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Linear Algebra, Theory And Applications, 2012a

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1.3. THE NUMBER LINE AND ALGEBRA OF THE REAL NUMBERS 13<br />

the following algebra properties, listed here as a collection of assertions called axioms. These<br />

properties will not be proved which is why they are called axioms rather than theorems. In<br />

general, axioms are statements which are regarded as true. Often these are things which<br />

are “self evident” either from experience or from some sort of intuition but this does not<br />

have to be the case.<br />

Axiom 1.3.1 x + y = y + x, (commutative law for addition)<br />

Axiom 1.3.2<br />

x +0=x, (additive identity).<br />

Axiom 1.3.3 For each x ∈ R, there exists −x ∈ R such that x +(−x) =0, (existence of<br />

additive inverse).<br />

Axiom 1.3.4 (x + y)+z = x +(y + z) , (associative law for addition).<br />

Axiom 1.3.5 xy = yx, (commutative law for multiplication).<br />

Axiom 1.3.6 (xy) z = x (yz) , (associative law for multiplication).<br />

Axiom 1.3.7 1x = x, (multiplicative identity).<br />

Axiom 1.3.8 For each x ≠0, there exists x −1 such that xx −1 =1.(existence of multiplicative<br />

inverse).<br />

Axiom 1.3.9 x (y + z) =xy + xz.(distributive law).<br />

These axioms are known as the field axioms and any set (there are many others besides<br />

R) which has two such operations satisfying the above axioms is called a field. Division and<br />

subtraction are defined in the usual way by x − y ≡ x +(−y) andx/y ≡ x ( y −1) .<br />

Here is a little proposition which derives some familiar facts.<br />

Proposition 1.3.10 0 and 1 are unique. Also −x is unique and x −1 is unique. Furthermore,<br />

0x = x0 =0and −x =(−1) x.<br />

Proof: Suppose 0 ′ is another additive identity. Then<br />

0 ′ =0 ′ +0=0.<br />

Thus 0 is unique. Say 1 ′ is another multiplicative identity. Then<br />

1=1 ′ 1=1 ′ .<br />

Now suppose y acts like the additive inverse of x. Then<br />

Finally,<br />

and so<br />

−x =(−x)+0=(−x)+(x + y) =(−x + x)+y = y<br />

0x =(0+0)x =0x +0x<br />

0=− (0x)+0x = − (0x)+(0x +0x) =(− (0x)+0x)+0x =0x<br />

Finally<br />

x +(−1) x =(1+(−1)) x =0x =0<br />

and so by uniqueness of the additive inverse, (−1) x = −x.

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