to prove it is always not true then you need to try and prove it on its own.
 
Can you think of 3 very simple sets, say with just two or three elements each, so that A U C= B U C but it is not true that A= B?
 
If I assign these elements

Set A = {1}
Set B = {1}
Set C = {2}

then assign it to my problem

A U C = B U C then B = C

A U C = {1, 2}
B U C = {1, 2}
which makes A U C = B U C true

But B does not equal C since 1 does not equal 2

Is there a way to prove this without assigning elements?
 
Sure, you're just proving that the original statement's negation is true. Of a statement S and its negation ~S, one is true and one is false. If ~S is true, then S is false. You're just proving that ~S is true. The negation of

1) For all A, B, and C, if A U C = B U C, then B = C

is

2) There exists some A, B, and C such that A U C = B U C and B ≠ C.

This is what your example shows. You can generalize your example. A = B and A ≠ C are conditions that satisfy (2).

And you can try to prove (2) however else you like.
 
If you want a general counter example consider what happens if A and B are subsets of C.
 
But to prove a general statement is NOT true only requires a single counter-example.
 
Absolutely, but sometimes understanding why something is false can be seen by considering how to make arbitrarily many counter examples. It might also prod the conscientious into working out what extra conditions might make the result true.

Actually, I may have misread the question. I assumed the question was:

show that if AuB=AuC then it is not necessarily true B=C.Actually what is written is:

show that if AuC=BuC then B=C is not necessarily true, which is surely typo, I mean, who would even think that that was true? At least the cancellation type problem is reasonable.
 
Last edited:
thanks all.. and also to HallsofIvy to the suggestion of making some simple sets. I submitted my homework and with the assigned variables and I received all points for this problem correctly.
 

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