Rules of Calculation with Inequalities

Elementary Proof

(related to "Rules of Calculation with Inequalities")

Proof of Rule 1) For any \(x\neq 0\) it is \(x \cdot x >0\).

If \(x > 0\), this rule follows from the third ordering axiom.
If \(x < 0\), then by definition it is equivalent with \(-x > 0\), where again, the rule follows from the third ordering axiom.

Proof of Rule 2) \(1 > 0 \).

From the uniqueness of 1 and the uniqueness of 0, it follows that \(1 \neq 0\). Therefore, according to the first ordering axiom, it must be either \(1 > 0\) or \(-1 > 0\). Since \(1\cdot 1 =1\) , it must be \(1 > 0 \), according to the rule Rule 1).

Proof of Rule 3) 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).

\(x > y\) means, by definition, that \(x - y > 0\). Therefore, it is \(x - y + a - a > 0\), and so it follows that \(x + a - (y + a) > 0\), and finally that \(x + a > y + a \).

Proof of Rule 4) 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).

By definition, it is \( x - y > 0 \) and \( a - b > 0 \). From the second ordering axiom, it follows that \( (x - y) + (a -b) > 0\), which by definition results in \(x + a > y + b\).

Proof of Rule 5) Transitivity of the “\( < \)” (analogously the “\( > \)”) relation

If \(x < y\) and \(y < z\), then by definition it is \(0 < y - x\) and \(0 < z - y\). From the second ordering axiom, it follows that \( 0 < y - x + z - y= z - x\), which by definition results in \(x < z\).

Proof of Rule 6) 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).

By definition, it is \(x -y > 0\). From the third ordering axiom, it follows that \( a(x-y) = ax - ay > 0\), which, by definition, results in \(ax > ay\).

Proof of Rule 7) 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 “\( > \)”).

By definition, it is \(x - y > 0\) and \(-a > 0\). From the third ordering axiom, it follows that \( (-a)(x-y) = -ax + ay > 0\), which, by definition, results in \(ay > ax\), and finally in \(ax < ay\).

Proof of Rule 8) If \(x > 0\), then \(x^{-1} > 0\).

It is \(x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x\), which is greater then \(0\), following the precondition, the Rule 1) and the third ordering axiom.

Proof of Rule 9) If \(x < 0\), then \(x^{-1} < 0\).

It is \(x^{-1}=x^{-1}\cdot 1=x^{-1}\cdot(x^{-1}\cdot x) =(x^{-1}\cdot x^{-1})\cdot x\), which is smaller then \(0\), following the precondition, the Rule 1) and the Rule 7).

Proof of Rule 10) If \(y > x > 0\), then \(x^{-1} > y^{-1} \).
(The rule is analogously valid and proven for the inequality \(0 < x < y\), resulting in \(y^{-1} < x^{-1}\)).

According to the precondition \(x\) and \(y\) are both positive. Therefore, it follows from the third ordering axiom that \(xy > 0\), and according to Rule 8) \((xy)^{-1} > 0\). Multiplying the inequality \(y > x\) by the number \((xy)^{-1}\) does not change the inequality by Rule 6), therefore \(y(xy)^{-1} > x (xy)^{-1}\). This is equivalent to \(x^{-1} > y^{-1}\).

Proof of Rule 11) If \(0 \le x < y\) and \(0 \le a < b\) then \(ax < by\).
(The rule is analogously valid and proven for \(by > ax\), following from \(y > x \ge 0\) and \(b > a \ge 0\)).

The argument is valid, if \(x=0\) or \(a=0\), since then \(0 < by\) follows from the third ordering axiom. So assume \(0 < x < y\) and \(0 < a < b\). In this case we have \(ax < ay\) and \(ay < by\), following in both cases from the Rule 6. The argument follows now from the transitivity (Rule 5)).

Contribute to BoP:

add a new Open Problem  N

add a new Comment (Branch)  N