Dividing by Zero: What's the Mistake?

  • Thread starter Thread starter Hernaner28
  • Start date Start date
  • Tags Tags
    Mistake
Hernaner28
Messages
261
Reaction score
0
What's the mistake here? hehe

http://a7.sphotos.ak.fbcdn.net/hphotos-ak-snc6/64102_341098562596296_100000884664248_973353_615500462_n.jpg
 
Last edited by a moderator:
Mathematics news on Phys.org
##2(a^2-ab) = 1(a^2 - ab)##
Therefore either ##2 = 1## or ##a^2 - ab = 0##
And ##a^2 -ab## is equal to 0, of course.
 
Say you wanted to solve the quadratic x^2=x. You wouldn't begin by dividing through by x because that assumes x\neq 0. If x=0 then you've just lost that solution.

In short, dividing by zero is bad.
 
Hmm I see... it was stupid from me not realising that.. lol :D

Thanks!
 
Don't try this at home...

spikedmath-089-dont-try-this-at-home.png
 
In the first line, you say a=b
and therefore a^2=ab. a^2-ab=0.
You are dividing by zero even though it's not defined in mathematics. That's the mistake... :D
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
2
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
1
Views
4K
Replies
47
Views
6K
Replies
2
Views
2K
Back
Top