Are these statements true or false? Prove...

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  • #1
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Homework Statement


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

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.)?
 

Answers and Replies

  • #2
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c.)false? Let. x=-y+8 ⇒|8|≠1 x can be any real number
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
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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?
no so it's false
 
  • #5
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Right.
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?
 
  • #7
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@mfb, are you sure? If y is any arbitrary real number, surely we can find a number x for which |x| > y.
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?
 
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  • #8
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I think I am misinterpreting d) as if it said ∀y∈R, ∃x∈R, :|x|>y
 
  • #9
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I think I am misinterpreting d) as if it said ∀y∈R, ∃x∈R, :|x|>y
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
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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.
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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