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    Laws of inequalities for $\mathbb{R}$

    E-MUN-4-2

    Area
    Topology
    Topics
    Integers and Real Numbers
    Status
    draft

    Prove the following “laws of inequalities” for \(\mathbb{R}\), using axioms (1)-(6) along with the results of Exercise 1:

    1. \(x > y\) and \(w > z \Rightarrow x + w > y + z\) .

    2. \(x > 0\) and \(y > 0 \Rightarrow x + y > 0\) and \(x \cdot y > 0\) .

    3. \(x > 0 \Leftrightarrow -x < 0\) .

    4. \(x > y \Leftrightarrow -x < -y\) .

    5. \(x > y\) and \(z < 0 \Rightarrow xz < yz\) .

    6. \(x \neq 0 \Rightarrow x^2 > 0\), where \(x^2 = x \cdot x\) .

    7. \(-1 < 0 < 1\)

    8. \(xy > 0 \Leftrightarrow x\) and \(y\) are both positive or both negative.

    9. \(x > 0 \Rightarrow 1 / x > 0\) .

    10. \(x > y > 0 \Rightarrow 1 / x < 1 / y.\)

    11. \(x < y \Rightarrow x < (x + y) / 2 < y\) .

    Solution

    Recall that \(x>y\) means \(x-y>0\), and that positive real numbers are closed under addition and multiplication.

    1. Since \(x-y>0\) and \(w-z>0\), \[ (x+w)-(y+z)=(x-y)+(w-z)>0, \] so \(x+w>y+z\).

    2. This is exactly closure of the positive reals under addition and multiplication: \[ x+y>0,\qquad xy>0. \]

    3. \[ -x<0\iff 0-(-x)>0\iff x>0. \]

    4. \[ -x<-y\iff (-y)-(-x)=x-y>0\iff x>y. \]

    5. Since \(x-y>0\) and \(-z>0\), \[ (x-y)(-z)>0. \] But this equals \(yz-xz\). Hence \(yz>xz\), i.e. \[ xz<yz. \]

    6. If \(x\ne0\), trichotomy gives either \(x>0\) or \(-x>0\). In the first case \(x^2>0\). In the second, \[ x^2=(-x)^2>0. \] Thus \(x^2>0\) whenever \(x\ne0\).

    7. Since \(1\ne0\), part (f) gives \(1=1^2>0\). Part (c) then gives \(-1<0\). Hence \[ -1<0<1. \]

    8. If \(x,y>0\), then \(xy>0\). If \(x,y<0\), then \(-x,-y>0\) and \[ xy=(-x)(-y)>0. \] Conversely suppose \(xy>0\). Neither factor is zero. If \(x>0\) and \(y<0\), then applying part (e) to \(x>0\) and the negative multiplier \(y\) gives \(xy<0\), a contradiction. Hence \(y>0\). Similarly, if \(x<0\), then \(y<0\). Thus \(xy>0\) exactly when the two factors have the same sign.

    9. Let \(x>0\). By E-MUN-4-1(p), \(1/x\ne0\). If \(1/x<0\), then a positive number times a negative number is negative, so \[ 1=x(1/x)<0, \] contradicting part (g). Therefore \[ 1/x>0. \]

    10. If \(x>y>0\), then \(xy>0\), and \[ \frac1y-\frac1x=\frac{x-y}{xy}>0. \] Hence \[ \frac1x<\frac1y. \]

    11. If \(x<y\), part (a) gives \[ 2x=x+x<x+y<y+y=2y. \] Also \(2=1+1>0\), so part (i) gives \(1/2>0\). Multiplication by the positive number \(1/2\) preserves inequalities, hence \[ x<\frac{x+y}{2}<y. \]

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