Prove pi as irrational.Prove it .....Full explanation ......
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pi is a number with value 3.14159265358979323846264338327950..... represented with a symbol π. Proof of pi as irrational number requires high level calculus and integration knowledge.It is a complex explanation. But still I'll try my level best to explain it to you.
To prove any number to be irrational, we need to prove that the number is not representable in a/b form.
If you have observed any theorem proof forms, we can prove a theorem by taking the assumption as the opposite of it. I.e., here we *assume* that pi is equal to a value a/b where a and b are integers
Step-1:
Consider a function f(x)= ((x^n)(a-bx)^n)/n!
we choose that function because
(i)f(x)=f(π-x)
(ii)f(0)=0
(iii) any highest( let us think it is kth order) derivative (higher order differentiation of f(x) ) when x=0 is always equal to an integer.i.e.,
fk(0)= some integer.
iv) for odd order derivatives,
fk(x)= -f(π-x)
for even order derivatives,
fk(x)= f(π-x)
so these are the properties of f(x)
Step-2:
now let's integrate the f(x) multiplied with sinx with variable x in the limits 0 to π.
This solution will always be an *integer*.(I could explain you why but that is too complex integration).
Step-3:
Now, let's go through other side.
we know, sinx rotates between 0 and 1 for 0 to π.
and for f(x),
when 0<x<π
=> 0<bx<bπ
=>0>-bx>-bπ
=>a>a-bx>a-bπ
therefore, (a-bx)< a
=>(a-bx)^n <a^n
now, (x^n)*(a-bx)^n<π^n*a^n
[let ((π^n)*(a^n))/n! be equal to some z]
so, 0<f(x)<z
=>0<f(x)sinx<z
=>0<integral with limits 0,π of f(x)sinx dx <z*x
integration means just adding small bits of answer values one by one to the small bits(let us think we divide x value into x small bits) of x values. that is why we multiplied z with x.(just so you know, here we are repeatedly adding the answer value x times. that is why i multiplied it with x. #repeated addition is called as multiplication)
for n~= infinity, by limits concept,
z*x becomes e^(aπ) which inturn is less than 1
so, totally,
=>0<answer<1
but in step 2 , we were known that the answer is always an integer value.
but in step 3, we get that the answer is in between 0 and 1.
since, there are no integers in between 0 and 1, we can conclude that our assumption ( pi is rational I.e., can be shown in a/b form) is false.
therefore, it is proved that pi is irrational.
To prove any number to be irrational, we need to prove that the number is not representable in a/b form.
If you have observed any theorem proof forms, we can prove a theorem by taking the assumption as the opposite of it. I.e., here we *assume* that pi is equal to a value a/b where a and b are integers
Step-1:
Consider a function f(x)= ((x^n)(a-bx)^n)/n!
we choose that function because
(i)f(x)=f(π-x)
(ii)f(0)=0
(iii) any highest( let us think it is kth order) derivative (higher order differentiation of f(x) ) when x=0 is always equal to an integer.i.e.,
fk(0)= some integer.
iv) for odd order derivatives,
fk(x)= -f(π-x)
for even order derivatives,
fk(x)= f(π-x)
so these are the properties of f(x)
Step-2:
now let's integrate the f(x) multiplied with sinx with variable x in the limits 0 to π.
This solution will always be an *integer*.(I could explain you why but that is too complex integration).
Step-3:
Now, let's go through other side.
we know, sinx rotates between 0 and 1 for 0 to π.
and for f(x),
when 0<x<π
=> 0<bx<bπ
=>0>-bx>-bπ
=>a>a-bx>a-bπ
therefore, (a-bx)< a
=>(a-bx)^n <a^n
now, (x^n)*(a-bx)^n<π^n*a^n
[let ((π^n)*(a^n))/n! be equal to some z]
so, 0<f(x)<z
=>0<f(x)sinx<z
=>0<integral with limits 0,π of f(x)sinx dx <z*x
integration means just adding small bits of answer values one by one to the small bits(let us think we divide x value into x small bits) of x values. that is why we multiplied z with x.(just so you know, here we are repeatedly adding the answer value x times. that is why i multiplied it with x. #repeated addition is called as multiplication)
for n~= infinity, by limits concept,
z*x becomes e^(aπ) which inturn is less than 1
so, totally,
=>0<answer<1
but in step 2 , we were known that the answer is always an integer value.
but in step 3, we get that the answer is in between 0 and 1.
since, there are no integers in between 0 and 1, we can conclude that our assumption ( pi is rational I.e., can be shown in a/b form) is false.
therefore, it is proved that pi is irrational.
pipi:
please mark my answer as brainliest
she can't understand the integration process make it easier...
yes I m new to these type of questions
as i told , unless you have a good understanding in differentiation , integration, limits concepts, i cannot make it easier... but if it helps you, i've edited the answer and kept the added text in bold. so, please refer to it. :)
ohk tysm
Answered by
2
so we have to prove π is irrational it required calculus .
so I m making it as simple .
we know that rational numbers are in form of p/q like 2/5 , 3/4 ..etc
rational number are either terminating or repeating like 1.44444... , 2.33333 etc...
so from above we can say that after seeing the value of ' π '
I.e. π = 3.14159265358979.... and soo on
so we can see value of π is neither terminating nor repeating so it doesn't satisfy the condition for to be a rational number ..
therefore " π " is irrational number.
______________________________
here I made it easier to understand but real proof of π is irrational requires high level calculus..
Hope it will help u.
so I m making it as simple .
we know that rational numbers are in form of p/q like 2/5 , 3/4 ..etc
rational number are either terminating or repeating like 1.44444... , 2.33333 etc...
so from above we can say that after seeing the value of ' π '
I.e. π = 3.14159265358979.... and soo on
so we can see value of π is neither terminating nor repeating so it doesn't satisfy the condition for to be a rational number ..
therefore " π " is irrational number.
______________________________
here I made it easier to understand but real proof of π is irrational requires high level calculus..
Hope it will help u.
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