Difference between revisions of "Division of Zero by Zero"
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== Detailed proof == | == Detailed proof == | ||
− | We will form two solution sets namely set(A) and set(B) | + | We will form two solution sets (namely set(A) and set(B)) |
Solution set(A): | Solution set(A): | ||
− | + | If we divide zero by zero then | |
<math>0/0</math> | <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>, | |
− | =<math>(1-1)/(1-1)</math> | + | =<math>(1-1)/(1-1)</math> equaling <math>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>, | |
Line 35: | Line 35: | ||
− | + | We can even write the 0 in the numerator as <math>( \infty- \infty) </math> and in the denominator as <math>(1-1)</math>, | |
− | =<math>( | + | =<math>( \infty-\infty)/(1-1) </math> |
− | = <math> | + | = <math> \infty(1-1)/(1-1) </math> [Taking <math> \infty</math> as common] |
− | = <math> | + | = <math> \infty</math> |
− | So, the solution set(A) comprises of all | + | So, the solution set(A) comprises of all real numbers. |
− | set(A) = <math>\{- | + | set(A) = <math>\{- \infty.....-3,-2,-1,0,1,2,3.... \infty\} </math> |
Solution set(B): | Solution set(B): | ||
− | + | If we divide zero by zero then | |
<math>0/0</math> | <math>0/0</math> | ||
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=<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>] | + | = 0^(1-1) [Laws of Indices, <math>a^m/a^n = a^{m-n} </math>] |
= <math>0^0 </math> | = <math>0^0 </math> | ||
− | =<math>1 </math> [ | + | =<math>1 </math> [https://brilliant.org/wiki/what-is-00| Already proven] |
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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. | + | 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> | Let the final solution set be <math>F </math> | ||
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<math>A\bigcap B </math> = <math>F </math> | <math>A\bigcap B </math> = <math>F </math> | ||
− | <math>\{- | + | <math>\{- \infty.....-3,-2,-1,0,1,2,3....\infty\} </math> <math>\bigcap </math> <math>\{1\} </math> |
<math>F </math> = <math>\{1\} </math> | <math>F </math> = <math>\{1\} </math> | ||
− | Hence proving | + | Hence proving <math>0/0 =1 </math> |
Latest revision as of 19:07, 4 February 2023
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):
If we divide zero by zero then
We can write the 0 in the numerator as and in the denominator as ,
= equaling
We can then write the 0 in the numerator as and in the denominator as ,
=
= [Taking 2 as common]
=
We can even write the 0 in the numerator as and in the denominator as ,
=
= [Taking as common]
=
So, the solution set(A) comprises of all real numbers.
set(A) =
Solution set(B):
If we divide zero by zero then
We know that the actual equation is
=
= 0^(1-1) [Laws of Indices, ]
=
So, the solution set(B) is a singleton set
set(B) =
Now we can get a finite value to division of by taking the intersection of both the solution sets.
Let the final solution set be
=
=
Hence proving