# Is my proof valid?

1. Oct 5, 2013

### kaos

Can someone check if my proof is correct.Please exscuse the bad notation, i've no idea how to type the symbols.
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)]

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.

2. Oct 5, 2013

### haruspex

Since you have not used any facts about rational numbers, it seems vanishingly unlikely that your proof is valid.
How about doing something really simple and obvious: given two rationals p/q and r/s construct a rational that lies between them.

3. Oct 5, 2013

### kaos

If i construct a rational in between p/q and r/s , i doesnt apply to any other rationals, so it doesnt really prove anything. Am i misinterpreting your statement ( im really bad at math so please excuse my lack of ability)?

4. Oct 5, 2013

### haruspex

P, q, r and s can be any integers (q, s nonzero). If you construct a rational between p/q and r/s then you will have provided a general construction for any given pair of rationals.

5. Oct 5, 2013

### Ibix

p/q and r/s are arbitrary rational numbers. Haruspex is suggesting that you construct an expression in terms of p, q, r and s that is rational and guaranteed to lie between the the two.

6. Oct 6, 2013

### kaos

Ah ok i see , thanks guys.