proof the .9(bar) =1

[MeatHead]

Since the moderators closed the last thread just as I was getting ready to post the proof, here goes. And this is straight from my old Calculus book.

.9999... = .9 + .09 + .009 + .0009 + ...
= 9/10 +9/100 +9/1000 +9/10000 +...
= sum of 9/(10^k) with k=1 to infinity
Using one of the convergence theory we find the partial sum

Sn = 9/10 +9/(10^2) + 9/(10^3) + ... + 9/(10^n)
multiply both side by 1/10 to get

1/10 Sn = 9/(10^2) + 9/(10^3) + 9/(10^4) + ... + 9/(10^(n+1))

Subtract the two equations to get

Sn -1/10 Sn = 9/10 - 9/(10^(n+1))

So solving for Sn we get

Sn = 10/9 (9/10 - 9/(10^(n+1))) = 9/9 (1 - 1/(10^n))

Taking the limit as n goes to infinity gives

S = 9/9 =1

So .9 repeating does in fact equal 1. If you are goes to prove it, you have to prove it. Not just state that .9 repeating equals 9/9.

But if you put 1 instead of .9 repeating on a test, I bet you will have a hard time getting your teacher to give you credit for the problem, unless you can work out the proof and show it.

By the way, I took many graduate level math classes for FUN when I was in college. Yes, thats right, they were my electives. I love math, I just haven't done anything with it in years. Thanks for making me get my books out again. It made be remember how much I like math.
When I took math classes for fun I was like this :D
My friends were like :huh

[tants]

thanks :huh

:huh :huh :huh what?:huh

im not up to calculus yet in school. lol i take pre calc next year. im screwed.

[MeatHead]

QuadMX, don't worry. Calculus is not that hard. It doesn't get hard until you get up to Ordinary Differential Equations. But that is also when it starts to get fun. Besides most colleges have pretty good math tutoring centers to help you through, because math is the one area that most people need help in.

[Ralph]

fat shicks + 3 beers= nightmares for life

the solution


fat chicks+ 3 beers+ lifted truck= safe sex when drunk

(fat chicks cant climb)

thats about all the formulas i know....

[kabd69667]

I don't see why everyone is arguing over this. I'm in algebra II as a freshman, so i haven't got to calc yet. I doubt these kids sit and make the **** up, but ya never know...Personally, i think its a pretty stupid thing to argue over. Oh well, not like i matter anyway. :confused:

[Ralph]

how are you in algebra 2 as a freshamn? shouldnt u be in algebra 1

[kabd69667]

no....i took algebra I and geometry last year. I don't know, i just pass the stupid classes

[MeatHead]

Not necessarily. My second term at college I took Second and Third term Calculus at the same time, but you were not suppose to be able to do that. Oh yeah, I also had 20 credits. But for some reason that was my best term in college, 3.6 GPA for the term. Go figure.

[Ralph]

Originally posted by kabd69667
no....i took algebra I and geometry last year. I don't know, i just pass the stupid classes

both in one year?

[kabd69667]

yeah...they offer it to certain kids. i guess its supposed to impress colleges that you can handle two math courses in one year.

[Bean]

it is possible to be put in geometry your freshman year, but you hav to be REFERRED to it, here, i am in ALG 1, and its pretty damn hard, but im dumb, so who cares

[kabd69667]

I liked algebra...fun compared to geometry. oh well....i still don't see why we're talking about math courses at exriders

[Ralph]

imin alg 1 and its easy

i think that that is a flaw in the math system becasue if you dont use forlumas or rounding its simple and not possibble i mean with 2 second of thought i came up with my answer of no. IM not doupting you i beleive you but i think that it is a flaw in your math system
killer

who cares in school they told me .59 was .6 so .99 would be 1

this is the gayest thing since that kid posted the pic of the fat guys junk

let me cool this up a bit

[Ralph]

...

lol, my freinds printed them last year and put em up in school

actually the one we printed said congratulations at the top and at the bottom it said u funkin retard

[phatswinn]

yea lets make this a non nerd thread with allitle dana creech

well this thread is pointless and should be eaten up and regergitated

[Bean]

and taken a load on


hey, i run in the retard olimpics, il hav u know

and then steped in

[toby400ex]

9/9 does equal one, since it is a whole, but i think it would be 9/10.:huh :confused:

[330exracer122]

hmm when i was freshmen i was alg2 go figure :D

[330exracer122]

well im also considered a genious... 160 iq

[Bean]

it is 9/10, but some kids sit at home to figure out loopholes in the problems, just like lawyers try to figure out loopholes to get guilty people out of jail

[RideRed04]

man, you went the hard way to prove that one. Look at it this way: 1/3=.3(bar). 2/3=.6(bar). 1/3+2/3=1. .3=.6=.9

[MeatHead]

You're right that is a bit easier. 1/3 + 2/3 = 3/3 = 1
What many people fail to realize is that any repeating number is not really a number it is a series. My mind got stuck on series for some reason. Oh well, same results slightly different method.

[Team Outlaw #34]

hahaha i have the proof why .99999999=.9999999

if you do this on your calc it will give you the answer....

6546465465464464646=654849741265132311132132+65423 165789641651++2349846512641111118641111321-313549841231-+-465456421679+16549845119*975467984531-95431567851679--98798413549814111679841897968456189765189798444444 4444987651898713249876351321657984516-9874567984+9876545648978975 = 0.99999999

see....now everyone needs to shut up....

math is dumb..

[BrianWrightR6]

Here is a good one for you and my answer.....

Here is what I wrote:

PROBLEM: Find the center of gravity of a quadrilateral.
Construction:

Given a quadrilateral ABCD, find the midpoint of each side (the
midpoints of AB, BC, CD, and DA are E, F, G, and H respectively).

Take two opposite vertices (A and C) and draw lines connecting each to
the midpoints of the two sides that meet at the opposite vertex: AF,
AG, CE, and CH.

Label the intersections of AF and CE, P, and of AG and CH, Q. Connect
them.

Repeat with the other two vertices (B and D); label the intersections
of BH and DE, R, and of BG and DF, S.

The intersection of PQ and RS is the centroid, O.

Discussion:

A tough problem, unless I'm missing something obvious! I found the
construction fairly easily, but I had to use what I remember about
centers of gravity from physics. Bringing what I know into the context
of geometry is the hard part, starting with finding a useable
geometrical definition of the center of gravity (centroid) of an
arbritrary figure.

First, the bonus: Prove that the six small triangles between the
medians have the same area.

PROOF: Consider triangle ABC in my figure.

C
/\
/ \ \
/ \
/ \ \
/ \
F / \ \ G
/ \ / \
/ \ \ / \
/ / \ \
/ / \ \ \
/ / \ \
/________________\________________\
A E B

The areas of triangles CAE and CEB are equal because they have the
same base (AE = EB) and height (distance of C from AB). Each is
therefore half the area of triangle ABC.

Likewise triangles ABF and AFC have the same area, again half the area
of triangle ABC.

Thus the areas of ABF and CEB are equal, and if we subtract the common
area BFPE, we have the areas of PAE and PFC equal.

Again, triangles PAE and PEB have the same area because their bases
and height are equal, and likewise for PBF and PFC.

So we have PEB = PAE = PFC = PBF, and in a similar fashion the
remaining two small triangles can be shown to be equal to the others.
The implication of the problem statement is that the definition of a
centroid has something to do with this fact, that is, lines passing
through the centroid dividing a figure into equal areas. But this is
not the case. The centroid is the mean (average) location of all the
points contained in a figure. Any line passing through the centroid
can serve as a pivot and the figure will "balance" (supposing it is
cut out of wood, for instance, of uniform thickness). In physics, this
means the "moments" on both sides are equal (and opposite). It does
not necessarily mean the areas on both sides are the same; a smaller
area farther from the pivot can balance a larger area closer to the
pivot.

In the case of the medians of a triangle, the figures on both sides
are triangles; therefore the area is distributed the same way, so
equal moments imply equal areas. Other lines through the centroid give
a triangle on one side and a quadrilateral on the other; the mass
distribution is different - the quadrilateral has more mass farther
out - so the triangle needs a larger area to balance it.

Now I will try to put the centroid of a general figure in a classical-
geometry context. I define the centroid by three postulates:

1. The centroid of a figure with a rotational symmetry is the rotation
axis.

2. If two figures a and b are similar, their centroids A and B form
similar figures with corresponding elements of a and b.

3. Given two figures a and b with centroids A and B, the centroid of
the two figures taken together is the point C on the line AB, such
that AC * Area(a) = CB * Area(b).

I don't know if these are all necessary or sufficient, but they are
consistent with the definition of a centroid as the mean location of
the points in a figure, and they are sufficient for my purpose. I have
one thing to prove before I can justify my construction: that this
definition is consistent with the centroid we know.

THEOREM: The centroid of a triangle is the intersection of its
medians. PROOF: Given any triangle ABC, suppose its centroid is P.
Construct the midpoints of AB, BC, and CA at D, E, and F respectively.
Triangle ABC can be divided into parallelogram BDFE and triangles ADF
and FEC. Call the centroids of these figures Q, R, and S respectively.

Triangles ADF and FEC are congruent, so they have the same area,
therefore by (1) above, the centroid T of the two triangles taken
together is at the midpoint of RS.

The centroid Q of parallelogram BDFE is on the diagonal BF by (1)
above.

Triangle ADF is similar to triangle ABC. Therefore by (2) above,
triangle ARF is similar to APC; angle RAF = angle PAC, so R lies on
line AP; and since F is the midpoint of AC, R is the midpoint of AP.
Similarly, S is the midpoint of CP.

Triangle RPS is similar to triangle APC. Therefore T, the midpoint of
RS, lies on PF.

Triangle ABC is the sum of parallelogram BDFE (with centroid Q) and
triangles ADF and FEC (taken together, with centroid T), so by (3)
above, the centroid P of ABC is on the line QT.

Since Since P, T, and F are collinear by (4), and P, T, and Q are
collinear by (5), then P, T, F, and Q are all collinear. Since Q, B,
and F are collinear by (2), we can add B to the list, so in
particular, P is on BF. But BF is a median of triangle ABC.

By similar arguments, P is on the other two medians of ABC as well. In
other words, the centroid of any triangle ABC is the intersection of
its medians.

Now I am finally ready to justify my construction of the centroid of a
quadrilateral. A quadrilateral can be divided into two triangles in
two ways: by cutting along either diagonal. If I make one such cut, I
know the centroid lies along the line joining the centroids of the two
triangles (P and Q), by definition part (3).

I could at this point find the centroid by choosing O on PQ such that
OP:OQ = Area(ACD):Area(ABC). But it is much easier with compass and
straightedge to take advantage of the other pair of triangles and find
another line (RS) on which the centroid lies. The centroid must of
course lie at the intersection of the two lines.

[JTRtrx250r]

man.....all I got was a headache just tryin' to comprehend anything you guys posted...............math was the worst class's I ever had,and still is............but I can ride:D

[motochris]

Well the CNC machine thinks diffrent... Not saying your wrong but the CNC machine see's it diffrent..............:D

[dhines]

No, the CNC machine simply can't calculate a series that repeats to infinity. If it could, it too would conclude that 0.9999... = 1. ;)

[exriderdude]

ok all you smarty pants, try this one :blah :

If a man walks up to you and says, " Everything i tell you is a lie,". Is he lieing or telling the truth??

[SlapNutz]

how doess all that works? what is goings on Time to fire upp my SHEE:macho