# The word equvilalent

1. Oct 5, 2011

### flyingpig

The word "equvilalent"

1. The problem statement, all variables and given/known data

Yes I know I spelled equivlaence wrong.

[PLAIN]http://img834.imageshack.us/img834/9285/unledkzo.png [Broken]

3. The attempt at a solution

What do they mean by equivlance? What is to show?

Last edited by a moderator: May 5, 2017
2. Oct 5, 2011

### SammyS

Staff Emeritus
Re: The word "equvilalent"

Statement (i) means the same thing as statement (ii) which means the same thing as statement (iii) .

They're equivalent logically speaking.

(A ⊆ B) iff (A ∩ B = A)

etc.

3. Oct 5, 2011

### Staff: Mentor

Re: The word "equvilalent"

Two statements P and Q are equivalent if P implies Q and Q implies P. To show that three statements, P, Q, and R, are equivalent you could show that P implies Q and Q implies P, then show that P implies R and R implies P, and finally, show that P implies R and R implies P.

A simpler way to show that statements P, Q, and R are equivalent is to show that these implications are true.

P ==> Q (that is, P implies Q)
Q ==> R
R ==> P

See above.
No, it's also true if A is a subset of B, which is what statement i says.
The three statements don't stand alone. What you are supposed to do is to show that they are all equivalent; loosely speaking, that they all say the same thing.

Last edited by a moderator: May 5, 2017
4. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

Before I even attempt this (don't want to head a long direction), this is my idea.

SHould I let some element x be in A and then deduct that since (i) A is a subset of B then x is also in B?

Then to go from there, I say that x is indeed (from what I just said) is in the intersection of A and B, i.e. x is in A ∩ B, which also means that x is in A (I am not sure about the whole A ∩ B = A (being only A?)

Then to go from there, I would say that from (i), since x is in A or in B, it follows that x is in either A or B?

That equal sign is messing me up

5. Oct 5, 2011

### Staff: Mentor

Re: The word "equvilalent"

No, you can assume that statement i is true, then deduce that statement ii is true.

IOW, assume that A $\subseteq$ B. Now let x be an element of A ∩ B. Show that x is necessarily in A.

And so on.

6. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

But that doesn't say if ii) is true, then i) is true. What if you assume ii) is true first? (follow below)

No that can't right

$$x \in A \cap B \iff x \in A$$ and $$x \in B$$

$$\therefore x \in A$$

But that doesn't say that $$A \subseteq B$$

And I just realize I did nothing at all...

Let's say now I do believe you (or the question) that $$A \subseteq B$$

(ii)
$$x \in A \cap B \iff x \in A$$ and $$x \in B$$

$$\therefore x \in A$$

But (i) is true, so that $$x \in B$$ is unnecessary, and that only $$x \in A$$ because A is a subset of B

(iii) $$x \in A \cup B \iff x \in A$$ or $$x\in B$$

Since (i) is true, i.e., then only $$x \in B$$ is necessary

How do I link (ii) and (iii) together? They give off the same results from agreeing on (i), but they don't have relations

Last edited: Oct 5, 2011
7. Oct 5, 2011

### Staff: Mentor

Re: The word "equvilalent"

Assume i is true.
Show that if i is true, then ii must be true.
Show that if ii is true, then iii must be true.
Show that if iii is true, then i must be true.

If you have done all this, then you're finished. How does this show that ii ==> i? It does so in a roundabout way.
Having done all the things I said in the first paragraph, we have
ii ==> iii ==> i

How do you link ii and iii together?
Assume ii is true (i.e., that A ∩ B = A). Show that this being true implies that A U B = B.

It would be helpful to your understanding to draw a simple Venn diagram, but you won't be able to use the diagram in your actual proof.

8. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

My example in post 6 says it is true, so i know they are true, just how to prove it.

So I just showed that

Assume i) is true and that ii) is true

Then I showed that

Assume i) is true, then iii) is true

Now I need to show that ii) and iii),

$$x \in A \cap B = x\in A$$

I don't know how to go from here.

My so far proof did

i) -> ii)
i -> iii)

Does ii) -> iii) complete the circle?

9. Oct 5, 2011

### SammyS

Staff Emeritus
Re: The word "equvilalent"

For two sets, C & D : To show (prove) they are equal, you must show that C ⊆ D and that D ⊆ C .

So for (i) implies (ii), show that A ∩ B = A is true by proving that A ∩ B ⊆ A and that A ⊆ A ∩ B , of course you prove these assuming that A ⊆ B is true.

10. Oct 5, 2011

### Staff: Mentor

Re: The word "equvilalent"

No, because you don't have a circle.
You have i --> ii, ii --> iii, and i --> iii. You don't have anything that shows i is true.

Did you not understand this?

11. Oct 5, 2011

### SammyS

Staff Emeritus
Re: The word "equvilalent"

fp,

If i --> ii and ii --> iii then you have i --> iii by default.

That along with iii --> i , gives i <--> iii .

12. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

Show that if (i) is true, then (ii) is true

This is a bit messy, but I think I did this part correctly.

$$x \in A \cap B \iff x \in A$$ and $$x\in B$$

$$x \in A \iff x \in B \iff x \in A \cap B$$

Assume $$A \subseteq B$$ is true

Then $$x \in A \iff x \in A \cap B$$

So I just showed

phew...

13. Oct 5, 2011

### SammyS

Staff Emeritus
Re: The word "equvilalent"

What do you mean by your two way arrows?

I know what Mark means by them, nut what do you mean?

14. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

if and only if? That seems to be used correctly in this problem

15. Oct 5, 2011

### SammyS

Staff Emeritus
Re: The word "equvilalent"

Some of the iff's shouldn't be there.

16. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

You mean this right...?

$$x \in A \iff x \in B \iff x \in A \cap B$$

So everything after this is wrong...

that is my guess because that looks the most suspicious.

I am going nowhere with this...

17. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

Actually wait, something hit me

$$x \in A \iff x \in B \implies x \in A \cap B$$

I just have to change iff to an "implies"

Which flows better

18. Oct 5, 2011

### Staff: Mentor

Re: The word "equvilalent"

The above makes no sense as you have written it.

$$x \in A ~\text{and}~ x \in B \iff x \in A \cap B$$

Definitions are often written as iff statements. Otherwise, P <==> Q means P ==> Q and Q ==> P.

Last edited: Oct 5, 2011
19. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

Sigh, this is due tomorrow and i have another midterm tomorrow.

But it does make sense because I assumed that $$A \subseteq B$$

20. Oct 5, 2011

### flyingpig

Re: The word "equvilalent"

I've attached a picture of where I was going with my algebra

[PLAIN]http://img198.imageshack.us/img198/91/unledctr.png [Broken]

Last edited by a moderator: May 5, 2017