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I feel a little bit stupid asking this;

I am asked to prove that, for all rational numbers if, x < y and y < z then x < z.

I have said this;

$ x + 0 < y $

$ x - z + z < y$

$ x - z < y- z $

but $ y - z < 0$

so $ x - z < y- z $ implies $ x - z < 0 $.

I had a little search before posting this just to make sure its not a duplicate, if it is i will delete it right away, sorry and thanks in advance.

ORDERING OF THE RATIONALS:

Let x and y be rational numbers. We say that x > y iff x - y is a positive rational number, and x < y iff x - y is a negative rational number.

I think this is the information that was missing.

vadim123
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    It's not correct. You used, in the last step, that $x-z<y-z$ and $y-z<0$ implies $x-z<0$. But this is again the original statement you want to prove. You cannot use a statement in its own proof. – Peter Franek Feb 01 '15 at 18:29
  • but the hypothesis says that if x < y and y < z, then... so surely i can use y < z thus y - z < 0, and since x - z < y - z < 0, its true that x - z < 0?? –  Feb 01 '15 at 18:32
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    This question is impossible to answer unless you specify what you are allowed to use in this proof. – vadim123 Feb 01 '15 at 18:35
  • all it gives me is that x,y and z are rational. the property is called "order is transitive" –  Feb 01 '15 at 18:40
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    So how do you define $x<y$? – Carsten S Feb 01 '15 at 18:46
  • I think my edit may answer your question? –  Feb 01 '15 at 18:53

2 Answers2

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We have $$z-x=(z-y)+(y-x)$$ We know $z-y$ is a positive rational number, since $y<z$. We also know that $y-x$ is a positive rational number, since $x<y$. We now need some sort of property that the sum of two positive rational numbers is again a positive rational number. With that property we know that $z-x$ is a positive rational number, and hence $x<z$.

vadim123
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As Peter Franek pointed out in his comment, your proof is flawed.

You used if $x-z<y-z$ and $y-z<0$ then $x-z<0$

This is the same as saying if $x<y$ and $y<z$ then $x<z$.

All you have to do to see that is set $x'=x-z,y'=y-z,z'=0$

As for how to prove this, use the axioms of the real number system.

By the axiom:

$x\leq y$ and $y\leq z$ imply $x\leq z$ for any real numbers $x,y,z$.

Now, say $x<y$ and $y<z$, it's the same as $x\leq y$ and $y\leq z$ and $x\neq y$ and $x\neq z$.

This implies that $x\leq z$. What we have left to prove is that $y\neq z$

Assume $y=z$, then $x\leq y$ and $y\leq x$ and $x\neq y$, that is $x=x$ and $x\neq x$ which leads to a contradiction. Thus, $x\neq z$.

This ends the proof.

Hasan Saad
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