Guide To Walter Rudin's Principles, 1.14 (Proof Details)
Proofs - Rudin Principles 1.14
It is useful to read about Abelian groups first. In particular, review inverse and identity elements. Any book on modern or abstract algebra should cover this. In the field of reals $\mathbb{R}$, prove:
1.14.a. if $ x+y = x+z$ then $y=z$
We need to prove $y=z$ under the condition given. Rudin starts with $y=y$ (although not shown in the text) $$ \begin{array}{l c} y = y & \\ y = 0+y & additive \ \ identity\\ y = (x-x) +y & additive \ \ inverse\\ y = x - (x+y) & associative \ \ property\\ y = x - (x+z) & equation \ \ given \ \ (substitute \ \ x+y = x+z )\\ y = (x - x)+z & associative \\ y = 0 + z & additive \ \ inverse \\ y=z \end{array} $$ 1.14.b. if $ x+y = x$ then $y=0$
Note: this proof asserts uniqueness of the additive identity element.
Why can we not substitute $y=0$ right away? We would get $x+0=x$ and prove $x=x$.
Rudin says substitute $z=0$ in 1.14.a and follow the same path. So starting with $y=y$ $$ \begin{array}{l c} y = y & \\ y = y + 0 & additive \ \ identity\\ y = y + (x - x) & \\ y = (y + x) - x & \\ y = x-x & substitute \ \ x+y = x\\ y=0 \end{array} $$ 1.14.c. if $ x+y = 0$ then $y=-x$
$$ \begin{array}{l c} y = y & \\ y = y + 0 & additive \ \ identity\\ y = y + (x - x) & \\ y = 0 -x & substitute \ \ x+y = 0\\ y = -x & \end{array} $$ 1.14.d: prove $-(-x)=x$
Rudin says start with (1.14.c) and put $-x$ where $x$ is:
if $ -x+y = 0$ then $y=-(-x)$ but we interpret this as ($y=x$)
if $-x + x = 0 $ then $x = -(-x)$
Additive inverse definition is $x + (-x) = 0 $ and putting $-x$ for each $x$ we get $-x + (-(-x)) = 0 $
$$ \begin{array}{l c} x = x & \\ x = x + 0 & additive \ identity \ x + 0 = x\\ x = x + (-x) - (-x) & additive \ inverse \ (-x) + (- (-x)) = 0\\ x = (x + (-x)) - (-x) & associativity \\ x = 0 - (-x) & additive \ inverse \ x + (-x) = 0 \\ x = -(-x) \end{array} $$
It is useful to read about Abelian groups first. In particular, review inverse and identity elements. Any book on modern or abstract algebra should cover this. In the field of reals $\mathbb{R}$, prove:
1.14.a. if $ x+y = x+z$ then $y=z$
We need to prove $y=z$ under the condition given. Rudin starts with $y=y$ (although not shown in the text) $$ \begin{array}{l c} y = y & \\ y = 0+y & additive \ \ identity\\ y = (x-x) +y & additive \ \ inverse\\ y = x - (x+y) & associative \ \ property\\ y = x - (x+z) & equation \ \ given \ \ (substitute \ \ x+y = x+z )\\ y = (x - x)+z & associative \\ y = 0 + z & additive \ \ inverse \\ y=z \end{array} $$ 1.14.b. if $ x+y = x$ then $y=0$
Note: this proof asserts uniqueness of the additive identity element.
Why can we not substitute $y=0$ right away? We would get $x+0=x$ and prove $x=x$.
Rudin says substitute $z=0$ in 1.14.a and follow the same path. So starting with $y=y$ $$ \begin{array}{l c} y = y & \\ y = y + 0 & additive \ \ identity\\ y = y + (x - x) & \\ y = (y + x) - x & \\ y = x-x & substitute \ \ x+y = x\\ y=0 \end{array} $$ 1.14.c. if $ x+y = 0$ then $y=-x$
$$ \begin{array}{l c} y = y & \\ y = y + 0 & additive \ \ identity\\ y = y + (x - x) & \\ y = 0 -x & substitute \ \ x+y = 0\\ y = -x & \end{array} $$ 1.14.d: prove $-(-x)=x$
Rudin says start with (1.14.c) and put $-x$ where $x$ is:
if $ -x+y = 0$ then $y=-(-x)$ but we interpret this as ($y=x$)
if $-x + x = 0 $ then $x = -(-x)$
Additive inverse definition is $x + (-x) = 0 $ and putting $-x$ for each $x$ we get $-x + (-(-x)) = 0 $
$$ \begin{array}{l c} x = x & \\ x = x + 0 & additive \ identity \ x + 0 = x\\ x = x + (-x) - (-x) & additive \ inverse \ (-x) + (- (-x)) = 0\\ x = (x + (-x)) - (-x) & associativity \\ x = 0 - (-x) & additive \ inverse \ x + (-x) = 0 \\ x = -(-x) \end{array} $$
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