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1986 IMO Longlists
19
f(x)=x forall x ∈ [0, 1]
f(x)=x forall x ∈ [0, 1]
Source:
August 29, 2010
function
induction
algebra proposed
algebra
Problem Statement
Let
f
:
[
0
,
1
]
→
[
0
,
1
]
f : [0, 1] \to [0, 1]
f
:
[
0
,
1
]
→
[
0
,
1
]
satisfy
f
(
0
)
=
0
,
f
(
1
)
=
1
f(0) = 0, f(1) = 1
f
(
0
)
=
0
,
f
(
1
)
=
1
and
f
(
x
+
y
)
−
f
(
x
)
=
f
(
x
)
−
f
(
x
−
y
)
f(x + y) - f(x) = f(x) - f(x - y)
f
(
x
+
y
)
−
f
(
x
)
=
f
(
x
)
−
f
(
x
−
y
)
for all
x
,
y
≥
0
x, y \geq 0
x
,
y
≥
0
with
x
−
y
,
x
+
y
∈
[
0
,
1
]
.
x - y, x + y \in [0, 1].
x
−
y
,
x
+
y
∈
[
0
,
1
]
.
Prove that
f
(
x
)
=
x
f(x) = x
f
(
x
)
=
x
for all
x
∈
[
0
,
1
]
.
x \in [0, 1].
x
∈
[
0
,
1
]
.
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