# basic algebra

Discussion in 'Undergraduate Math' started by Chergarj, Jun 23, 2003.

1. ### ChergarjGuest

if x<y and z<q can I assert that x+z<y+q?
No.

My answer is based on intuition, and sometimes intuition by itself could be
wrong.

G C

Chergarj, Jun 23, 2003

2. ### Rich CarreiroGuest

Yes.

A quickie proof
x<y
x - y < y - y
x - y < 0

z<q
z - z < q - z
0 < q - z

Since x - y < 0 and 0 < q - z, then x - y < q - z
x - y < q - z
x - y + z < q - z + z
x - y + z < q
x - y + z + y < q + y
x + z < q + y
x + z < y + q QED

Rich Carreiro, Jun 23, 2003

3. ### VigualGuest

if x<y and z<q can I assert that x+z<y+q?

Vigual, Jun 23, 2003
4. ### George CoxGuest

It depends on what rules govern "<" and "+". In a ring a set P of
so-called positive elements has the property that x and y in P implies x
+ y in P, and x <y is defined to mean y - x in P. So you have

y - x in P and q - z in P,
hence

(y - x) + (q - z) in P

Now ring manipulations give

(y + q) - (x + z) in P.

Though rings are the usual setting for such things, you can do the same
in an additive group. But can it be done in _your_ system? I don't
know.

GC

George Cox, Jun 23, 2003
5. ### Stan BrownGuest

Yes, and what is more, your assertion will be correct. Stan Brown, Jun 24, 2003