1. Sep 15, 2008

### PhysicsHelp12

Are these TRue or FALSE Im trying to figure it out for 3 hours now and Im really stuck

1) For all x and all y, there exists a z such that x+y=z

I know is True

2) For all x there exists a y such that for all z, x+y=z

I think this is true too since y=z-x and then For all x, there exists a y
=z-x such that for all z, x+z-x=z...?

3) There exists a x such that for all y, there exists a z such that xz=y.

TRUE?

4)For all x and all y, there exists z such that yz=x

TRUE because z=x/y ..always works

5) For all x there exists y such that for all z, z>y implies that z>x+y

This one took me a long time ...TRue? because if y=abs(x)+1
would work ?

6) For all x and all y, there exists a z such that z>y implies that z>x+y.

I couldn figure this one out ..but I think it oculd be false

2. Sep 16, 2008

### tiny-tim

Hi PhysicsHelp12!

Why the question mark? Can't you find an x for which it works?
No … assuming we're talking about ordinary real numbers, when does x/y not exist?

3. Sep 16, 2008

### nicksauce

I believe that the way this is phrased, y should not depend on z, so y = z-x is not a valid y.

Suppose x is 2, and z = 2. Then you would need y = 0. Now suppose x is 2 and z = 3, and y=0 (since y=0 is supposed to be good for all z). Do you see the problem?

4. Sep 16, 2008

### soumyashant

I think all your doubts will be cleared if you pick up some good book of analysis and study about the field axioms....

Its not clear what you mean by x,y and '+'...

If you do mean "real numbers" and 'addition', just keep in mind that all the questions you have asked follow from the definition of R
R is a complete ordered field (and the only one such field)...
By the field axioms, your statements (though some of them are imprecisely framed) are trivially true.... and the lines of argument you have mentioned are correct too....

But on the whole I think you better read from a textbook, as this needs a good formal approach...

5. Sep 16, 2008

### soumyashant

I think you are confusing between "all" and "every"... You should use the latter instead of the former in most cases above..
There is a subtle difference...

"every" tells you that you are considering a particular case now and then will scan your argument over each 'x' and thus conclude for 'all' x...
"all" suggests that you are using the argument on the whole set at once...

I hope you get me..

6. Sep 16, 2008

### Tac-Tics

One way you can sometimes check these kinds of questions is by replacing the universal qualifier "for all/for every" with a particular element.

"For all x there exists a y such that for all z, x+y=z"

If this formula works for all x, then we can drop the quantifier and replace x with a number of our choice. Say 0 to keep things easy. So we get:

There exists a y such that for all z, 0 + y = z", or
There exists a y such that for all z, y = z"

Is it possible to find a number y which is equal to all numbers? No (unless your universal set is trivial). From that last statement, we can derive:

"There exists a y such that y = 1 and y = 2"

Whenever your statement starts with an existential quantifier, look for a concrete example which satisfies the predicate.

In this case, let's make a random guess that it holds true for x=0. If it does, then the statement is true. If it doesn't, we need to keep looking (or look for a proof that no such value exists).

So, we get:

"for all y, there is a z such that 0*z = y", or
"for all y, there is a z such that 0 = y"

Replacing y with a value other than 0 clearly leads to a false statement (note that when we have a situation like "there exists z such that 0 = y" we can drop the quantifier).

So the value 0 for x doesn't work. Try again... maybe with x=1.

"for all y, there is a z such that 1*z = y", or
"for all y, there is a z such that z = y"

This statement is true. I'll let you think about why (note that since z is quantified "in the scope" of the variable y, it can be written in *terms* of y).

Thus, we shown the existence of an x that satisfies that property.

If you're working with the full set of integers or the real numbers, consider the case y=0.

If what is meant is universal quantification, then "all" and "every" are equivalent. However, in proofs, mathematicians are not always very strict with how they word things, and this can cause some ambiguity. I see this all the time in epsilon-delta definitions:

"The limit of a function at a point a is L iff for every e > 0, there exists a d > 0 such that |f(x) - L| < e whenever |x - a| < d."

What is meant by that is really:

"The limit of a function at a point a is L iff for all e > 0, there exists a d > 0, such that for all x, |x - a| < d implies |f(x) - L| < e".

Quantifiers are really quite confusing!

7. Sep 18, 2008

### Robokapp

Are these TRue or FALSE Im trying to figure it out for 3 hours now and Im really stuck

1) For all x and all y, there exists a z such that x+y=z

ya. obviously.

2) For all x there exists a y such that for all z, x+y=z

obviously.

3) There exists a x such that for all y, there exists a z such that xz=y.

2 numbrs multiplied can produce anything. pick x=y and z=1. it's true.

4)For all x and all y, there exists z such that yz=x

no...if y=0 and x =/=0 you have no z that'll save you.

5) For all x there exists y such that for all z, z>y implies that z>x+y

no. take z=x=100 and y=99. 100>99 does not imply that 100>99+100. the 'implies' part makes it false in my eyes. Hoever it would be true if you had the choice of y. say -200. It can be true but it doesn't imply in my oppinion.

second look after doing #6: all x. so i'll pick a big option. 100. y is questinoable so i'll leave it at y. z>y so i'll pick it as y+1. We have y+1>y+100. there's no y that does that.

6) For all x and all y, there exists a z such that z>y implies that z>x+y.

take x=100 y=0 z=1. once again z>y does not imply z>X+y that should be obvious for all x>(z-y). however such a z exists. but is not implied.

edit: try not to divide by stuff labeled "all". the particularity of 0 on denominator kills the 'all'.

Last edited: Sep 18, 2008