Mathematical Proofs

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  • Thread starter Rob Hal
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  • #1
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Hi all,

If I have these two statements given to me, and I have to determine whether they are true or not.

a) [tex] \forall x \epsilon R [/tex] [tex]\exists y \epsilon R [/tex] [tex](y^2 = x^2 + 1)[/tex]
b) [tex]\exists y \epsilon R [/tex] [tex]\forall x \epsilon R [/tex] [tex](y^2 = x^2 + 1) [/tex]

Now, to me, they both mean exactly the same thing, and both can be shown to be false by setting x = 2, then y is not a real number.

However, seeing that the question specifically asks to prove just those two statements, I'm wondering if perhaps I am interpreting them wrong and they actually mean two different things.

Thanks in advance for any advice,
Robbie
 

Answers and Replies

  • #2
TD
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Rob Hal said:
Now, to me, they both mean exactly the same thing, and both can be shown to be false by setting x = 2, then y is not a real number.
If x = 2 then [itex]y^2 = 2^2 + 1 \Leftrightarrow y = \pm \sqrt 5 [/itex]. Those are real numbers, no?
 
  • #3
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lol... yeah...
I was thinking I was looking for rationals only... whoops...

Still, is there any difference in the two statements themselves?
 
  • #4
honestrosewater
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Try a simpler one to see how the order of the quantifiers makes a difference:

[tex]\forall x \in \mathbb{R} \ \exists y \in \mathbb{R} \ (x = y)[/tex]

This says that for every real x that I choose, I can find at least one real y that is equal to that x. This is obviously true, since x = x.

[tex]\exists y \in \mathbb{R} \ \forall x \in \mathbb{R} \ (x = y)[/tex]

This says that I can find at least one real y that is equal to every real x. Well, there's more than one real number, so this is false.
 
Last edited:

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