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| − | Division of '''Zero by Zero''', is an '''unexplained mystery''', since decades in field of Mathematics and is refereed as undefined. This is been a great mystery to solve for any mathematician and rather to use '''limits''' to set value of '''Zero by Zero''' in '''[https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwij6oLv_OvwAhVT83MBHc1LCzQQFjAHegQIDhAD&url=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FDifferential_calculus%23%3A~%3Atext%3DIn%2520mathematics%252C%2520differential%2520calculus%2520is%2Cthe%2520area%2520beneath%2520a%2520curve.&usg=AOvVaw1YROgVEzpqoR0TXuAWa-Ju 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]]. |
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| − | == About Zero and it's Operators == | + | == Proof of Indeterminacy == |
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| − | === 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
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| − | === Operators ===
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| − | "'''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.
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| − | == Detailed proof ==
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| − | We will form two solution sets namely set(A) and set(B)
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| − | Solution set(A):
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| − | Now if we divide zero by zero then
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| − | <math>0/0</math>
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| − | we can write the 0 in the numerator as <math>(1-1) </math> and again other in the denominator as <math>(1-1)</math>,
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| − | =<math>(1-1)/(1-1)</math> and that equals to be <math>1</math>
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| − | Now we can write the 0 in the numerator as <math>(2-2) </math> and again other in the denominator as <math>(1-1)</math>,
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| − | =<math>(2-2)/(1-1) </math>
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| − | = <math>2 (1-1)/(1-1) </math> [Taking 2 as common]
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| − | = <math>2 </math>
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| − | Now we can write the 0 in the numerator as <math>( infinity- infinity) </math> and again other in the denominator as <math>(1-1)</math>,
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| − | =<math>( infinity-infinity)/(1-1) </math>
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| − | = <math> infinity (1-1)/(1-1) </math> [Taking <math> infinity </math> as common]
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| − | = <math> infinity </math>
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| − | So, the solution set(A) comprises of all the values of real numbers.
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| − | set(A) = <math>\{- infinity.....-3,-2,-1,0,1,2,3.... infinity \} </math>
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| − | Solution set(B):
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| − | Now if we divide zero by zero then
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| − | <math>0/0</math>
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| − | We know that the actual equation is <math>0^1/0^1 </math>
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| − | =<math>0^1/0^1 </math>
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| − | = 0^(1-1) [Laws of Indices, <math>a^m/a^n = a^m-n </math>]
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| − | = <math>0^0 </math>
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| − | =<math>1 </math> [Already proven<ref>https://brilliant.org/wiki/what-is-00/</ref>]
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| − | So, the solution set(B) is a singleton set
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| − | set(B) =<math>\{1\} </math>
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| − | Now we can get a finite value to division of <math>0/0 </math> by taking intersection of both the solution sets.
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| − | Let the final solution set be <math>F </math>
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| − | <math>A\bigcap B </math> = <math>F </math>
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| − | <math>\{- infinity.....-3,-2,-1,0,1,2,3....infinity \} </math> <math>\bigcap </math> <math>\{1\} </math>
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| − | <math>F </math> = <math>\{1\} </math>
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| − | Hence proving and deriving value of <math>0/0 =1 </math>
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. Rearranging, we get 
there are infinite solutions for this.
