# Definitions

##### 3 minutes • 508 words

## Table of contents

## Definitions

- “Commensurable” are magnitudes measured by the same measure.

“Incommensurable” are magnitudes which have no common measure.

- Two straight-lines are commensurable when their bounding area are measured by the same area.

They are incommensurable when their area has no common measure.*

## Superphysics Note

- There is an infinite number of straight-lines which are either commensurable or incommensurable.

- Some are commensurable or incommensurable in length only.
- Others are commensurable or incommensurable in area also, with an assigned straight line.

‘Rational’ lines are:

- the assigned straight line
- those straight lines which are commensurable with the assigned straight line, whether in length and in square, or in square only

‘Irrational’ lines are those which are incommensurable with the assigned straight line.*

## Superphysics Note

- The [binding] square on the assigned straight line is called ‘rational’

Those areas which are commensurable with it are rational.

But those areas which are incommensurable with it are irrational

- The straight lines which produce those areas are irrational

## Proposition 1

Assume there are 2 unequal lines.

- The longer line has more than half of it subtracted
- More than half of the remainder is subtracted

If this happens continually, then some magnitude will eventually be left. It will be less than the lesser laid out magnitude.

Let:

`AB`

and`C`

are unequal magnitudes`AB`

is the longer or greater

We subtract more than half from `AB`

. From the remainder, we also subtract more than half. If this happens continually, there will be some magnitude left. But it will be less than the magnitude `C`

.

If we keep on multiplying `C`

, it will eventually be greater than `AB`

.

Let it be multiplied, and let `DE`

be a multiple of `C`

, and greater than `AB`

.

Divide `DE`

into the parts `DF`

, `FG`

, `GE`

equal to `C.

Subtract:

- from
`AB`

,`BH`

greater than its half - from
`AH`

,`HK`

greater than its half

Repeat this process continually until the divisions in `AB`

are equal in multitude with the divisions in `DE`

.

Then, let `AK`

, `KH`

, `HB`

be divisions which are equal in multitude with `DF`

, `FG`

, `GE`

.

The remainder `GD`

is greater than the remainder `HA`

because:

`DE`

is greater than`AB`

- from
`DE`

there has been subtracted`EG`

less than its half - from
`AB`

,`BH`

greater than its half

`GD`

is greater than `HA`

The following were subtracted:

- from
`GD`

, the half`GF`

- from
`HA`

,`HK`

greater than its half

This is why 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`

.

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