Corollary: Rules of Calculations with Inequalities
(related to Definition: Order Relation of Real Numbers)
Let \(x,y,z,a,b\) be (if nothing else is stated) arbitrary real numbers. The following rules are valid for manipulating inequalities:
 $x^2 > 0$ for all \(x\neq 0\).
 \(1 > 0 \).
 From \(x > y\) it follows that \(x+a > y+a\) for any \(a\in\mathbb R\). (The rule is analogously valid and proven for any inequalities with "\( < \)" in between).
 If \(x > y\) and \(a > b\), then it follows that \(x+a > y+b\). (The rule is analogously valid and proven for any inequalities with "\( < \)" in between).
 Transitivity of the "\( < \)" (analogously the "\( > \)") relation
 The inequality \(x > y\) does not change, if it is multiplied by any number \(a > 0\), i.e. it is \(ax > ay\). (The rule is analogously valid and proven for an inequalities with "\( < \)" in between).
 The inequality \(x > y\) changes, if it is multiplied by any number \(a < 0\), i.e. it is \(ax < ay\). (The rule is analogously valid and proven for an inequality with "\( < \)" in between, which then changes into "\( > \)").
 If \(x > 0\), then \(x^{1} > 0\).
 If \(x < 0\), then \(x^{1} < 0\).
 If \(y > x > 0\), then \(x^{1} > y^{1} \). (The rule is analogously valid and proven for the inequality \(0 < x < y\), resulting in \(y^{1} < x^{1}\)).
 If \(y > x > 0\), then \(x^{1} > y^{1} \). (The rule is analogously valid and proven for the inequality \(0 < x < y\), resulting in \(y^{1} < x^{1}\)).
Table of Contents
Proofs: 1
Mentioned in:
Parts: 1
Proofs: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Propositions: 17 18 19
Sections: 20
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
 Modler, F.; Kreh, M.: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition