What's the rule regarding 'ones' in algebra?

Can you use ones [1's] indescriminately in algebra? If not why not...

Chergarj wrote:



Sadly, it was.

Don sHead is a retired civil service worker with an eighth-grade education. He
has no understanding of math beyond grade-school arithmetic. There is no point
in attempting to provide any meaningful answer to any question he may ask,
concerning math or physics. No matter how simplified your reply, it will be
utterly beyond his comprehension.

"Mark Mallory" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ]...
your
Yes it is serious: Many years ago I overheard a H.S. algebra teacher tell
his class _something to the effect_ that the number one [1] was such that it
didn't change the value (of an equation) when it was inserted in an
equation. That made a lasting impression, since I didn't understand then,
and don't now:

In particular: Writing that acceleration [a] is _inversely_ proportional to
the mass [m] of a body, as [a is proportional to 1/m]. That somehow doesn't
look 'copesthetic' to me; especially if 'm' is a variable.

I thought the rule might be simple enough for me to understand; but it's
not, according to 'The Ghost in the Machine'.
Cut< The rest of Mark's [ignorant; prejudicial jealous, and cutting]
comentary(;^)

"Donald G. Shead" <[Only registered users see links. ]> wrote in message news:<NWnbb.4795$[Only registered users see links. ].prodigy.c om>...

Yes, of course. For example, start with: a + b = c, then we can put
in as many ones as we like: a + b + 1 + 1 + 1 + 1 ... = c. You can be
as indiscriminate as you like with the ones, and you'll be as right as
you always are.

"Paul Cardinale" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ] ...
news:<NWnbb.4795$[Only registered users see links. ].prodigy.c om>...

Please, tell me that this was a joke...

Kostas.

"Paul Cardinale" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ] ...
news:<NWnbb.4795$[Only registered users see links. ].prodigy.c om>...

Got any real ideas Paul?

"Constantine" <[Only registered users see links. ].uk> wrote in message
news:bkng02$c17$[Only registered users see links. ].ac.uk...
'Fess-up Paul: That you did it because you are scared as all get out that
Shead's simplicity will upset the applecart: Maybe even shake-up the
gravyboat and a few gravytrains too(:-)

On Mon, 22 Sep 2003 20:41:13 GMT, "Donald G. Shead" <[Only registered users see links. ]>
wrote:


You were supposed to learn something from that,
ShitHead--specifically, that you _cannot_ use ones indiscriminately in
algebra, which was your original premise. But once again, you have
proved you are incapable of learning the simplest concepts in
mathematics or physics, and have no real interest in learning
anything. All you want is attention. Must be terrible to be not only
old and forgetful, but so terribly lonely on top of it.

Gene Nygaard
"Life's tough. But it's tougher if you're stupid." - John Wayne

"Donald G. Shead" <[Only registered users see links. ]> wrote in message news:<KLBbb.6128$[Only registered users see links. ].prodigy.c om>...

If you multiply 1 by something in the equation, it will
not change the meaning of the equation, because any number
multiplied by 1 is still the same number. For example,

xy - st^2 = sx + 3 + sqrt(y-456)

is the same as

1xy - 1st^2 = 1sx + 1*3 + 1sqrt(1y-1*456)

This is probably the wrong newsgroup to be asking math
questions. Future math questions, unless they relate
directly to physics, should probably be posted to
sci.math.

"Donald G. Shead" <[Only registered users see links. ]> wrote in message news:<KLBbb.6128$[Only registered users see links. ].prodigy.c om>...

I think you've just about got it.

You can multiply or divide any term or variable in an equation by 1
and still keep the equation in balance.

In your paricular case, if you have

a = F/m

then multiplying the variable F by 1 we have

a = (F*1)/m

which is the same as

a = F * 1/m

so a is directly proportional to F and inversely proportional to m.

Double-A