91Ó°ÊÓ

If \(f(x)=0\) for all \(x,\) then \(f\) satisfies \(f(x+y)=f(x)+f(y)\) for all \(x\) and \(y\) and also \(f(x \cdot y)=f(x) \cdot f(y)\) for all \(x\) and \(y\). Now suppose that \(f\) satisfics these two properties, but that \(f(x)\) is not always \(0 .\) Prove that \(f(x)=x\) for all \(x,\) as follows: (a) Prove that \(f(1)=1\) (b) Prove that \(f(x)=x\) if \(x\) is rational. (c) Prove that \(f(x)>0\) if \(x>0 .\) (This part is tricky. but if you have been paying attention to the philosophical remarks accompanying the problems in the last two chapters, you will know what to do.) (d) Prove that \(f(x)>f(y)\) if \(x>y\) (e) Prove that \(f(x)=x\) for all \(x\). I lint: Use the fact that between any two numbers there is a rational number.

Step by step solution

01

Prove that \(f(1)=1\)

Given that \(f(x + y) = f(x) + f(y)\), substitute \(x = 1\) and \(y = 1\) to get \(f(2) = f(1) + f(1) = 2f(1)\). Also for \(f(x \times y) = f(x) \times f(y)\), substitute \(x = 1\) and \(y = 1\) to get \(f(1 \times 1) = f(1) \times f(1)\). This simplifies to \(f(1) = f(1)^2\). The solutions to this equation are \(f(1) = 0\) or \(f(1) = 1\). Since \(f(x)\) is not always zero, \(f(1) = 1\).

02

Prove that \(f(x) = x\) if \(x\) is rational

Consider any rational number \(q\) that can be written as \(\frac{a}{b}\) where \(a\) and \(b\) are integers. Using the identity \(f(b) = bf(1) = b\) and \(f(a) = af(1) = a\), examine \(f\left(\frac{a}{b}\right)\). Using the additive property, we get \(f(a) = f(b \cdot \left(\frac{a}{b}\right)) = b f\left(\frac{a}{b}\right)\). This simplifies to \(a = b f\left(\frac{a}{b}\right)\) and solving gives \(f\left(\frac{a}{b}\right) = \frac{a}{b}\). Thus \(f(x) = x\) for all rational \(x\).

03

Prove that \(f(x) > 0\) if \(x > 0\)

Assume \(x > 0\). For any positive rational number \(r\), \(f(r) = r > 0\). For irrational \(x\), observe that any irrational number \(x\) can be approached by a sequence of rational numbers \(\{r_n\}\) such that \(r_n \to x\). By continuity (or limits) arguments and properties of \(f\), \(\forall r_n > 0 \Rightarrow f(r_n) = r_n > 0\) for all \(n\), hence \(f(x)\) must also be positive.

04

Prove that \(f(x) > f(y)\) if \(x > y\)

Assuming \(x > y\), express \(x\) as \(y + k\) for some positive \(k\). By the additive property: \(f(x) = f(y + k) = f(y) + f(k)\). Since \(f(k) > 0\) when \(k > 0\), it implies \(f(x) = f(y) + f(k) > f(y)\). Therefore, \(f(x) > f(y)\) if \(x > y\).

05

Prove that \(f(x) = x\) for all \(x\)

Using the result from Step 2 that \(f\) matches rational numbers and positive continuous behavior established in Steps 3 and 4, consider any real number \(x\). Any \(x\) can be approached by rational approximations. Due to the properties of continuity and order preservation combined with rational-frame matching, \(f(x) = x\) for all \(x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Additive Functions

Additive functions are essential in understanding many mathematical properties. These functions satisfy the equation: \( f(x + y) = f(x) + f(y) \).

This means that applying the function to a sum of two inputs is the same as adding the function values of the individual inputs. This property can help us simplify many complex calculations and explore further properties about the function.

Multiplicative Functions

Multiplicative functions work differently. They obey the rule: \( f(x \cdot y) = f(x) \cdot f(y) \).

In other words, applying the function to the product of two numbers is the same as multiplying the function values of those numbers. This trait often interplays with the additive property to unveil deeper insights into the function's behavior.
In the given problem, combining both properties leads us to recognize that if we have both additive and multiplicative properties, we can converge to a unique solution.

Rational Numbers

Rational numbers are numbers that can be written as a fraction of two integers, where the denominator is not zero. Examples include \( \frac{1}{2}, 2, \text{and} -\frac{4}{3} \).

Rational numbers are crucial in proving that \( f(x) = x \) for all rational \( x \). By expressing rationals as a fraction and breaking them down using additive and multiplicative properties, we can easily figure out that \[ f\bigg(\frac{a}{b}\bigg) = \frac{a}{b}\bigg. \], which implies \( f(x) = x \) for any rational \( x \).

Continuity

Continuity ensures that small changes in inputs result in small, predictable changes in outputs. A function \( f \) is continuous if, for any sequence of numbers \( x_n \to x \), the sequence of function values \( f(x_n) \to f(x) \).

In this problem, continuity is used to extend the property \( f(x) = x \) from rational numbers (which are dense in the reals) to all real numbers. Therefore, if \( f(x) = x \) for all rational \( x \) and \( f \) is continuous, it implies that \( f(x) = x \) for any real number \( x \) because we can approximate any real number by a sequence of rational numbers.