Mixed Quantifers confusion Descrete Math

[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.

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!

 

[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.

 

[HallsofIvy]

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] ?? <br /> If you meant \forall y then y= x-1 works, doesn&#039;t it?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I wrote the following:<br /> 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.<br /> <br /> I really don&#039;t know if i did this right or not but there was an example but slighty different and the book had the following:<br /> \forall x \in Z, \exists y \in Z such that x = y + 1. </div> </div> </blockquote> 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.<br /> <br /> [/quote]There answer was:<br /> 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.<br /> <br /> I&#039;m really confused on how to go about tackling these problems. Any help would be great! thanks![/QUOTE]

 

[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.

 

[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?