Question

Is it possible to prove that \(a_1+b_1i=a_2+b_2i\) if and only if \(a_1=a_2\) and \(b_1=b_2\) or is it a definition?

Answer 1

i ws sure that its a property , but wanted tomake sure from my txt book this is exactly what it said :-
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Answer 2

So it is a definition and it is impossible to prove that it is true?

Answer 3

wait i missed something hehe
since its a properties , that means its axiom sys assumtion :)
and its not def , its prop
and this is what i ment to post ealiear
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Answer 4

there is a deffrence btw defenition and a proparty

Answer 5

How do you define equality? I looked up the axioms of field on Wikipedia but the axioms assume a equality relation.

Answer 6

equality is definition , also imaginary part and real part
but saying iff reals part are equal , and imaginary part are equal is a property :D

Answer 7

got it nw ?
or still confused ?
i can explain axiom system to you if you want :)

Answer 8

Maybe it is? haha
Suppose \[\Large a_1 + \color{blue}ib_1 = a_2 +\color{blue} ib_2\]

Answer 9

and I don't know what I'm doing lol

Answer 10

lolz tere , well what happen is this
1_ we assumed that this statment is true in complex plane system
" that a1+b1i=a2+b2i if and only if a1=a2 and b1=b2 " ( property)

2_and we can use this in other proofs in the same system , how ever this property it self is stick xD no way to prove or disprove it

Answer 11

Okay... idea :D
The above assumption implies that
\[\Large (a_1 - a_2) + \color{blue}i(b_1 - b_2) = 0\]

Answer 12

terez lol ur going on circle proof :P

Answer 13

Hear me out first...lol

Answer 14

We can then can take the modulus of both sides... and they should still be equal, obviously.
\[\Large (a_1- a_2)^2 + (b_1+b_2)^2 = 0\]

Answer 15

ok lol im all ears :)

Answer 16

And squares of real numbers are nonnegative.

Answer 17

And yet we have
\[\Large (a_1 - a_2)^2 = -(b_1 - b_2)^2\]

Answer 18

We can say with confidence that the left-side is nonnegative.
We can say with confidence that the right-side is nonpositive.

Answer 19

That can only be true if
\[\Large a_1 - a_2 = 0 = b_1 - b_2\]

Answer 20

Sound good? ^_^

Answer 21

haha and ?

Answer 22

Well, \[\Large a_1 - a_2 =0 \qquad \rightarrow \qquad a_1 = a_2\]
And same goes for the b's XD

Answer 23

mmm cool xD

Answer 24

next, thomas is going to ask us to prove : \(\large z_1=z_2 \implies |z_1| = |z_2|\)

Answer 25

Heaven help us D:

Answer 26

Can we define equality in the first place? If not, this is a wild goose chase.

Answer 27

Oh noes... philosophy...
Nothing to do here haha

Answer 28

wow , ur going into equality definition which is undefined term , right @ganeshie8

Answer 29

i never took analysis before, so im bit clueless on where exactly im going lol

Answer 30

"In modern first-order logic equality is primitive. We don't define it, we just know that two things are equal if and only if they are the same thing. So two numbers are equal if and only if they are the same number."

http://math.stackexchange.com/questions/855903/how-do-we-define-equality-in-real-numbers

Answer 31

so undefined term

Answer 32

Okay. Assuming that equality on real numbers is defined, can you derive the equality in complex number?

Answer 33

me ? , no :P
i stick with axioms and stay in the safe side

Answer 34

It seems like we need Zermelo–Fraenkel set theory to solve this simple question...
._.

Answer 35

yep , so you want it ?
i think i have it in my note

Answer 36

i guess all complex reasearch / books mintion that operation are the same in real or exaclty they say
The comple number system is , a natural extention of real number system

Answer 37

This question looks ridiculously complex now. First we have to use ZFC. Then we have to define real numbers. Thirdly, we can now use something called Cayley–wingspanson construction to construct the complex. Finally, we can (hopefully) derive this definition of equality.

Too much work... I give up...

Answer 38

OpenStudy is dumb. D!ckson becomes wingspanson...

Answer 39

haha lolz u dnt have to do this work , its already exist
only small lovely paper of axioms on complex u need :)

Answer 40

Where is that small and lovely piece of paper?