# Is this enough to show these statements are true or false, it seems too easy to be T

1. Sep 19, 2006

### mr_coffee

Hello everyone. I have a series of statments i'm suppose to indicate which of the following statments are true and which are false. I am then suppose to justify my anser the best i can...

Here are some of the ones i did, can you verify they are correct? They seemed to easy for them to be right

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Thanks!

Last edited by a moderator: Apr 22, 2017 at 12:55 PM
2. Sep 19, 2006

### AKG

Your answers to c, d, and h are wrong, and your reasons for your answers to g and i are faulty. For g, you shouldn't say "z is an integer", you should say, "x - y is an integer because x and y are integers and differences of integers are integers, so let z = x-y, and that proves the claim." The idea is to find an integer z, not say that z is an integer; do you see the difference? For i, the questions asks something about ALL v, so you should not be proving things about a particular v, like v=1.

Last edited: Sep 19, 2006
3. Sep 19, 2006

### mr_coffee

Thanks for the responce!

I do see the difference but i'm having troubles getting it for all cases. Can i say somthing like, let u and v be positve real numbers such that u < v, then (u)(v) < v ?

Is C false becuase of counter example, if u let y = x, then x cannot equal y+1?

For part D, i don't see why it isn't false, doesn't that show a counter example, when xy is not equal to 1, given 2 year postive numbers?

4. Sep 20, 2006

### matt grime

You are not paying attention to your quantifiers.

For all means you need to do it for all, there exists mean you need to find one thing.

For instance when showing

for all v in R+ there is a u in R+ such that uv<=v

you cannot just pick *one* v and say it is true becuase it is true for one v.

Similarly, for the uv=1 question, look at the quantifiers: for all u inR+ there is a v in R+ with uv=1. your 'counter' example of u=1 v=3 doesn't matter. A counter example would be: there exists a u in R+ such that for all v in R+ uv=/=1, which is obviously false.

5. Sep 20, 2006

### HallsofIvy

Staff Emeritus
"There exist x such that for all y, x+ y= 1."

Can the same make that true for both y= 1 and y= 2?
That is, x+1= 1 and x+ 2= 1.

6. Sep 20, 2006

### mr_coffee

I remember the professor saying you can disprove somthing if you find 1 counter example, but to prove somthing you must prove it for all cases. If i have somthing like #40 part h, Where i have the existance quantifier first, doesn't that mean, I only have to prove 1 counter example to make that whole statment false which i supplied?

There exists a real postive number u such that For all real postive numbers v has the property uv < v. This is false because if you let u = 1, v = 3. Then 3 is not less than 3 which makes this statment invalid. Becuase they just said There exists 1 real u, for ALL postive numbers v which is a strong statement to make isn't it?

And if i have the For all quanitifer first in a statement, and then an existance statement like matts exampe:

THen I must suppply a counter example for all cases to make the statement false, is this correct? I must hae misunderstood what the professor ment when he said 1 counter, or it must be for only a certian case.

Thanks for the help guys.

7. Sep 20, 2006

### CRGreathouse

You showed that for some v and u, but you need to show it for all v (and some u).

A counterexample would be some v for which there exists no u (but if the statement is true, you can't find this).

8. Sep 20, 2006

### matt grime

You have to show what the quantifiers tell you.

If you are asked to show

for all x blah blah blah

then you have to show it for all x. If you need to disprove it a single counter example for one x will suffice.

there is a y such that....

then you need to show at least one y exists, possibly by writing it down, so that the 'such that.....' is satisfied. If you need to disprove it then you need to show that *no* possible choice of y will work, not just something you pick arbitrarily.

The moral: look at the quantifiers, they tell you precisely what you need to show.

9. Sep 20, 2006

### HallsofIvy

Staff Emeritus
To put what matt grime said in more technical terms:

$$\~(\forall p)= \exists(\~p)$$
and
$$\~(\exists p)= \forall (\~p)$$

10. Sep 20, 2006

### matt grime

Are there supposed to be some 'not' symbols there?

11. Sep 20, 2006

### HallsofIvy

Staff Emeritus
Actually, there are. There is a little "tilda" above each. I wanted a tilda (~) in front but couldn't figure out the LaTex.

12. Sep 20, 2006

### matt grime

It is \sim. What is there doesn't show up very clearly on my screen.

$$\sim(\forall x)= \exists \sim x$$

the logical symbol is

$$\neg (\forall x) = \exists \neg x$$

13. Sep 20, 2006

### mr_coffee

Ahh thanks for the explanation guys!! I think in time i will get the hang of it hopefully!