# Mixed Quantifers confusion! Descrete Math

1. Sep 17, 2006

### mr_coffee

THe directions say< indicate which fo the following statements are true and which are false, Justify your answers as best you can.

Here is the question:
$$\exists$$ x $$\in$$ R such that $$\forall$$ $$\in$$ R, x = y + 1.

I wrote the following:
There exists a real number x such that given any real number y the property x=y+1 will be true. True. y = x-1. Then y is a real number, and y + 1 = (x-1)+1 = x.

I really don't know if i did this right or not but there was an example but slighty different and the book had the following:
$$\forall$$ x $$\in$$ Z, $$\exists$$ y $$\in$$ Z such that x = y + 1.

Given any integer, there is an integer such that tthe first inteer is one more than the second integer. this is true. Given any integer x, take y = x-1. Then y is an integer, and y + 1 = (x-1) + 1 = x.

I'm really confused on how to go about tackling these problems. Any help would be great! thanks!

2. Sep 17, 2006

### StatusX

For the first one, there needs to be a single x that works for all y. Note how this is different from the second one.

3. Sep 17, 2006

### HallsofIvy

Staff Emeritus
Doesn't make sense. Did you mean $$\for all y[/itex] ?? If you meant [tex]\forall y$$ then y= x-1 works, doesn't it?

What is true in Z (set of all integers) is not necessarily true in R (set of all real numbers) but the difference is usually a matter of multiplication or division, not addition.

Given any integer, there is an integer such that tthe first inteer is one more than the second integer. this is true. Given any integer x, take y = x-1. Then y is an integer, and y + 1 = (x-1) + 1 = x.

I'm really confused on how to go about tackling these problems. Any help would be great! thanks![/QUOTE]

4. Sep 17, 2006

### mr_coffee

I thought it was odd that I could solve them exactly the same way. For the first one, if i had to find a single x for all y, u would think i would have to write it differently than if i was finding for all x there exists a y.

5. Sep 17, 2006

### 0rthodontist

The order of the quantifiers is switched between the example and your problem. Your problem says, (as you correctly interpreted):
"There exists a real number x such that given any real number y the property x=y+1 will be true."
Another way of putting it is:
"There exists a real number x such that no matter what real number y I choose, x = y+1."
You should intuitively convince yourself that these are the same.

So let's try an example--say that x = 2. Is it true that no matter what real number y I choose, x = y + 1? No, because if I choose y = 100, then x does not equal 100 + 1 = 101. What if I chose x = 5. Could you find a y that makes the equation false? Is there ANY x that wouldn't have a y that would make the equation false?