Prove the following “laws of algebra” for \(\mathbb{R}\), using only axioms (1)-(5):
If \(x + y = x\), then \(y = 0\) .
\(0 \cdot x = 0\) . [Hint: Compute \((x + 0) \cdot x\) .]
-0 = 0.
\(-(-x) = x\) .
\(x(-y) = -(xy) = (-x)y.\)
\((-1)x = -x\) .
\(x(y - z) = xy - xz.\)
\(-(x + y) = -x - y; -(x - y) = -x + y.\)
If \(x \neq 0\) and \(x \cdot y = x\), then \(y = 1\) .
\(x / x = 1\) if \(x \neq 0\) .
x/1 = x.
- \(x \neq 0\) and \(y \neq 0 \Rightarrow xy \neq 0\) .
\((1 / y)(1 / z) = 1 / (yz)\) if \(y, z \neq 0\) .
\((x / y)(w / z) = (xw) / (yz)\) if \(y, z \neq 0\) .
\((x / y) + (w / z) = (xz + wy) / (yz)\) if \(y, z \neq 0\) .
\(x \neq 0 \Rightarrow 1 / x \neq 0\) .
\(1 / (w / z) = z / w\) if \(w, z \neq 0\) .
\((x/y)/(w/z)=(xz)/(yw)\) if \(y, w, z \neq 0\) .
\((ax)/y = a(x/y)\) if \(y \neq 0\) .
\((-x) / y = x / (-y) = -(x / y)\) if \(y \neq 0\) .
Solution
We use only the field axioms, together with earlier parts as they are proved.
From \(x+y=x=x+0\), add \(-x\) to both sides to obtain \(y=0\).
Since \(x+0=x\), \[ (x+0)x=x^2. \] By distributivity the left side is \(x^2+0x\). Hence \(x^2+0x=x^2\), so part (a) gives \[ 0x=0. \]
Both \(-0\) and \(0\) satisfy \(0+u=0\). By part (a), \(-0=0\).
Since \[ (-x)+x=x+(-x)=0, \] \(x\) is an additive inverse of \(-x\). Additive inverses are unique, so \[ -(-x)=x. \]
We have \[ x(-y)+xy=x((-y)+y)=x0=0, \] so \(x(-y)\) is the additive inverse of \(xy\), hence \[ x(-y)=-(xy). \] By commutativity, \[ (-x)y=y(-x)=-(yx)=-(xy). \]
Taking \(x=1\) in part (e) gives \[ (-1)x=-(1x)=-x. \]
Since \(y-z=y+(-z)\), distributivity and part (e) give \[ x(y-z)=xy+x(-z)=xy-xz. \]
The element \(-x-y\) is the additive inverse of \(x+y\), since \[ (x+y)+(-x-y)=0. \] Thus \[ -(x+y)=-x-y. \] Then, using parts (d) and the first identity, \[ -(x-y)=-(x+(-y))=-x-(-y)=-x+y. \]
Suppose \(x\ne0\) and \(xy=x\). Multiply by \(1/x\): \[ (1/x)(xy)=(1/x)x, \] so \(y=1\).
By definition \(1/x\) is the multiplicative inverse of \(x\), hence \[ x/x=x(1/x)=1. \]
Since \(1\cdot1=1\), the multiplicative inverse of \(1\) is \(1\). Therefore \[ x/1=x(1/1)=x. \]
Suppose \(x,y\ne0\). If \(xy=0\), multiplying by \((1/x)(1/y)\) gives \[ 1=0, \] contrary to the field axiom \(1\ne0\). Hence \(xy\ne0\).
By part (l), \(yz\ne0\). Moreover \[ (yz)(1/y)(1/z)=(y(1/y))(z(1/z))=1. \] Thus \((1/y)(1/z)\) is the inverse of \(yz\), so \[ (1/y)(1/z)=1/(yz). \]
Using the definition of quotient and part (m), \[ (x/y)(w/z)=xw(1/y)(1/z)=\frac{xw}{yz}. \]
Since \(z/z=y/y=1\), \[ \frac{x}{y}+\frac{w}{z} =\frac{xz}{yz}+\frac{wy}{yz} =\frac{xz+wy}{yz}. \]
If \(1/x=0\), then \[ 1=x(1/x)=x0=0, \] a contradiction. Hence \(1/x\ne0\).
Since \(w,z\ne0\), part (l) implies \(w/z\ne0\). Also \[ (w/z)(z/w)=\frac{wz}{zw}=1, \] so \(z/w\) is the inverse of \(w/z\). Therefore \[ \frac1{w/z}=\frac zw. \]
Using part (q), \[ \frac{x/y}{w/z}=\frac{x}{y}\frac{z}{w}=\frac{xz}{yw}. \]
Associativity gives \[ \frac{ax}{y}=(ax)(1/y)=a(x(1/y))=a(x/y). \]
By part (e), \[ \frac{-x}{y}=(-x)(1/y)=-\bigl(x(1/y)\bigr)=-(x/y). \] Also \((-y)(-1/y)=1\), so \(1/(-y)=-(1/y)\). Hence \[ \frac{x}{-y}=x\bigl(-(1/y)\bigr)=-(x/y). \] Thus \[ \boxed{\frac{-x}{y}=\frac{x}{-y}=-\frac{x}{y}}. \]