Are these statements true or false? Prove...

1. Sep 13, 2015

lep11

1. The problem statement, all variables and given/known data
Are these statements true or false? Prove/reason why/why not.
a.)∃x∈R , ∃y∈R : (x2-2)2+y2=1
b.)∀x∈R , ∀y∈R : y >x2-1
c.)∀x∈R , ∃y∈R : |x+y|=1
d.)∃x∈R , ∀y∈R :|x|>y

3. The attempt at a solution
a.) true, for example (x=1 and y=0)
b.) false, because if I choose (x=1 and y=0) I get 0>(1-1) which cannot be true so the statement cannot be always true
c.)false? Let. x=-y+8 ⇒|8|≠1 x can be any real number
d.)?

2. Sep 13, 2015

Staff: Mentor

You cannot choose an x that depends on y, that would reverse the logic.

d) Can you find an x such that ∀y∈R :|x|>y?

3. Sep 13, 2015

lep11

no so it's false

4. Sep 13, 2015

Staff: Mentor

Right.

5. Sep 14, 2015

lep11

Okay. The remaining question is: how do I prove c and d are false?

d.)∃x∈R , ∀y∈R :|x|>y

its negation: ∀x∈R , ∃y∈R :|x|≤y is true but how i prove it is true?

6. Sep 14, 2015

Staff: Mentor

@mfb, are you sure? If y is any arbitrary real number, surely we can find a number x for which |x| > y.

7. Sep 14, 2015

lep11

You need to re-read the statement d. It says: 'there exists an x∈R such that for all y∈R, |x|>y'.

It is obvious that there isn't an x of which absolute value is greater than every real number, isn't it?

Last edited: Sep 14, 2015
8. Sep 14, 2015

Staff: Mentor

I think I am misinterpreting d) as if it said ∀y∈R, ∃x∈R, :|x|>y

9. Sep 14, 2015

nuuskur

Tempting, isn't it?

I am at a loss too, though. How are we supposed to read d as? "There exists an x and for every y..?
I think what is meant that there is such an x we can couple with Any y such that the implication holds.

10. Sep 14, 2015

Staff: Mentor

I'm now convinced that the correct reading is: "There is a number x such that, for any real y, |x| > y." It's not difficult to show that this is not true.

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