# Sets and Relations - quick one

1. Oct 28, 2005

### Natasha1

What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

Is it that AnBnC ?

Can someone explain if they are different and why?

Now If A = {irrationals}, B= {integers} and C={reals} does the equality from above hold in this case?

I answered yes the equality holds as AcBcC.

Any suggestions?

Last edited: Oct 28, 2005
2. Oct 28, 2005

### hypermorphism

The irrationals are not a subset of the integers.

3. Oct 28, 2005

### Natasha1

so for the first part of the question

What are the conditons on A, B and C for (AuB)nC = Au(BnC) ?

Is it that AnBnC ? This is correct

And the second part we get AcC only right?

4. Oct 28, 2005

### hypermorphism

$A\cap B\cap C$ is not a complete sentence. Did you mean the condition for the equality to be true is that $A\subseteq B \subseteq C$ ?
An easy way to check is to look at the condition of elements of each set. For example, if $S = ((A\cup B)\cap C)$, then S = {x: ($x\in A$ and $x\in C$) or ($x\in B$ and $x\in C$)} which can be written more concisely (or draw a Venn diagram). Do the same for the other side.
For the second part, yes, $A\subset C$, but you didn't answer the question posed.

5. Oct 28, 2005

### HallsofIvy

Staff Emeritus
AnBnC isn't even a "condition"! It's a set and you have to say something about it to get a "condition". (AuB)nC consists of all things that are in either A or B and in C. Au(BnC) consists of all things that are in A or in both B and C. Drawing Venn diagrams might help. Suppose "z" is in A but not in C. then it would certainly be in Au(BnC) since it is in A. But since it is not in C it is not[\b] in (AuB)nC. So one possible condition is that A is a subset of C. Is that sufficient?

6. Oct 28, 2005

### Natasha1

I did mean AcBcC actually not AnBnC sorry.

But actually the conditions for (AuB)nC = Au(BnC) in real numbers not our examble. Is that AcC and BcC right?

And in our example which is A= irrationals B= integers and C= reals is satisfies as AcC and BcC right?