# Proof

Let $$x,y,z$$ be natural numbers with $$z\neq 0$$.

### $$(i)$$ "$$\Rightarrow$$"

According to the definition of order relation of natural numbers, we there exists a natural number $$u\neq 0$$ with $$y=x+u$$. By virtue of the distributivity law for natural numbers we get $yz=(x+u)z=xz+uz.$

Since $$uz\neq 0$$ by hypothesis, it follows, $$xz < yz$$.

### $$(ii)$$ "$$\Leftarrow$$"

Assume, $$xz < yz$$, but not $$x < y$$. According to the trichotomy of the order relation for natural numbers, we must have otherwise $$x = y$$ or $$x > y$$. If $$x = y$$, it would follow from the cancellation law for multiplying natural numbers that $$xz = yz$$, which is a contradiction to the assumption $$x z < y z$$. If $$x > y$$, then it would exist a natural number $$v\neq 0$$ with $$x=y+v$$. Then we would get $$(y+v)\cdot z < y z$$, or equivalently (applying the distributivity law for natural numbers) $$yz+vz < yz$$. This is again a contradiction, since $$yz+vz > yz$$.Thus, we must have $$x < y$$.

#### $$(1a)$$ Proof of $$x < y\Longleftrightarrow zx < zx$$

Follows from $$(1)$$ and the commutativity of multiplying natural numbers.

#### $$(2)$$ Proof of $$x > y\Longleftrightarrow x z > y z$$ and of $$x > y\Longleftrightarrow z x > z x$$

The proof is analogous to the proof of $$(1)$$ and $$(1a)$$, for symmetry reasons.

#### $$(3)$$ Proof of $$x \le y\Longleftrightarrow x z \le y z$$ and of $$x \le y\Longleftrightarrow z x \le z y$$

In the "$$<$$" case, the proof is identical to the proof $$(1)$$ or $$(1a)$$, for symmetry reasons. For the "$$=$$" case, the proof is identical to the proof of the cancellation law for multiplying natural numbers.

#### $$(4)$$ Proof of $$x \ge y\Longleftrightarrow x z \ge y z$$ and of $$x \ge y\Longleftrightarrow z x \ge z y$$

The proof is analogous to the proof of $$(3)$$, for symmetry reasons.

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### References

#### Bibliography

1. Landau, Edmund: "Grundlagen der Analysis", Heldermann Verlag, 2008