proove that pi is irrational
Answers
Suppose
π
=
a
/
b
π=a/b. Define
f
(
x
)
=
x
n
(
a
−
b
x
)
n
n
!
f(x)=xn(a−bx)nn!
and
F
(
x
)
=
f
(
x
)
−
f
(
2
)
(
x
)
+
f
(
4
)
(
x
)
−
.
.
.
+
(
−
1
)
n
f
(
2
n
)
(
x
)
F(x)=f(x)−f(2)(x)+f(4)(x)−...+(−1)nf(2n)(x)
for every positive integer
n
n.
First note that
f
(
x
)
f(x) and its derivatives
f
(
i
)
(
x
)
f(i)(x) have integral
values for
x
=
0
x=0, and also for
x
=
π
=
a
/
b
x=π=a/b since
f
(
x
)
=
f
(
a
/
b
−
x
)
f(x)=f(a/b−x).
We have
d
d
x
(
F
′
(
x
)
s
i
n
x
−
F
(
x
)
c
o
s
x
)
=
F
′′
(
x
)
s
i
n
x
+
F
(
x
)
s
i
n
x
=
f
(
x
)
s
i
n
x
ddx(F′(x)sinx−F(x)cosx)=F″(x)sinx+F(x)sinx=f(x)sinx
whence
∫
π
0
f
(
x
)
s
i
n
x
d
x
=
[
F
′
(
x
)
s
i
n
x
−
F
(
x
)
c
o
s
x
]
π
0
=
F
(
π
)
+
F
(
0
)
∈
Z
∫0πf(x)sinxdx=[F′(x)sinx−F(x)cosx]0π=F(π)+F(0)∈Z
But for
0
<
x
<
π
0<x<π, we have
0
<
f
(
x
)
s
i
n
x
<
π
n
a
n
n
!
0<f(x)sinx<πnann!
which means we have an integer that is positive but tends to zero as
n
n approaches infinity, which is a contradiction.
Answer:
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Step-by-step explanation:
in in 1760s, Johann Heinrich Lambert proved that π(pi) is irrational i.e. it cannot be it cannot be expressed as a fraction a/b, where where is a integer and b is a non zero integer.