Mixed Quantifers confusion! Descrete Mathby mr_coffee Tags: confusion, descrete, math, mixed, quantifers 

#1
Sep1706, 02:07 PM

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: [tex] \exists [/tex] x [tex]\in[/tex] R such that [tex]\forall[/tex] [tex]\in[/tex] 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 = x1. Then y is a real number, and y + 1 = (x1)+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: [tex]\forall[/tex] x [tex]\in[/tex] Z, [tex]\exists[/tex] y [tex]\in[/tex] 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 = x1. Then y is an integer, and y + 1 = (x1) + 1 = x. I'm really confused on how to go about tackling these problems. Any help would be great! thanks! 



#2
Sep1706, 02:57 PM

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.




#3
Sep1706, 03:06 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

If you meant [tex]\forall y[/tex] then y= x1 works, doesn't it? [/quote]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 = x1. Then y is an integer, and y + 1 = (x1) + 1 = x. I'm really confused on how to go about tackling these problems. Any help would be great! thanks![/QUOTE] 



#4
Sep1706, 05:01 PM

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.




#5
Sep1706, 10:39 PM

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 examplesay 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? 


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