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The question was prove that between any 2 rational number , there is a third rational.

x,y ,z are elements of Q

(for all x ) (for all y) (there exist z)[x>z>y] <->

(for all x ) (for all y) (there exist z)[(x>z) ^ (z>y)]

Proof by contradiction:

Suppose its false that for any x and y , there exists a z between x and y

~((for all x ) (for all y) (there exist z)[x>z>y])

(there exists x) (there exists y)( for all z)[ (x< or = z) V (z < or = y)]

There is no x that is smaller than or equals to any z.

There is no y that is larger than or equals to any z.

Both are false, the disjunction is false.

Therefore the statement (there exists x) (there exists y)( for all z)[ (x< or = z) V (z < or = y)]

is false and the statement (for all x ) (for all y) (there exist z)[x>z>y] is true.