# A^2=B^2, C^2=D^2......does (A+C)^2=(B+D)^2?

• Apr 21st 2010, 08:40 AM
sebko
A^2=B^2, C^2=D^2......does (A+C)^2=(B+D)^2?
Given
A^2=B^2, C^2=D^2

......does (A+C)^2=(B+D)^2?

I got down to this

AC=BD

I then divived both sides by "A"

C=BD/A

My proof goes as following....

It is possible that B=A.....eg A=6....B=6.....A^2=B^2.......36=36

So if this is possible. this equation breaks down to

C=D.... However it is clear that C is not always equal to D

Can anyone recommend a better proof?

Thankyou
• Apr 21st 2010, 08:49 AM
Defunkt
• Apr 21st 2010, 09:57 AM
sebko
extra note:

i ended up with

AC=BD

+-A/+-B=+-D/C
+-1=+-1

Which is not always TRUE.... all i have to do is show one case where it is not true.....

Defunkt. For A=B=1,C=-1,D=1

it becomes

1=1/-1....which is false. Do you think this proof is ok?

All im proving is that it doesnt work for all numbers of A and B... shouldnt all
i have to show is one FALSE? thanks for the input :)
• Apr 21st 2010, 12:45 PM
Defunkt
Quote:

Originally Posted by sebko
extra note:

i ended up with

AC=BD

+-A/+-B=+-D/C
+-1=+-1

Which is not always TRUE.... all i have to do is show one case where it is not true.....

Defunkt. For A=B=1,C=-1,D=1

it becomes

1=1/-1....which is false. Do you think this proof is ok?

All im proving is that it doesnt work for all numbers of A and B... shouldnt all
i have to show is one FALSE? thanks for the input :)

That is correct! :) (It is false because \$\displaystyle A^2 = B^2 = 1 \ , \ C^2 = D^2 = 1\$ but \$\displaystyle (A+C)^2 = 2^2 = 4 \neq 0 = (1 + (-1))^2 = (B + D)^2\$)