Guide To Walter Rudin's Principles, 1.15, 1.16 (Proof Details)

Rudin Principles 1.15.a. Prove:
$x \not = 0 \land xy = xz \implies y=z$
$$ \begin{array}{l c} y =y & \\ y = y \times 1 & multiplicative \ identity\\ y = y \times (x \times x^{-1}) & multiplicative \ inverse \ x \times x^{-1} = 1 \\ y = (y \times x) \times x^{-1} & associativity \\ y = (z \times x) \times x^{-1} & substitute \ given \ xy=xz \\ y = z \times (x \times x^{-1}) & associative \ given \ xy=xz \\ y = z \times 1 & multiplicative \ inverse \\ y = z & multiplicative \ identity \end{array} $$ 1.15.b. Prove:
$x \not = 0 \land xy = x \implies y=1$
$$ \begin{array}{l c} y = y & \\ y = y \times 1 & multiplicative \ identity\\ y = y \times (x \times x^{-1}) & multiplicative \ inverse \ x \times x^{-1} = 1 \\ y = (y \times x) \times x^{-1} & associativity \\ y = x \times x^{-1} & substitute \ given \ xy=x \\ y = 1 & multiplicative \ inverse \end{array} $$ 1.15.c. Prove:
$x \not = 0 \land xy = 1 \implies y=x^{-1}$
$$ \begin{array}{l c} y =y & \\ y = y \times 1 & multiplicative \ identity \\ y = y \times (x \times x^{-1}) & multiplicative \ inverse \ x \times x^{-1} = 1 \\ y = (y \times x) \times x^{-1} & associativity \\ y = 1 \times x^{-1} & substitute \ given \ xy=1 \\ y = x^{-1} & multiplicative \ identity \end{array} $$ 1.15.d.
Given $x \not = 0$ Prove: $(x^{-1})^{-1} = x$
$$ \begin{array}{l c} x = x & \\ x = x \times 1 & multiplicative \ identity \\ x = x \times (x^{-1} \times (x^{-1})^{-1}) & mult \ inverse \ (x^{-1} \times (x^{-1})^{-1} = 1) \\ x = (x \times x^{-1}) \times (x^{-1})^{-1} & associativity \\ x = (1) \times (x^{-1})^{-1}) & mult \ inverse \ x \times x^{-1} = 1 \\ x = (x^{-1})^{-1} & multiplicative \ identity \end{array} $$
1.16.a. Prove
$0x=0$ in the field of reals $\mathbb{R}$
We recognize this is a multiplication with the additive identity $x+0=0$ with $x=0$ so $0+0=0$. We have the distributive property of multiplication over addition [see field axioms]. $a(b+c) = ab+ac$ $$ \begin{array}{l c} 0x = 0x & \\ 0x = (0+0)x & additive \ identity \ 0 + 0 = 0 \\ 0x = 0x + 0x & distributive \\ \hline 0x = 0x + 0 & \\ 0x = 0x + (0x - 0x) & additive \ inverse \ 0x + (-0x) = 0 \\ 0x = (0x + 0x) - 0x & associative \\ 0x = 0x - 0x & using \ result \ above \ line \ 0x = 0x + 0x \\ 0x = 0 & additive \ inverse \ 0x + (-0x) = 0 \end{array} $$ 1.16.b. Prove
$x \not = 0 \land y \not = 0 \implies xy \not = 0$
Do proof by contradiction with $x \not = 0, y \not = 0$ and $xy=0$ $$ \begin{array}{l c} 1 \times 1 = 1 & mult \ identity \\ (x \times x^{-1}) \times (y \times y^{-1}) = 1 & mult \ inverse \ x \times x^{-1} = 1\\ xy (xy)^{-1} = 1 & if \ 1 \ here, cannot \ be \ 0\\ \hline 0 (xy)^{-1} = 0 & by \ 0x=0 \ and \ xy=0 \end{array} $$ A contradiction, therefore, the equation is not not true [=true].

1.16.c. Prove
$(-x)y=-(xy)=x(-y)$ $$ \begin{array}{l c} 0y = 0 & \\ (-x + x)y = 0 & \\ (-x)y + xy = 0 & \\ \hline (-x)y = -xy & 1.14.c \ \ \ x+y = 0 \implies y=-x \end{array} $$ 1.16.d. Prove
$(-x)(-y)=xy$
$$ \begin{array}{l c} xy = -(-(xy)) & 1.14.d \\ xy = -(x(-y)) & 1.16.c \\ xy = -(xz) & z = (-y)\\ xy = (-x)z & 1.16.c\\ xy = (-x)(-y) & z = (-y) \ \\ \end{array} $$

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