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This page contains central results which are well-known at senior high
school level but not always proved. In order that a proof make sense it
is important that axioms and other fundamental assumptions are stated explicitly.
This is so, in particular, when the proved results seem self-evident.
commutative law for addition
commutative law for multiplication
associative law for addition
associative law for multiplication
the number 0
the number 1
a·(1/a)=1 om a<>0
Usually the multiplikation dot is, of course, not needed.
a/b=a·(1/b) if b<>0
Multiplication and division have higher priority than addition and
Operators with the same prioriy are applied from left to right.
The order of the operations can be changed by parentheses.
The priority rules can be expressed by means of a set of formulas,
ab + cd = (ab) + (cd) , a -
b + c = (a - b) + c , -abc
= -((ab)c) .
By a formula we mean an identity, ie an equation which holds
for all allowed values of the variables.
The axioms, definitions, and priority formulas above are formulas in
If all occurences of a variable in a formula are replaced by another
variable a new formula is obtained.
Likewise, replacements of a variable by an expression enclosed in parentheses
or by a digits results in a formula.
If a subexpression within parentheses, identical to one side of a formula,
is replaced by the other side of the formula, then the two expressions
constitute the two sides of a new formula.
If a subexpression (u) within parentheses is replaced by a variable
A, not occuring before, then an identity with the new expression
at one side gives a formula on back-substituting (u) for A.
Examples: The first three types of substitution, appled
to a + b = b + a , give e g
Rewriting (simplifying) an expression means, in principle, the formation
of a series of identities by means of earlier known ones and substitution.
The correctness of the statements about substitutions is obvious since
a substitution means inserting a value for one or several variables, and
identities hold for all values of the variables.
c + b = b + c , a + (b
- c) = (b - c) + a , a + 2 = 2
+ a .
The fourth type of substitution, applied to the expression (a
- d)(b + c) , gives e g the formula
(a - d)(b + c) = (a - d)(c
+ b) , since b + c = c + b
is a formula.
As an example of the fifth type, replace (a - 2b)
in (a - 2b)(c + d) by A
, giving A(c + d) .
Then the formula A(c + d) = Ac +
Ad gives, on back-substitution, the formula
(a - 2b)(c + d) = (a - 2b)c
+ (a - 2b)d .
-(-a) = a
Proof: -(-a) = -(-a) + (-a) + a =
0 + a = a
a - a = 0
Proof: a - a = a + (-a) = 0
-a + b = b - a
Proof: -a + b = b + (-a) = b
-(a + b) = -a - b
Proof: -(a + b) = -(a + b) + 0 +
0 = 0 + 0 - (a + b) = a - a + b - b
- (a + b) = (a + b) - (a + b)
- a - b =
a + (b - c) = a + b - c
= 0 - a - b = -a - b + 0 = -a
Proof: a + (b - c) = (b - c)
+ a = b - c + a = a + b - c
a + b + c = c + a + b
Proof: a + b + c = (a + b)
+ c = c + (a + b) = (c + a) +
b = c + a + b
a + b - c = b - c + a =
-c + a + b , etc
Proof: a + b - c = (a + b)
- c = -c + (a + b) = -c + a +
b = b + (-c) + a = (b + (-c))
+ a =
a - (b - c) = a - b + c
= (b - c) + a = b - c + a
Proof: a - (b - c) = a + (-(b
+ (-c))) = a + (-b - (-c)) = a + (-(-c)
- b) = a + (c - b) = a + c -
b = a - b + c
abc = cab
Proof: abc = (ab)c = c(ab)
= (ca)b = cab
a·0 = 0
The nine axioms are exactly those defining a field in abstract algebra.
So, they hold for the rational numbers, the real numbers, and the complex
numbers separately. Some examples of the implications of the above formulas
Proof: a·0 = a·0 + ab + (-ab)
= a·(0 + b) + (-ab) = a·b
+ (-ab) = 0
a·(-b) = -ab
Proof: a·(-b) = a·(-b)
+ ab + (-ab) = a·(-b + b) + (-ab)
= a·0 + (-ab) = 0 + (-ab) = -ab
(-a)·(-b) = ab
Proof: (-a)·(-b) = -(-a)b
= -(b·(-a)) = -(-ba) = ba = ab
(1/a)·(1/b) = 1/(ab)
Proof: (1/a)·(1/b) = 1·(1/a)·(1/b)
If ab = 0 then a = 0 or
b = 0
Proof: Suppose a <> 0 . Then b
= 1·b = a·(1/a)·b = ab·(1/a)
= 0·(1/a) = 0
(a/b)·(c/d) = (ac)/(bd)
Proof: (a/b)·(c/d) = a·(1/b)·c·(1/d)
= (ac)·(1/(bd)) = (ac)/(bd)
a/b + c/b = (a + c)/b
Proof: a/b + c/b = a·(1/b)
+ c·(1/b) = (a + c)·(1/b) = (a
a·(b - c) = ab - ac
Proof: a·(b - c) = a·(b
+ (-c)) = a·b + a·(-c)
= ab + (-ac) = ab - ac
a/b - c/b = (a - c)/b
Proof: a/b - c/b = a·(1/b)
- c·(1/b) = (a - c)·(1/b)
= (a - c)/b
(ac)/(bc) = a/b
Proof: (ac)/(bc) = ac·(1/(bc))
= ac·(1/b)·(1/c) = a·(1/b)·c·(1/c)
= (a/b)·1 = a/b
a/1 = a
Proof: a/1 = a·(1/1) = a·(1·(1/1))
= a·1 = a
(ab)/b = a
Proof: (ab)/b = (ab)/(1·b)
= a/1 = a
a·(b/c) = (ab)/c
Proof: a·(b/c) = (a/1)·(b/c)
= (ab)/(1·c) = (ab)/c
(a/b)/(c/d) = (a/b)·(d/c)
Proof: (a/b)/(c/d) = ((a/b)bd)/((c/d)bd)
= (((ad)b)/b)/(((cb)d)/d) = (ad)/(cb)
= (ad)/(bc) = (a/b)·(d/c)
a/(b/c) = (ac)/b
Proof: a/(b/c) = (a/1)/(b/c)
= (a/1)·(c/b) = (ac)/(1·b)
(a/b)/c = a/(bc)
Proof: (a/b)/c = (a/b)/(c/1)
= (a/b)·(1/c) = (a·1)/(bc)
1/(a/b) = b/a
Proof: 1/(a/b) = (1·b)/a =
1/(1/a) = a
When the values take (non-negative) integer values, some of the above
formulas give the usual rules for calculation with fractions. The usual
technique for addition and subtraction, finding the least common denominator
etc, is not among the theorems above. Its correctness follows, however,
from a combination of some theorems.
The solution of the equation ax = b is, of
course, b/a for any non-zero a and any
b. The usual way to find this is to divide both sides of the equation
by a. But why is this allowed? Well, it follows from the non-existence
of zero-divisors, ie the theorem ab = 0 implies a
= 0 or b = 0. To see this, notice that b = a·b/a
, and use the equation. Then
Since a is non-zero this gives x - b/a
= 0 , x = x - b/a + b/a
= 0 + b/a = b/a .
ax - b = b - b = 0 , ax -
a·b/a = 0 , a·(x
- b/a) = 0 .
Another method: x = (ax)/a = b/a
. However, the first method can be use more generally in rings of a certain
kind (domains) provided b is a multiple of a. Then the ninth
axiom does not hold and is not needed.
The complex numbers can be defined as the field with elements
a + b·i , a, b real,
and i² = -1.
We will use the non-existence of zero-divisors to solve the equation
x² = -1 . Of course, i and -i are solutions
but we must show that these are the only ones. The equation implies
From this x - i = 0 or x + i
= 0 , i e x = ±i .
x² - (-1) = 0 , x² - i² =
0 , (x - i)(x + i) = 0 .
© Copyright Bengt Månsson 1997, email@example.com.
Last updated March 13, 2006.