◀ ▲ ▶Branches / Number-systems-arithmetics / Proof

Proof

(related to Proposition: Transitivity of the Order Relation of Natural Numbers)

\( (i)\) Proof of the transitivity law \(x < y\wedge y < z\Longrightarrow x < z\).

Assume that for any natural numbers \(x,y,z\in\mathbb N\), the order relations * \(x < y\) and \(y < z\) are fulfilled. Then there exist natural numbers \(u\neq 0\) and \(v\neq 0\) with \(y=x+u\) and \(z=y+v\). From the associativity of adding natural numbers, it follows

\[z=y+v=(x+u)+v=x+(u+v).\] Since \(u+v\neq 0\), it follows \(x < z\).

\( (ii)\) Proof of the transitivity law \(x \le y\wedge y \le z\Longrightarrow x \le z\).

It follows from \((i)\) and the transitivity of the equality relation "\(=\)".

\( (iii)\) Proof of the transitivity law \(x > y\wedge y > z\Longrightarrow x > z\).

Analogously to \((i)\), there exist natural numbers \(u\neq 0\) and \(v\neq 0\) with \(x=y+u\) and \(y=z+v\). From the associativity of adding natural numbers, it follows

\[x=y+u=(z+v)+u=z+(v+u).\] Since \(v+u\neq 0\), it follows \(x > z\).

\( (iv)\) Proof of the transitivity law \(x \ge y\wedge y \ge z\Longrightarrow x \ge z\).

It follows from \((iii)\) and the transitivity of the equality relation "\(=\)".

Thank you to the contributors under CC BY-SA 4.0!

Github:

References

Bibliography

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