Disprove the nested quantifier

[albert1992]

Homework Statement

∃x∀y(y̸=0→xy=1) in the real numbers universe.

Homework Equations

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 can't find any y to prove this correct

 

[Hurkyl]

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.

 

[albert1992]

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

 

[Mark44]

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.

 

[albert1992]

i have to disprove the statement since it is false

 

[Dick]

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?

 

[albert1992]

Exactly why the statement is false, but i have to prove that it is false
by negating the whole expression