A doctor in Michigan has a sister who lives in California but the sister does not have any brother that lives in Michigan, how come??
:P

yes you got it ratamakai.

substitute (1-1) by a, Then

(1+1)a = a

(1+1) = 1

2 = 1

Can be solved.......but the question itself is 0 = 0.

You can't write (1-1) for 1^2. It should be 2-1.

(1^2 - 1^2) = 1 - 1
I don't see any mistakes slackdemic, 1^2 = 1 rt?

Okay,

1^2 - 1^2 = 1^2 - 1^2

(1-1) = (1-1)

Now, how the heck did you get (1+1) on LHS..........???

This had been quite a story when I was in class 8. It took me quite a while to know where something was wrong. In case you wanted to refresh yourself with class 7-8 mathematics.

Q. Prove 2=3.

Sol. -6 = -6
=> 4 - 10 = 9 - 15
=> 4 - 10 + 25/4 = 9 - 15 + 25/4
=> 2^2 - 2x2x(5/2) + (5/2)^2 = 3^2 - 2x3x(5/2) + (5/2)^2
=> (2 - 5/2)^2 = (3 - 5/2)^2
=> 2 - 5/2 = 3 - 5/2
=> 2 = 3 //Proved//

:-)

a^2 - b^2 = (a + b) (a - b)
from formula ....... nuke

can u take square root of negative number ??? doesn't it become imaginary number ? I have never seen this one, i had seen yhe one I posted in school days. If I am wrong what is the answer galaab? Good one.

There are three people and they are blindfolded. Now, the are given one cap each, two red caps and one black cap, which they wear, unware what color it is. However, they are told that either they have a black cap or a red cap and they can see two other's cap once the blindfold is removed. Also, they are told that once the blindfold is removed, and if they see any red cap, they will have raise their hand.

Now they open their eyes. One can't see what one is wearing, but can see two other's cap. One of the people their would be able to tell "I am wearing red cap!" logically. How?

Here it is wrong --->

(4-5)/2 = (6-5)/2

-1/2 = -1/2

1 = 1

Solved //

if he sees other two people both wearing red cap, he knows that he is the third one and the only one with black cap.

Typo: "Now, the are given..."

To be read: "Now, they are given..."

Ritthe, there is no square of negative number in Galaab's question. He just expanded -6 to 4 - 10 and 9 - 15. And, you can manipulate those numbers to be in (a-b)^2 = a^2 + 2.a.b + b^2.

Well, no, Ritthe.

The question says: They know that either they are given red or black. It could be all red or all black, but very true one can see two others' caps.

Also, one is able to tell he is wearing RED cap, not black.

oops...I meant (a-b)^2 = a^2 - 2.a.b + b^2.

Ok guys, see you tomorrow. I am off to bed now. Good night y'all! :)

since they were told to raise the hand when they see red, X and Y had red and Z had black, X realised he had red cap because Y was raising hand when Z had black cap, he was the one left with red cap which Y saw.

2 - 5/2 is a negative number it is like

4 = 4
(-2)^2 = 2^2
-2 = 2

Going back to Republican's milk and water problem... hopefully you can follow it:



X cups of milk------>xm
X cups of wate----->xw

Container 1: xm Container 2: xw


Step 1: xm-1m, xw+1m
(x-1)m xw+1m

Step 2: (x-1)m+ (xw+1m)/(x+1) xw+1m-(xw+1m)/(x+1)
{(x^2-1)m+xw+1m}/(x+1) {x(x+1)w+(x+1)m-xw-1m}/(x+1}

Final: (x^2m+xw)/(x+1) (x^2w+xm)/(x+1)

Since, coefficient of m in container 1= coefficient of w in container 2;
They are equal...

Correction

(2 - 5/2)^2 = (3 - 5/2)^2
=> 2 - 5/2 = -(3 - 5/2)
=> -1/2 = -1/2

Ignore last post:

going back to Republican's milk and water problem... hopefully you can follow it: