Problem 5

Let

\mathbb{Q}_{>0}

be the set of all positive rational numbers. Let

f:\mathbb{Q}_{>0}\to\mathbb{R}

be a function satisfying the following three conditions: (i) for all

x,y\in\mathbb{Q}_{>0}

f(x)f(y)\ge f(xy)

; (ii) for all

x,y\in\mathbb{Q}_{>0}

f(x+y)\ge f(x)+f(y)

; (iii) there exists a rational number

a>1

such that

f(a)=a

. Prove that

f(x)=x

for all

x\in\mathbb{Q}_{>0}

Step 5 of 6: Trap any x between 1 and a^n − x

Previous step Next step

a^n = f(a^n) \ge f(x) + f(a^n - x) \ge x + (a^n - x - 1) = a^n - 1

Detailed analysis

Fix $x>1$ and pick $n$ large enough that $a^n-x>1$ . Applying (ii) to $x$ and $a^n-x$ and using $f(a^n)=a^n$ together with the lower bound $f(y)\ge y$ from two steps above (for both $x$ and $a^n-x$ , both exceeding $1$ ) gives $a^n=f(a^n)\ge f(x)+f(a^n-x)\ge f(x)+(a^n-x)$ , so $f(x)\le x$ ; combined with $f(x)\ge x$ already shown, $f(x)=x$ .