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Disprove the nested quantifier

  • Thread starter albert1992
  • Start date
11
0
1. Homework Statement
∃x∀y(y̸=0→xy=1) in the real numbers universe.


2. Homework Equations



3. The Attempt at a Solution
Since the given statement is false I negated the whole statement to become

∀x∃y(y̸!=0^xy!=1) (!= means not equal to)

then I would have to prove this correct by setting y to something except zero
I cant find any y to prove this correct
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
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I can't make sense of your formulae. I think you messed up the typesetting. Also, it may help to use words instead of symbols... especially if you have to improvise to make the symbols.
 
11
0
there exists an x,for every y (if y does not equal to zero then x*y=1)


*real numbers universe
 
32,843
4,562
there exists an x,for every y (if y does not equal to zero then x*y=1)


*real numbers universe
How about this?

For every nonzero y in R, there exists an x in R such that xy = 1.
 
11
0
i have to disprove the statement since it is false
 

Dick

Science Advisor
Homework Helper
26,258
618
there exists an x,for every y (if y does not equal to zero then x*y=1)


*real numbers universe
If y does not equal 0, then there is only one value x such that x*y=1. That's x=1/y. How can there be an x that has an infinite number of solutions to x*y=1?
 
11
0
Exactly why the statement is false, but i have to prove that it is false
by negating the whole expression
 

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