a²-a²=a²-a²
a(a-a)=(a+a)(a-a)
a=a+a
1a=2a
1=2
2006-09-02 00:17:01 · 20 answers · asked by Jack S 1 in Science & Mathematics ➔ Mathematics
yu cant cancel (a-a) in this because any common term can be cancelled only if it is non zero. here (a-a) = 0, for any value of a.
so this assumption is wrong & hence 1 isn't equal to 2.
2006-09-02 00:24:00 · answer #1 · answered by joycyrus83 2 · 0⤊ 0⤋
a²-a²=a²-a²
a(a-a)=(a+a)(a-a)
a=a+a
1a=2a
1=2
False. a-a = 0 so...
a(a - a) = (a + a)(a - a)
a(0) = (a + a)0
0 = 0
When you divided by (a - a) you actually divided by 0. Division by 0 is impossible (you can't divide something into no parts. So that's why it comes out 1 = 2, or "no solution". But the property that says a number minus itself equals 0 (forget what it's called) makes the equation work.
The reflexive property of equality says already that a²-a²=a²-a²
. So if you say 1 = 2 (which is false), then logical reasoning requires the statement to be false.
2006-09-02 10:11:02 · answer #2 · answered by j 4 · 0⤊ 0⤋
1=0
2006-09-02 07:20:08 · answer #3 · answered by Diamond in the Rough 6 · 0⤊ 0⤋
I'll proof to you that 4=5
Theorem: 4 = 5
Proof:
-20 = -20
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5
2006-09-02 08:33:35 · answer #4 · answered by Edmond 2 · 0⤊ 0⤋
Depends on what you believe in...
Your answer can also be 1 = 0 or 2 = 0
in the first place you could subtract already the two a^2 in the right and do what you did in the the other two questions...
back to the question...
False!!!
yet
True!!!
2006-09-02 07:25:30 · answer #5 · answered by Hi-kun 2 · 0⤊ 0⤋
False ofcourse. Is 1=2??
2006-09-02 07:20:31 · answer #6 · answered by keerthu_93 2 · 0⤊ 0⤋
1) True! They BOTH equal zero!
2) True
3) False (a+a=a+a)
4) False ( 1a=a and 2a= 2*a)
5) False
2006-09-02 07:29:23 · answer #7 · answered by seraphimpatriot 1 · 1⤊ 0⤋
The only reason it works the first two times is because 0 is equal to zero. When you divide by a-a, you are effectively dividing by 0, which is not allowed. Therefore, your proof, though convincing, has a fundamental logical error.
2006-09-02 11:02:34 · answer #8 · answered by Anonymous · 0⤊ 0⤋
The last result is false, but where the wrong step?
a(a-a)=(a+a)(a-a) true
a=a+a false
u divided over (a-a)
but a-a = 0 while you can't divide over zero
2006-09-02 07:25:51 · answer #9 · answered by phantom_man17 4 · 0⤊ 0⤋
An oldie but a goodie. . .
I used this way back in 9th grade algebra to trick the class (I got the class but my teacher caught it).
You are dividing by zero, a big no no in mathematics.
2006-09-02 07:51:00 · answer #10 · answered by Walking Man 6 · 0⤊ 0⤋