hey all of u are play tricks any one can see it and say its correct but its sure wrong first 2=1 u cant divide by a variable number and 1=-1 there are 2 solutions 1 refused which is 1=-1 and the other 1=1 and u cant take it with another way which is the square of square root is the moduls so |-1|=1 and if we are talking about tricks could any one prove that 0=-1 and 927=23 ????
i wait for reply ..........sherif.el-sayed@student.guc.edu.eg
When replying to a thread/post please quote the post/s you are responding to.
As it is very few people reading your post will have any idea what you are
refering to.
Also read the site rules they are the second item on the first dark tool bar on
every page on this site. In particular don't use text speak abreviations.
RonL
heh, i didnt think my silly math proof i posted over 2 years agao would turn into a "i know how to make things that obviously are not equal equal" thread.
you get funny stuff like that all the time. x-1 = x+1 try to get the x's on one side, it wont work. you should just say "No Solution" rather than bragging about what you can appear true..
though i like the 1 = -1 one, that was nice.
Invalid proof - Wikipedia, the free encyclopedia
There are quite a few to see here,
Paul
Everything is fine until the step marked in red. The problem with your “proof” is in writing $\displaystyle \sqrt{-1}$ for the imaginary number i. The square-root function $\displaystyle \sqrt{x}$ is only defined for $\displaystyle x\geq0$; for negative x, it makes no sense at all!
The imaginary number i is defined by $\displaystyle \mathrm{i}^2=-1$. It is not denoted by $\displaystyle \mathrm{i}=\sqrt{-1}$!
In fact, it is my opinion that writing $\displaystyle \mathrm{i}=\sqrt{-1}$ (as you did there) is just as mathematically idiotic and meaningless as writing $\displaystyle \infty=\frac{1}{0}$!
Defined here is a slightly odd term, since as a definition it seems to have "defined" $\displaystyle i=-i$ (since once you have an $\displaystyle i$, you find that $\displaystyle -i$ also satisfied the definition).
We extend the reals by introducing an ideal element $\displaystyle i$ with the property that
$\displaystyle i^2=-1$,
then it is not important that $\displaystyle (-i)^2=-1$.
(note that the above is the method of getting to a definition of complex numbers is not unique, nor necessarily the best method).
RonL
I completely agree with you, $\displaystyle i^2 = -1$ and not $\displaystyle i=\sqrt{-1}$. The second is how it is done in high schools, but it does not make it correct. I made this point many time on this forum.
However, once you define $\displaystyle \log z$ in complex analysis you can define $\displaystyle \sqrt{z} = |z|^{1/2} e^{i\arg z/2}$ (for $\displaystyle z\not =0$). But still the definition $\displaystyle i$ is not $\displaystyle \sqrt{-1}$.
I agree, I doubt it was meant to be taken seriously, why be upset with someone for writing a joke?
And even if it was meant to be taken seriously, I would think we would welcome when people try to disprove mathematics as if their proof is correct, then we have something we really need to think about, and if it's false, then math has stood up to one more proof, it's nature of being objectively true shines through.
If questioning something isn't allowed, then that thing isn't worth believing in because it hasn't tested or verified, it hasn't stood up to any criticism. Math however, does stand up to such criticism, which is why it shouldn't be discouraged, every failed attempt to disprove it only proves it further.
That is not why I do not like these questions. I dislike them because sometimes people do not agree with the answers they are given, as if they know what they are talking about. While in reality they know very little math and treat it as if they argument is worth listening to.
This is Mine 89th Post!!!
i suppose it was a kind of joke.
the purpose of this thread is to provide "proofs" that seem to prove something contradictory, but is invalid because of one or more invalid steps. and so, it is an exercise to identify what steps are incorrect. hence, i do not think anyone here actually believes the "proof" that they have posted, they are just teasing. so perhaps calling someone retarded for posting is a bit harsh...but i can't stay mad at Jane