Two proofs needed help

1) Let x,y, and z be real numers...prove the following

a) (-x)*y = -(xy) and (-x)*(-y) = xy
c) if x is not equal to 0, then (1/x) is not equal to zero and 1/(1/x) = x
d) if x>1, then x^2>x
e) if 0<x<1, then x^2<1

2) prove this
if x >=0 and x<= E for all E > 0, then x = 0

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Originally Posted by slowcurv99

1) Let x,y, and z be real numers...prove the following

a) (-x)*y = -(xy) and (-x)*(-y) = xy

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This holds in any ring. Since the reals are a ring, it certainly works.

Proof.
Use use the left-right distributive laws:
a(b+c)=ab+ac
(b+c)a=ba+ca

We know that (xy)+(-xy)=0 by definition.
That is -(xy) is an additive inverse.
We need to show that
xy+(-x)y=0
And hence an additive inverse of xy.
By uniqueness we conclude that -(xy)=(-x)y.
In order to show that, meaning,
xy+(-x)y=0
We use distrivutive laws,
xy+(-x)y=(x+(-x))y=0y=0.
Q.E.D.
(The other proofs are anagolus).

Quote:

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c) if x is not equal to 0, then (1/x) is not equal to zero and 1/(1/x) = x

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Let us argue by contradiction.

If 1/x=0
Then x(1/x)=x(0)=0
Thus, 1=0
Which is false/

Hey everyone, since everyone on this forum has been so good today to me I thought I would try and take a stab at helping someone else....please check my work and make sure I did the right thing because I do not want to lead slowcurv in the wrong direction

d) if x>1, then x^2>x
e) if 0<x<1, then x^2<1


d) x(x) > x
so
x>1

e) x(x) < 1

x < 1/x for every 0<x<1

I do not know how to do the second problem....do not assume this is right until someone else checks it because I am lost as anyone else, I just wanted to try and help out.

Thanks luckyc1423 for trying to help, does anyone know if he is right on this one. The proof he did makes since but seems like it is to simple to be the true proof. And as for the last problem I am still completely lost.