# Math Help - Discrete Maths [Confused]

1. ## Discrete Maths [Confused]

Sorry about posting up for help all the time I am just struggling with some equations, as it seems to be asking to divide where you cant divide in this equation as how do you divide names (I'm going to look stupid) so the key is as follows;

Workforce = {Tesco, Asda, Lidl, M&S, Coop}
A = {Tesco, Asda}
B = {Lidl, M&S, Asda}
C = {Coop, Asda, Lidl}

and the equations are like .......

Workforce \ B
&
(Workforce \ C) \ A

The problem is the only time I have ever seen a symbol like \ in maths was a divide symbol but I am taking a wild stab in the dark and going to say that it isnt divide in this situation. So any help would be really helpful too me

Sw!ft.

2. ## Set Notation

Hello Swift
Originally Posted by Swift
Sorry about posting up for help all the time I am just struggling with some equations, as it seems to be asking to divide where you cant divide in this equation as how do you divide names (I'm going to look stupid) so the key is as follows;

Workforce = {Tesco, Asda, Lidl, M&S, Coop}
A = {Tesco, Asda}
B = {Lidl, M&S, Asda}
C = {Coop, Asda, Lidl}

and the equations are like .......

Workforce \ B
&
(Workforce \ C) \ A

The problem is the only time I have ever seen a symbol like \ in maths was a divide symbol but I am taking a wild stab in the dark and going to say that it isnt divide in this situation. So any help would be really helpful too me

Sw!ft.

You're right. In this context \ denoted the difference of two sets, and is defined as follows:

A\B is the set of elements that are in A but not in B.

So, for example, Workforce \ B are
all the stores except Lidl, M&S and Asda; in other words {Tesco, Coop}

OK now?

3. And you would do the one in brackets before the one outside of the brackets?

4. Hello Swift
Originally Posted by Swift
And you would do the one in brackets before the one outside of the brackets?
Yes, just as you would expect.

5. Sorry about this last question, [promise] haha. If I had the equation - A\(B\C) I would work it out as B\C then A\B and coming out with the answer as = Tesco??. I really apprieciate this Grandad, you are very helpful

6. Hello Swift
Originally Posted by Swift
Sorry about this last question, [promise] haha. If I had the equation - A\(B\C) I would work it out as B\C then A\B and coming out with the answer as = Tesco??. I really apprieciate this Grandad, you are very helpful
To work out $A\setminus(B\setminus C)$ you do work out $B\setminus C$ first, and then use the result of this, $D$ say, to work out $A\setminus D$.

So $D = B\setminus C =$ the things that are in $B$ but not in $C = \{ \text{M and S} \}$

And then $A\setminus D$ = the things that are in $A$ but not in $D = \{ \text{Tesco, Asda}\}$

So, not right with $\{ \text{Tesco} \}$.

The reason that $A\setminus B$ is called the set difference is that another way of looking at this is to think of $A\setminus B$ as taking away from $A$ any elements that are also in $B$ - similar to (but not the same as) subtraction.

Like subtraction, $A\setminus B$ is not equal to $B\setminus A$: just as $x - y$ is not equal to $y -x$.

And, just as $x - (y - z)$ is not the same as $(x - y) - z$, $A\setminus (B\setminus C)$ is not the same as $(A\setminus B)\setminus C$.

Try this last one with these sets: $(A\setminus B)\setminus C = \{\text{Tesco}\}\setminus C=\{\text{Tesco}\}$, whereas (as we've just seen) $A\setminus (B\setminus C) = \{\text{Tesco, Asda}\}$.

One final thing worth noting is this: $A\setminus B$ is the set of elements that are in $A$ but not in $B$. In other words, they're the elements that are in $A$ and in the complement of $B$. So $A\setminus B = A\cap B'$.