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Mixed Quantifers confusion! Descrete Math

by mr_coffee
Tags: confusion, descrete, math, mixed, quantifers
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mr_coffee
#1
Sep17-06, 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 = 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:
[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 = 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!
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StatusX
#2
Sep17-06, 02:57 PM
HW Helper
P: 2,567
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.
HallsofIvy
#3
Sep17-06, 03:06 PM
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Quote 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:
[tex] \exists [/tex] x [tex]\in[/tex] R such that [tex]\forall[/tex] [tex]\in[/tex] R, x = y + 1.
Doesn't make sense. Did you mean [tex]\for all y[/itex] ??
If you meant [tex]\forall y[/tex] 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:
[tex]\forall[/tex] x [tex]\in[/tex] Z, [tex]\exists[/tex] y [tex]\in[/tex] 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.

[/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 = 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]

mr_coffee
#4
Sep17-06, 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.
0rthodontist
#5
Sep17-06, 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 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?


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