|
|
| (One intermediate revision by the same user not shown) |
| Line 1: |
Line 1: |
| − | '''Division of Zero by Zero''', is an unexplained mystery, since decades in the field of mathematics and is [[indeterminate]]. This is been a great mystery to solve for any mathematician and rather to use '''limits''' to set value of Zero by Zero in [[differential calculus]] one of the Indian-Mathematical-Scientist [[Jyotiraditya Jadhav]] has got correct solution set for the process with a proof. | + | '''Division of Zero by Zero''', is a mathematical concept and is [[indeterminate]]. |
| | | | |
| − | == About Zero and its Operators == | + | == Proof of Indeterminacy == |
| | | | |
| − | === Discovery === | + | We let <math>x=\frac{0}{0}</math>. Rearranging, we get <math>x\cdot0=0</math> there are infinite solutions for this. |
| − | The first recorded '''zero''' appeared in Mesopotamia around 3 B.C. The Mayans invented it independently circa 4 A.D. It was later devised in India in the mid-fifth century, spread to Cambodia near the end of the seventh century, and into China and the Islamic countries at the end of the eighth
| |
| | | | |
| − | === Operators ===
| |
| − | "'''Zero''' and its '''operation''' are first '''defined''' by [Hindu astronomer and mathematician] Brahmagupta in 628," said Gobets. He developed a symbol for '''zero''': a dot underneath numbers.
| |
| − |
| |
| − | == Detailed proof ==
| |
| − | We will form two solution sets (namely set(A) and set(B))
| |
| − |
| |
| − |
| |
| − | Solution set(A):
| |
| − |
| |
| − | If we divide zero by zero then we have <math>0/0</math>.
| |
| − |
| |
| − | We can write the 0 in the numerator as <math>(1-1)</math> and in the denominator as <math>(1-1)</math>, which yields <math>(1-1)/(1-1)=1</math>.
| |
| − |
| |
| − | We can then write the 0 in the numerator as <math>(2-2)</math> and in the denominator as <math>(1-1)</math>, which yields <math>(2-2)/(1-1)=2(1-1)/(1-1)</math> [Taking 2 as common]
| |
| − | <math>=2</math>
| |
| − |
| |
| − |
| |
| − | We can even write the 0 in the numerator as <math>( \infty- \infty) </math> and in the denominator as <math>(1-1)</math>,
| |
| − |
| |
| − |
| |
| − | =<math>( \infty-\infty)/(1-1) </math>
| |
| − |
| |
| − | = <math> \infty(1-1)/(1-1) </math> [Taking <math> \infty</math> as common]
| |
| − |
| |
| − | = <math> \infty</math>
| |
| − |
| |
| − |
| |
| − | So, the solution set(A) comprises of all real numbers.
| |
| − |
| |
| − |
| |
| − | set(A) = <math>\{- \infty.....-3,-2,-1,0,1,2,3.... \infty\} </math>
| |
| − |
| |
| − |
| |
| − | Solution set(B):
| |
| − |
| |
| − | If we divide zero by zero then
| |
| − |
| |
| − | <math>0/0</math>
| |
| − |
| |
| − | We know that the actual equation is <math>0^1/0^1 </math>
| |
| − |
| |
| − |
| |
| − | =<math>0^1/0^1 </math>
| |
| − |
| |
| − | = 0^(1-1) [Laws of Indices, <math>a^m/a^n = a^{m-n} </math>]
| |
| − |
| |
| − | = <math>0^0 </math>
| |
| − |
| |
| − | =<math>1 </math> [https://brilliant.org/wiki/what-is-00| Already proven]
| |
| − |
| |
| − |
| |
| − | So, the solution set(B) is a singleton set
| |
| − |
| |
| − |
| |
| − | set(B) =<math>\{1\} </math>
| |
| − |
| |
| − |
| |
| − | Now we can get a finite value to division of <math>0/0 </math> by taking the intersection of both the solution sets.
| |
| − |
| |
| − | Let the final solution set be <math>F </math>
| |
| − |
| |
| − |
| |
| − | <math>A\bigcap B </math> = <math>F </math>
| |
| − |
| |
| − | <math>\{- \infty.....-3,-2,-1,0,1,2,3....\infty\} </math> <math>\bigcap </math> <math>\{1\} </math>
| |
| − |
| |
| − | <math>F </math> = <math>\{1\} </math>
| |
| − |
| |
| − |
| |
| − | Hence proving <math>0/0 =1 </math>
| |
| | | | |
| | {{stub}} | | {{stub}} |
. Rearranging, we get 
there are infinite solutions for this.
