# Mixed Quantifers confusion! Descrete Math

by mr_coffee
Tags: confusion, descrete, math, mixed, quantifers
 P: 1,629 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. There answer was: 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!
 HW Helper P: 2,566 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.
Math
Emeritus
Thanks
PF Gold
P: 38,456
 Quote by 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.
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?

 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.
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]

P: 1,629

## Mixed Quantifers confusion! Descrete Math

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.
 Sci Advisor P: 1,253 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?

 Related Discussions Calculus & Beyond Homework 2 Calculus & Beyond Homework 4 Calculus & Beyond Homework 2 Calculus & Beyond Homework 3 Beyond the Standard Model 2