Difference between revisions of "Division of Zero by Zero"

Revision as of 06:26, 29 May 2021

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 differential calculus one of the Indian-Mathematical-Scientist Jyotiraditya Jadhav has got correct solution set for the process with a proof.

About Zero and it's Operators

Discovery

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):

Now if we divide zero by zero then

$0/0$

we can write the 0 in the numerator as $(1-1)$ and again other in the denominator as $(1-1)$ ,

= $(1-1)/(1-1)$ and that equals to be $1$

Now we can write the 0 in the numerator as $(2-2)$ and again other in the denominator as $(1-1)$ ,

= $(2-2)/(1-1)$

= $2 (1-1)/(1-1)$ [Taking 2 as common]

= $2$

Now we can write the 0 in the numerator as $( infinity- infinity)$ and again other in the denominator as $(1-1)$ ,

= $( infinity-infinity)/(1-1)$

= $infinity (1-1)/(1-1)$ [Taking $infinity$ as common]

= $infinity$

So, the solution set(A) comprises of all the values of real numbers.

set(A) = $\{- infinity.....-3,-2,-1,0,1,2,3.... infinity \}$

Solution set(B):

Now if we divide zero by zero then

$0/0$

We know that the actual equation is $0^1/0^1$

= $0^1/0^1$

= 0^(1-1) [Laws of Indices, $a^m/a^n = a^m-n$ ]

= $0^0$

= $1$ [Already proven<ref>https://brilliant.org/wiki/what-is-00/</ref>]

So, the solution set(B) is a singleton set

set(B) = $\{1\}$

Now we can get a finite value to division of $0/0$ by taking intersection of both the solution sets.

Let the final solution set be $F$

$A\bigcap B$ = $F$

$\{- infinity.....-3,-2,-1,0,1,2,3....infinity \}$ $\bigcap$ $\{1\}$

$F$ = $\{1\}$

Hence proving and deriving value of $0/0 =1$

@@ Line 9: / Line 9: @@
 "'''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.
-== Jyotiraditya Jadhav Proof for Zero by Zero ==
+== Detailed proof ==
+We will form two solution sets namely set(A) and set(B)
-=== Solution Set: A : ===
-<math>0/0 </math>
-= <math>(1-1)/(1-1)  </math>
+Solution set(A):
-= 1
+Now if we divide zero by zero then
-Also,
+<math>0/0</math>
-<math>0/0  </math>
+we can write the 0 in the numerator as <math>(1-1) </math> and again other in the denominator as <math>(1-1)</math>,
-= <math>(2-2)/(1-1) = 2(1-1)/(1-1)
-  </math>
-= <math>2  </math>
+=<math>(1-1)/(1-1)</math> and that equals to be <math>1</math>
-Also,
-<math> 0/0 </math>
+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>,
-=<math>(Infinity - Infinity) / (1-1) = Infinity(1-1)/(1-1)   </math>
-= <math>Infinity  </math>
+=<math>(2-2)/(1-1)  </math>
-So, Solution set of : A is <math>\{ 1,2,3.......Infinity\}  </math>
+= <math>2 (1-1)/(1-1) </math>                                                               [Taking 2 as common]
-=== Solution Set :B: ===
+= <math>2 </math>
-<math>0/0
-= 0^1/0^1
-= 0^1-1 (a^m/a^n= a^m-n)
-= 0^0
-=1  </math>
-So, Solution set of : B is <math>\{1\}  </math>
-=== Conclusion ===
+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>,
-Intersection of both the sets will be :
-<math>A\bigcap B   </math>= <math>1  </math>
-So, we can conclude that the '''division of 0/0 is 1'''.
+=<math>( infinity-infinity)/(1-1) </math>
-__INDEX__
+= <math> infinity (1-1)/(1-1) </math>                                                             [Taking <math> infinity  </math> as common]
+= <math> infinity  </math>
+So, the solution set(A) comprises of all the values of real numbers.
+set(A) = <math>\{- infinity.....-3,-2,-1,0,1,2,3.... infinity \}  </math>
+Solution set(B):
+Now 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>                                                                                        [Already proven<ref>https://brilliant.org/wiki/what-is-00/</ref>]
+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 intersection of both the solution sets.
+Let the final solution set be <math>F </math>
+<math>A\bigcap B </math> = <math>F </math>
+<math>\{- infinity.....-3,-2,-1,0,1,2,3....infinity \}  </math> <math>\bigcap </math> <math>\{1\} </math>
+<math>F </math> = <math>\{1\} </math>
+Hence proving and deriving value of  <math>0/0 =1 </math>

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