Proof: By Euclid

Let $AB$ and $C$ be two unequal magnitudes, of which (let) $AB$ (be) the greater.
I say that if (a part) greater than half is subtracted from $AB$, and (if a part) greater than half (is subtracted) from the remainder, and (if) this happens continually, then some magnitude will (eventually) be left which will be less than the magnitude $C$.

fig001e

For $C$, when multiplied (by some number), will sometimes be greater than $AB$ [Def. 5.4] .
Let it have been (so) multiplied.
And let $DE$ be (both) a multiple of $C$, and greater than $AB$.
And let $DE$ have been divided into the (divisions) $DF$, $FG$, $GE$, equal to $C$.
And let $BH$, (which is) greater than half, have been subtracted from $AB$.
And (let) $HK$, (which is) greater than half, (have been subtracted) from $AH$.
And let this happen continually, until the divisions in $AB$ become equal in number to the divisions in $DE$.
Therefore, let the divisions (in $AB$) be $AK$, $KH$, $HB$, being equal in number to $DF$, $FG$, $GE$.
And since $DE$ is greater than $AB$, and $EG$, (which is) less than half, has been subtracted from $DE$, and $BH$, (which is) greater than half, from $AB$, the remainder $GD$ is thus greater than the remainder $HA$.
And since $GD$ is greater than $HA$, and the half $GF$ has been subtracted from $GD$, and $HK$, (which is) greater than half, from $HA$, the remainder $DF$ is thus greater than the remainder $AK$.
And $DF$ (is) equal to $C$.
$C$ is thus also greater than $AK$.
Thus, $AK$ (is) less than $C$.
Thus, the magnitude $AK$, which is less than the lesser laid out magnitude $C$, is left over from the magnitude $AB$.
(Which is) the very thing it was required to show.
- (The theorem) can similarly be proved even if the (parts) subtracted are halves.
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