MATH 2200 { Problem Set #1 Solutions Question 3: prove the following

MATH 2200 ? Problem Set #1 Solutions

3 Feb. 2017

Question 3: prove the following...

a) -(-x) = x Proof.

-x + -(-x) = 0 - x + -(-x) + x = 0 + x (-x + x) + -(-x) = x + 0

0 + -(-x) = x

[A.5] [A.1] [A.2, A.3] [A.5]

-(-x) = x [A.4]

Alternatively:

Proof. (Using uniqueness of the additive inverse.) x R we have both

(-x) + x = 0 [A.5]

and (-x) + -(-x) = 0 [A.5 applied to y = -x.]

So both -(-x) and x are additive inverses for -x. Thus, by uniqueness of the additive inverse, we must have

x = -(-x).

b) (i) (-x) ? y = -(x ? y)

Proof.

-x + x = 0 (-x + x) ? y = 0 ? y (-x) ? y + x ? y = 0 ((-x) ? y + x ? y) + - (x ? y) = 0 + - (x ? y) (-x) ? y + (x ? y + -(x ? y)) = - (x ? y) (-x) ? y + 0 = - (x ? y)

[A.5] [M.1] [D.L., Thm. ( x ? 0 = 0)] [A.1] [A.2, A.3, A.4] [A.5]

(-x) ? y = - (x ? y) [A.4]

1

(ii) (-x) ? (-y) = x ? y

Proof.

(-x) ? (-y) = - (x ? (-y)) (-x) ? (-y) = - ((-y) ? x) (-x) ? (-y) = (-(-y)) ? x

(-x) ? (-y) = y ? x (-x) ? (-y) = x ? y

by (i) above with y -y M.2 by (i) again, in reverse [3a] M.2

c)

(i)

If

x

=0

then

1 x

=

0.

Proof.

By

contradiction. . . Suppose

to

the

contrary

that

x=0

and

1 x

= 0.

Then

1 ?x=1

x 0?x=1

[M.5] [by assumption x = 0]

0 = 1 [Thm. (0 ? x = 0)]

This contradicts [M.4]. Thus

1 x = 0 = = 0

x

(ii)

If

x

=

0

then

1/

1 x

=

x.

Proof.

11 1/ ? = 1

xx 11 1/ ? ? x = 1 ? x xx 11 1/ ? ? x = x xx

1 1/ ? 1 = x

x 1

1/ = x x

[M.5] [M.1] [M.3, M.4] [M.5] [M.4]

2

Alternatively:

Proof. (Using uniqueness of the multiplicative inverse.) x R, x = 0, we have both

1 ? x = 1 [M.5]

x

and 11 ? = 1 [M.5 applied to y = 1/x]. x 1/x

So

both

x

and

1 1/x

are

multiplicative

inverses

for

1 x

(i.e.

numbers

which

multiplied

by

1 x

give

1).

Thus, by uniqueness of the multiplicative inverse, we must have

1 x= .

1/x

d) If x ? z = y ? z and z = 0 then x = y.

Proof.

Suppose

x?z

=y?z

and

z

= 0.

Then

1 z

exists

and

1

1

(x ? z) ? = (y ? z) ?

z

z

1

1

x? z? =y? z?

z

z

[M.1, M.5] [M.3]

x ? 1 = y ? 1 [M.5]

x = y [M.4]

e) If x = 0 then x2 > 0.

Proof. Suppose x = 0. Then either x > 0 or x < 0 [O.1]. There are two cases to consider: case x > 0:

x > 0 [by assumption] x ? x > 0 ? x [O.4]

case x < 0:

x2 > 0 [Thm. (0 ? x = 0)]

x < 0 [by assumption] x ? x > 0 ? x [Thm. 3.2.2g]

x2 > 0 [Thm. (0 ? x = 0)]

3

f) 0 < 1 Proof.

0 = 1 [M.4] 0 < 1 ? 1 [Thm. 3(e) above]

0 < 1 [M.4]

Alternatively:

Proof. (By contradiction.) Suppose on the contrary that 0 1. Since 0 = 1 [M4] this implies 0 > 1 [O1]. Suppose x R and x > 0. Then

1 ?x=1 1 [by assumption] x ? x > 1 ? x [O.4]

x2 > x [M.4]

h) If 0 < x < 1 then x2 < 1.

4

Proof. Suppose 0 < x < 1. Then

0 ................
................

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