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# 2=1, can you counter this?

speeDemon
Well, I was looking at some methods to prove impossible things.. and here is something I found on a forum
 Quote: Everyone knows this: -1/1= 1/-1 Now we will square root both sides: √-1/1 = √1/-1 Now we break up the roots: √-1 √1 --- = --- √1 √-1 The square root of a negative 1 is i and the square root of 1 is 1. In other words: i/1 = 1/i Now we multiply the entire thing by 1/2: i/2 = 1/2i Now let's add 3/(2i) to this to make the math easier. i/2 + 3/2i = 1/2i + 3/2i Now we can multiply the entire thing by i: i(i/2 + 3/2i) = i(1/2i + 3/2i) So now we expand this beast: 1^2/2 + 3i/2i = i/2i + 3i/2i We know that the square root of -1 is i, so i^2 must be -1.: -1/2 + 3i/2i = i/2i + 3i/2i Now we simplify the i's -1/2 + 3/2 = 1/2 + 3/2 Let's calculate this thing: 2/2 = 4/2 And so 1 = 2

Can anyone counter this... where did the person go wrong?
metalfreek
http://www.frihost.com/forums/vt-104254.html
Afaceinthematrix
√-1/1 = √1/-1

Now we break up the roots:

√-1 √1
--- = ---
√1 √-1

You cannot get from the first line to the second line mathematically correctly.

sqrt(a/b)=sqrt(a)/sqrt(b) if and only if a > 0 and b >= 0. So in other words, that does not work with negative numbers and, of course, b must not equal 0 or else you have a whole other set of issues that can come across...

A different set of rules come out when you're using complex numbers (numbers in the form a + bi with the assumption b=/=0).

Nice try, though.
speeDemon
well, I understood later on that you can make sqrRoot(a/b) into root a/root b, only if both are positive... though root -1 is iota (i), which is correct, but you cant turn both a and b as a root unless both are positive...

if you use this method, then right in the second step, when you get 1/i=i/1, if you multiply with i on both sides, you will get 1=i^2 so, 1=-1!!! so that's the wrong step....

well, anyway.....here is another method..

we all know that -2=-2
so, 4-6=1-3
on adding 9/4 on both sides,
4-6+9/4=1-3+9/4
if you work it out, you can apply the identitiy (a-b)^2 = a^2 + b^2 - 2ab,
so,
(2-3/2)^2=(1-3/2)^2
on sqr. rooting,
2-3/2=1-3/2
so,
2=1
infinisa
 speeDemon wrote: (2-3/2)^2=(1-3/2)^2 on sqr. rooting, 2-3/2=1-3/2 so, 2=1

Same sort of problem again. When square rooting both sides of an equation, you must take the positive square root of each side, which you didn't: the right hand side, 1-3/2 = -1/2, is the negative square root of its square.
Flarkis
a poor misconception about the result of a square root...this hurts me deep down inside
weableandbob
There's another way you can do it by dividing by zero, but I forgot the exact steps
soljarag
stuff like this reminds me how much math I have forgotten since i graduated college....
Joshwa
Guh.. this is like working with variances.

Never forget the negative root, it will come back to haunt you in multivariable calculus.
This is one of those things that if you study Maths long enough, you will find so-called contradictory proofs like this.

Almost invariably, there is some small, hidden flaw in the logic.
eday2010
2 only equals 1 when you include the rest of the equation. Like, why are you multiplying everything by 1/2? For what reason? To get to 2=1. Otherwise I can just say that 5=23 or that 4=pi of a pie.
Bikerman
Why not multiply by 1/2? If it is within the rules then you are free to do it. The point is to test the rigour of maths, not to do a practical problem using it.
eday2010
Which makes it a complete waste of time because you can make anything equal anything else. In the real world, 2 doesn't equal one. When you are making up equations, sure, 2=1. But then everything equals everything else.

Completely pointless.
Bikerman
 eday2010 wrote: Which makes it a complete waste of time because you can make anything equal anything else. In the real world, 2 doesn't equal one. When you are making up equations, sure, 2=1. But then everything equals everything else. Completely pointless.

You obviously don't understand the basics. Maths is a rigorous language and you cannot just make things up. The previous postings do not simply throw in terms illogically - there is a point (which is to test the consistency of the language).
You SAY that 2 does not equal 1, but since you cannot prove it then that is just blind faith. Maths needs more than faith, it requires proof.
eday2010
 Bikerman wrote: The previous postings do not simply throw in terms illogically

Yes they do. Look at the opening post. After stating that -1/1 =1/-1, they squareroot both sides. Why? Who said to do that? There is no reason or logic behind it. They just decided to do it. It hardly seems logical to me. UNLESS you are out to prove that 2=1 or some other fallacy. Then you can concoct any equation you want, and it can be as illogical - if not moreso - than the opening post.
Bikerman
eday2010 wrote:
 Bikerman wrote: The previous postings do not simply throw in terms illogically

Yes they do. Look at the opening post. After stating that -1/1 =1/-1, they squareroot both sides. Why? Who said to do that? There is no reason or logic behind it. They just decided to do it. It hardly seems logical to me. UNLESS you are out to prove that 2=1 or some other fallacy. Then you can concoct any equation you want, and it can be as illogical - if not moreso - than the opening post.

No, you have to stick to the rules. If you square root one side, you must do the same to the other. You might wish to square root one side for a variety of reasons...see the other thread and you might understand. Try the little problem I set for you.