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

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SUMMARY

The discussion centers on proving the statement "If A U C = B U C then B = C" as false without using a Venn Diagram. The participants demonstrate that by providing a counter-example, such as A = {1}, B = {1}, and C = {2}, the equality holds while B does not equal C. The key takeaway is that to disprove a universal statement, it suffices to find a single counter-example, which validates the negation of the original statement. The discussion also emphasizes the importance of understanding the conditions under which the statement may hold true.

PREREQUISITES
  • Understanding of set theory concepts, specifically union and equality of sets.
  • Familiarity with logical statements and their negations.
  • Ability to construct and analyze counter-examples in mathematical proofs.
  • Basic knowledge of mathematical notation and terminology.
NEXT STEPS
  • Explore the concept of set operations in depth, focusing on union and intersection.
  • Study logical proofs and the structure of universal and existential quantifiers.
  • Learn about counter-example techniques in mathematical reasoning.
  • Investigate conditions under which set equalities hold true, including subset relationships.
USEFUL FOR

This discussion is beneficial for students of mathematics, particularly those studying set theory, logic, and proof techniques. It is also valuable for educators seeking to enhance their understanding of teaching mathematical concepts related to set operations and logical reasoning.

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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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