Diophantus of Alexandria/Arithmetica/Book 1/Problem 1

Problem

To divide a given number into two having a given difference.

Let $N$ be the given number.

Let $d$ be the given difference.

Let $x$ be the smaller of the two numbers into which $N$ is to be divided.

Thus the greater of the two numbers is $x + d$.

We also have that the greater of the two numbers is $N - x$.

Hence the smaller of the two numbers is found by:

$\ds N$

$=$

$\ds x + \paren {x + d}$

$\ds $

$=$

$\ds 2 x + d$

simplifying

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \dfrac {N - d} 2$

rearranging

$\blacksquare$

Given number $100$, given difference $40$.

c. 250: Diophantus of Alexandria: Arithmetica ... (next): Book $\text I$: Problem $1$
1910: Sir Thomas L. Heath: Diophantus of Alexandria (2nd ed.) ... (next): The Arithmetica: Book $\text {I}$: Problems: $1.$