solve this it is clear challenge if you do it first of all you will get10points
2007-06-13 05:31:19 · 20 answers · asked by chums12000 2 in Science & Mathematics ➔ Mathematics
4-10 = 9-15
4-10+25/4 = 9-15+25/4
Factor: (2-5/2)(2-5/2) = (3-5/2)(3-5/2)
Take square root: 2-5/2 = 3-5/2
2 = 3
The fallacy is that you are taking the square root of a negative number which is imaginary. √(-6)
2007-06-13 05:33:14 · answer #1 · answered by Barkley Hound 7 · 2⤊ 1⤋
3=2?
2007-06-13 13:16:51 · answer #2 · answered by Virgilio C 2 · 0⤊ 0⤋
Clarification regarding Barkley Hound's excellent answer:
He says, "The fallacy is that you are taking the square root of a negative number which is imaginary. â(-6)"
Actually, the fallacy doesn't involve imaginary numbers. Instead it occurs when you take the square root of 1/4; and on one side of the equation you use the positive square root (1/2) and on the other side you use the negative square root (-1/2). And 1/2 is not equal to -1/2; that's the fallacy.
Here's how it happens:
One of the steps in the "proof" is:
Factor: (2-5/2)(2-5/2) = (3-5/2)(3-5/2)
If you analyze this equation, you have:
(-1/2)(-1/2) = (1/2)(1/2)
That is perfectly true. Both sides are equal to 1/4.
However, in the next step you take the square root of both sides. On the left side you get the negative square root (-1/2), and on the right side you get the positive square root (1/2). And the two square roots are NOT equal to each other; but if you pretend that they are, then you can "prove" that 3=2, which Barkley Hound does as follows:
Take square root: 2-5/2 = 3-5/2
Add 5/2: 2 = 3
The fallacy is that you are taking the square root of a negative number which is imaginary. â(-6)
2007-06-13 13:02:44 · answer #3 · answered by Anonymous · 0⤊ 0⤋
-6 = -6
9-15 = 4-10
adding 25/4 to both
sides:
9-15+(25/4) =
4-10+(25/4)
Changing the order
9+25/4-15 = 4+25/4-10
(this is just like a
square plus b square minus
two a b = a-b whole
square.)
Here a = 3,2 b=5/2
so it can be expressed
as follows
(3-5/2)(3-5/2)
= (2-5/2)(2-5/2)
taking positive square
root on both sides:
3 - 5/2 = 2 - 5/2
3 = 2
2007-06-13 12:36:15 · answer #4 · answered by M&M 5 · 0⤊ 0⤋
Quote:
Assume 3=2. The by the commutative law, 2=3. Add the two equations, and we get 5=5. Subtract 5 from both sides, and we get 0=0, which I happen to know is true
how can you assume 3=2 is true when you try to prove it
2007-06-13 12:40:03 · answer #5 · answered by Amanda 2 · 0⤊ 1⤋
Assume 2 = 1. Add 1 to each side. Then, 3 = 2.
2007-06-13 13:31:26 · answer #6 · answered by Anonymous · 0⤊ 0⤋
Assume 3=2. The by the commutative law, 2=3. Add the two equations, and we get 5=5. Subtract 5 from both sides, and we get 0=0, which I happen to know is true.
2007-06-13 12:34:31 · answer #7 · answered by Anonymous · 2⤊ 1⤋
we can only prove things that are true, 3=2 is not. you can contruct a "proof" but it will definitely have some bad math in it...more than likely division by zero, which makes no sense.
2007-06-13 13:02:46 · answer #8 · answered by Ervin C 1 · 0⤊ 0⤋
The only way to do this is thru bad math. I have seen some "proofs" that deal with it but they usually have some sort of error in there that is hard to notice. Most of them involve dividing by zero.
Here is a site that shows the classic "proof" of this to be false and shows why
http://www.math.utoronto.ca/mathnet/falseProofs/first1eq2.html
2007-06-13 12:34:21 · answer #9 · answered by A.Mercer 7 · 2⤊ 1⤋
My highschool math teacher made 4=7 one day just for kicks. but in reality it is not true. it' sjust the equivelant of a magic trick, you don't see the one little error int he process.
2007-06-13 12:39:47 · answer #10 · answered by Anonymous · 0⤊ 0⤋