: b ∷ c : d , and if 2 aaa + 3 aab and bbb + abb be chosen, which are both homogeneous with respect to a and b , and both of the third degree; and if the corresponding expressions 2 ccc + 3 ccd and ddd + cdd be formed, which are made from c and d precisely in the same manner as the two former ones from a and b , then will
2aaa + 3aab : bbb + abb ∷ 2ccc + 3ccd : ddd + cdd
| To prove this, let | a | be called x . | ||||
|---|---|---|---|---|---|---|
| b | ||||||
| Then, since | a | = x , and | a | = | c | , |
| b | b | d | ||||
| it follows that | c | = x . | ||||
| d |
But since a divided by b gives x, x multiplied by b will give a, or a = bx. For a similar reason, c = dx. Put bx and dx instead of a and c in the four expressions just given, recollecting that when quantities are multiplied together, the result is the same in whatever order the multiplications are made; that, for example, bxbxbx is the same as bbbxxx.
Hence, 2aaa + 3aab = 2bxbxbx + 3bxbxb
= 2bbbxxx + 3bbbxx
which is bbb multiplied by 2xxx + 3xx
or bbb (2xxx + 3xx)4
Similarly, 2ccc + 3ccd = ddd (2xxx + 3xx)
Also, bbb + abb = bbb + bxbb
= bbb multiplied by 1 + x
or bbb(1 + x)
Similarly, ddd + cdd = ddd (1 + x)
Now, bbb : bbb ∷ ddd : ddd
Whence (186), bbb(2xxx + 3xx): bbb(1 + x) ∷ ddd(2xxx + 3xx): ddd(1 + x), which, when instead of these expressions their equals just found are substituted, becomes 2aaa + 3aab: bbb + abb ∷ 2ccc + 3ccd: ddd + cdd.
The same reasoning may be applied to any other case, and the pupil may in this way prove the following theorems:
If
a : b ∷ c : d
2 a + 3 b : b ∷ 2 c +