---
book: 10
number: 1
id: "X.1"
kind: "theorem"
uses: []
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:10.prop_1.1"
license: "CC-BY-SA-4.0"
---

# X.1

*Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out*.

## Proof

Let *AB, C* be two unequal magnitudes of which *AB* is the greater: I say that, if from *AB* there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude *C*.

For *C* if multiplied will sometime be greater than *AB*. cf. [[book-5/definitions#Definition 4|v. Def. 4]]

Let it be multiplied, and let *DE* be a multiple of *C*, and greater than. *AB*; let *DE* be divided into the parts *DF*, *FG*, *GE* equal to *C*, from *AB* let there be subtracted *BH* greater than its half, and, from *AH*, *HK* greater than its half, and let this process be repeated continually until the divisions in *AB* are equal in multitude with the divisions in *DE*.

Let, then, *AK*, *KH*, *HB* be divisions which are equal in multitude with *DF*, *FG*, *GE*.

Now, since *DE* is greater than *AB*, and from *DE* there has been subtracted *EG* less than its half, and, from *AB*, *BH* greater than its half, therefore the remainder *GD* is greater than the remainder *HA*.

And, since *GD* is greater than *HA*, and there has been subtracted, from *GD*, the half *GF*, and, from *HA*, *HK* greater than its half, therefore the remainder *DF* is greater than the remainder *AK*.

But *DF* is equal to *C*; therefore *C* is also greater than *AK*.

Therefore *AK* is less than *C*.

Therefore there is left of the magnitude *AB* the magnitude *AK* which is less than the lesser magnitude set out, namely *C*. Q. E. D.

And the theorem can be similarly proved even if the parts subtracted be halves.