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Iran MO (2nd Round)
1991 Iran MO (2nd round)
3
The old Cauchy function [Iran Second Round 1991]
The old Cauchy function [Iran Second Round 1991]
Source:
November 30, 2010
function
algebra proposed
algebra
Problem Statement
Let
f
:
R
→
R
f : \mathbb R \to \mathbb R
f
:
R
→
R
be a function such that
f
(
1
)
=
1
f(1)=1
f
(
1
)
=
1
and
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
f(x+y)=f(x)+f(y)
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
And for all
x
∈
R
/
{
0
}
x \in \mathbb R / \{0\}
x
∈
R
/
{
0
}
we have
f
(
1
x
)
=
1
f
(
x
)
.
f\left( \frac 1x \right) = \frac{1}{f(x)}.
f
(
x
1
)
=
f
(
x
)
1
.
Find all such functions
f
.
f.
f
.
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