Question

I found a mistake in an article, i was wondering if someone could check it

Answer 1

There is a mistake (I think) in this article
http://www.learner.org/courses/mathilluminated/units/3/textbook/03.php#infinite

where it says
"(for if they were, it would indicate that the common unit could instead be 1/2 u, and we would simply make that adjustment)."

Answer 2

it should say, if the common unit should be 2u, not 1/2 u

Answer 3

That is actually interesting, I only comment on this to stay up to date on answers, I will have to read this tomorrow, pardon me. This article requires some attention, so I want to be more awake when I tackle it :-)

Answer 4

ok :)

Answer 5

Yeah, I think it should be 2u.

Take 12 and 16
It can be 2*6 and 2*8 with 2 as the common unit. m = 6 and n = 8. Both are even
But we can double u and write it as 4*3 and 4*4 making u = 4 and m = 3 and n = 4.
Now m and n are not both even.

Answer 6

ok lets assume we have two lengths one is 12 inches and one is 16 inches (here i am using the common unit as an 'inch' )
but 12 inches = 4*3 inches, 16 inches = 4 * 4 inches. so i can make a new unit , call it 'kinches' , and 1 kinche = 4 inches.
now 12 inches = 3 kinches, 16 inches = 4 kinches

Answer 7

here we quadrupled the common unit, (or we can double it twice).

Answer 8

Yes. As long as m and n are even, we can keep pulling the common factor 2 out until it is reduced to the lowest form where they are not both even. So u will keep doubling.

Answer 9

right :)

Answer 10

to make it more general if
a = 12 u
b = 16 u
then
a = 3*4u
b = 4* 4u
Now let u' = 4u
then
a = 3*u'
b = 4*u' , now m , n are not both even

Answer 11

or to use easier numbers
a = 4u
b = 6u
then
a = 2 * 2u
b = 3 * 2u

Let u' = 2u
a = 2u'
b = 3u'

Answer 12

Yes. We are trying to pull the greatest common factor out of a and b.
a = GCF * m
b = GCF * n

If m and n are both even, then we can keep pulling the factor 2 out and so the earlier GCF is really not the GCF.

Answer 13

Yes.

Answer 14

i probably made a mistake, but you get the idea : )

Answer 15

ok one more time
Suppose that a and b have a common unit length u such that
a = m*u
b = n*u
Let
u' = GCF(m,n)*u.
m' = m / GCF(m,n)
n' = n / GCF(m,n)
Then we can rewrite a,b as
a = m' * u'
b = n' * u'
where m',n' are not both even (at least one is odd) .

Answer 16

we can prove writing it out this way, m', n' are not both even

Answer 17

but doing a proof is beyond the scope of my capabilities :)

Answer 18

an interesting side note, I will prove a commensurability corollary.

if it is true that any two lengths are commensurable (i know this hypothesis false , but assuming this is true) , then it follows that *all* lengths have a common unit.
proof :
starting with any two lengths, a & b which by the hypothesis is commensurable , have common unit length u. Then for any third length c, since any pair of lengths are commensurable , c is commensurable with a or b, and without loss of generality I will assume that c is commensurable with a. So far we have
a = m*u
b = n*u

a = p*u'
c = q*u'

where m,n,p,q are positive integers, and u,u' are lengths.

since a = m*u = p*u' , then u'= m*u/p
Then c = q*u' = q * (m*u/p) = q*m/p * u
now you can divide u until q*m/p is an integer.

Answer 19

anyways, this underscores the difficulty in doing greek mathematics :)