SOLUTIONS TO A FEW OF THE ADDITIONAL EXERCISES ON THE ALGEBRAIC
PROPERTIES OF THE REAL NUMBER SYSTEM
Additional Exercises: Some more Basic Algebraic Properties of R
Prove each of the following propositions.
(1) Zero is its own additive inverse; that is, −0 = 0. (Hint: This really is trivial, but there is something to say! What
is the defining property of −0?)
Proof. 0 + 0 = 0, by the defining property of 0; therefore, 0 is its own additive inverse (by the defining property
of an additive inverse).
Alternative proof:
Proof. −0 = (−1) · 0, by the result of Exercise 7 on the previous handout. (−1) · 0 = 0, by the result of Exercise
5 from the previous handout.
Comment: Clearly we need to use the defining property of 0 to prove this! That is all we know about 0.
(2) One is its own multiplicative inverse; that is, 11 = 1.
Proof. 1 · 1 = 1 by the defining property of 1; therefore, 1 is its own multiplicative inverse.
Alternative proof:
Proof.
1
1
= 1 · 11 , by the defining property of 1. 1 ·
1
1
= 1, by the definition of a multiplicative inverse.
Comment: Clearly we need to use the defining property of 1 somewhere!
(3) −(−x) = x.
Proof. Let x ∈ R. x + (−x) = 0, by the definition of additive inverse. But, by the definition of additive inverse,
the equation above means that x is the additive inverse of −x. In other words, x = −(−x).
(7) x/1 = x.
Proof. x/1 = x( 11 ), by the definition of division. By the result of exercise (2) above,
x/1 = x( 11 ) = x · 1 = x, by the defining property of 1.
1
1
= 1.
Therefore,
y
1
(8) If x 6= 0 and y 6= 0, then “ ” = .
x
x
y
y x
Proof. Assume x 6= 0 and y 6= 0. By the definition of division, x
· y = y · x1 · x · y1 . (These multiplicative inverses
exist because x 6= 0 and y 6= 0.) From the commutativity of multiplication, the definition of multiplicative inverse,
y x
y
and the defining property of 1, it follows that x
· y = 1. Thus, by definition, x
= “ x1 ” , the multiplicative inverse
of
y
x
.
y
(9) If y 6= 0, (−x)/y = x/(−y) = −(x/y).
Proof.
x
y
+
(−x)
y
=
x+(−x)
,
y
x+(−x)
y
x
.
y
by the result of exercise (9) on the previous handout. But
“ ”
(−x)
(You fill in the justifications!) Thus, by definition, y = − x
, the additive inverse of
y
=
0
y
=0·
1
y
= 0.
Alternatively, you can use the fact that −x = (−1)x. The other equation takes a few more steps, but the
1
proof is similar. Alternatively, you can first prove that −1
= −1.
1