Section 4: Problem 1 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Prove the following "laws of algebra" for $\mathbb{R}$ , using only axioms (1)-(5):

(a) If $x+y=x$ , then $y=0$ .

(b) $0\cdot x=0$ . [Hint: Compute $(x+0)\cdot x$ .]

(d) $-(-x)=x$ .

(e) $x(-y)=-(xy)=(-x)y$ .

(f) $(-1)x=-x$ .

(g) $x(y-z)=xy-xz$ .

(h) $-(x+y)=-x-y$ ; $-(x-y)=-x+y$ .

(i) If $x\neq0$ and $x\cdot y=x$ , then $y=1$ .

(j) $x/x=1$ if $x\neq0$ .

(k) $x/1=x$ .

(l) $x\neq0$ and $y\neq0$ $\Rightarrow$ $xy\neq0$ .

(m) $(1/y)(1/z)=1/(yz)$ if $y,z\neq0$ .

(n) $(x/y)(w/z)=(xw)/(yz)$ if $y,z\neq0$ .

(o) $(x/y)+(w/z)=(xz+wy)/(yz)$ if $y,z\neq0$ .

(p) $x\neq0$ $\Rightarrow$ $1/x\neq0$ .

(q) $1/(w/z)=z/w$ if $w,z\neq0$ .

(r) $(x/y)/(w/z)=(xz)/(yw)$ if $y,w,z\neq0$ .

(s) $(ax)/y=a(x/y)$ if $y\neq0$ .

(t) $(-x)/y=x/(-y)=-(x/y)$ if $y\neq0$ .

(a) $y=0+y=((-x)+x)+y$ $=(-x)+(x+y)=(-x)+x=0$ .

(b) Using (a), $x\cdot x=(x+0)\cdot x=x\cdot x+0\cdot x$ implies $0\cdot x=0$ .

(d) $-(-x)=-(-x)+((-x)+x)$ $=(-(-x)+(-x))+x=x$ .

(e) Using (b), $x(-y)=x(-y)+(xy+(-(xy)))$ $=(x(-y)+xy)+(-(xy))$ $=x((-y)+y)+-(xy)=-(xy)$ . Similarly for $(-x)y$ .

(f) This follows from (e).

(g) This follows from the distributive law and (e).

(h) Using the associative and commutative laws, we can omit parentheses and regroup terms in addition: $-(x+y)=-(x+y)+(x-x)+(y-y)$ $=-(x+y)+(x+y)+(-x)+(-y)$ $=-x-y$ , and similarly for $-(x-y)$ .

(i) Similar to (a), $y=((1/x)\cdot x)\cdot y$ $=(1/x)\cdot(x\cdot y)=(1/x)\cdot x=1$ .

(j) By definition, $x/x=x\cdot(1/x)=1$ .

(k) By definition, $1/1=1$ as $1\cdot1=1$ , hence, $x/1=x\cdot(1/1)=x\cdot1=x$ .

(l) From (b) it follows that there is no $z$ such that $0\cdot z=1$ . However, if $x\neq0$ and $y\neq0$ , then $(xy)\cdot((1/x)\cdot(1/y))$ $=((yx)\cdot(1/x))\cdot(1/y)$ $=(y\cdot(x\cdot(1/x)))\cdot(1/y)$ $=y\cdot(1/y)$ $=1$ , hence, $xy\neq0$ .

(m) See the second part of (l) that shows that $1/(xy)=(1/x)\cdot(1/y)$ for $x,y\neq0$ .

(n) Using associativity and commutativity we can regroup terms and then use (m): $(x/y)\cdot(w/z)=x\cdot w\cdot((1/y)\cdot(1/z))$ $=(xw)\cdot(1/(yz))$ $=(xw)/(yz)$ .

(o) Using (j) and (n), $x/y=(x/y)\cdot(z/z)$ $=(xz)/(yz)$ $=(xz)\cdot(1/(yz))$ , and similarly $w/z=(wy)\cdot(1/(yz))$ , so that we can use the distributive law.

(p) Again, from (b) it follows that if $y\cdot z=1$ for some $y$ and $z$ , then neither $y$ nor $z$ equals $0$ , but we have $x\cdot(1/x)=1$ .

(q) Using (n) and (j), $(z/w)\cdot(w/z)$ $=(zw)/(zw)$ $=1$ .

(r) Using (q) and (n), $(x/y)/(w/z)=(x/y)\cdot(z/w)$ $=(xz)/(yw)$ .

(s) Using (n) and (k), $(ax)/y=(ax)/(1\cdot y)$ $=(a/1)\cdot(x/y)$ $=a\cdot(x/y)$ .

(t) Using (f), (s) and then (f) again, we immediately get $(-x)/y=-(x/y)$ . Also, using (f) and (d), $(-1)\cdot(-1)=-(-1)=1$ , hence, $1/(-1)=-1$ , and, using (f), (n) and (f) again, $x/(-y)=(1/(-1))\cdot(x/y)$ $=(-1)\cdot(x/y)=-(x/y)$ .

Subpages

Section 4: Problem 1 Solution

Section 4: Problem 1 Solution