Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## Post a reply

Write your message and submit
|
Options

## Topic review (newest first)

{7/3}
2013-07-03 16:50:09

thanks.

anonimnystefy
2013-07-02 18:06:07

I do not think so.

a+bi=c+di
implies
a-c=(b-d)*i

If (b-d)!=0, then a-c is complex, which is not possible. So, b-d must be equal to 0, i.e. b must equal d. That further implies that a-c=0, i.e. a=c.

bob bundy
2013-07-02 18:02:15

I expect {7/3} meant

a = c and b = d

I think it is axiomatic rather than provable ... but as ever, it depends on the initial axioms chosen.

Bob

researching .....

LATER:

Looks like it isn't an axiom.  hhmmm.

So try a-c = i(d-b) => i = (a-c)/d-b)  =><= i is not real unless d = b

anonimnystefy
2013-07-02 16:59:49

That is not a true statement.

{7/3}
2013-07-02 15:44:01

I can't see any way to solve this
prove,a+bi=c+di implies a=b and c=d

## Board footer

Powered by FluxBB