Can You Prove These Two Questions About a One-to-One Function f: R -> R?

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This discussion focuses on proving properties of a one-to-one function \( f: \mathbb{R} \to \mathbb{R} \) under specific conditions. The first question requires demonstrating that if \( f(y) = y + r \) for some \( y \), then \( f(y - nr) = (y - nr) + r \) holds for all positive integers \( n \). The second question asserts that if \( f \) is continuous and satisfies \( f(0) = 0 \) and \( f(2x - f(x)) = x \), then \( f(x) = x \) for all \( x \in \mathbb{R} \).

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Realli need help on those two questions! any1 can help with that, or post the solution. Thank you guys!


1. Suppose f : R -> R is a function such that
f(2x - f(x)) = x
for all x and let r be a fixed real number.
(a) Prove that if there exists y such that f(y) = y + r, then f(y - nr) =
(y - nr) + r for all positive integers n.
(b) Prove that if, in addition to the assumptions in part a), f is also oneto-
one, then f(y - nr) = (y - nr) + r is true for all integers n.

2. Suppose that f: R -> R is a one-to-one function such that the following
are true
f is continuous at all points a 2 R; that is to say that
lim f(x) = f(a) for all a 2 R.
x->a

* f(0) = 0, and
* f(2x - f(x)) = x for all x 2 R.

Prove that f(x) = x for all x 2 R. (Hint: use the precise definition of limits
and the result of question 1.)
 
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For 1a) notice that if f(y) = y + r, then f(2y - f(y)) = y = f(2y - (y + r)) = f(y - r). Can you repeat this process?
 

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