Does point 999 repeating equal 1?

[Cogito Ergo Sum]

You look mighty purdy!

 

Thanks!

 

I can only assume that by your statements, you did not have Integral Calculus or Differential Equations? Lucky You!

 

Therefore we do not have a common vocabulary.

 

f(x)=0 is a function expression and ummm...without a lot of ground work here, I would be at a disadvantage to explain it to you because you are looking at my equation in an algebraically frame of mind, when in fact, it is Calculus. Ummm...I know...Google Search, Newton's Method. I'm sure you will find a zillion hits with better explinations than I can offer!

 

"Roots" are values where the function of f(x) are ZERO.

 

In the attached pic, look at the dots on the X asis. They are "roots" of the function 1tan(x)=0

 

f(x) and f '(x) are Calculus terms and are used in calculations regarding the plotting of functions.

 

Suffice it to say that even with the substitution of .999 Ad Infinitum the resulting differences in plotting points of the function of both f(x) and f '(x) would still be clearly evident, thereby proving that .999 Ad Infinitum does not equal 1. At least in Calculus. Apologies to Sir Issac Newton for my feeble attempt at understanding his great math, but I try!

 

I guess in a type of mathematics of which I know nothing, where one was "theorizing", perhaps Steven Hawkins could provide "theoretical" argument in defense of the claim, but under strict scrutiny in terms of functions within Calculus, I believe that I am correct.

[Mohammed_Rots_In_Hell]

Thanks!

 

I can only assume that by your statements, you did not have Integral Calculus or Differential Equations? Lucky You!

 

 

You said...

 

Hmmm...

 

Put's on Professor's Cap along with reading glasses on the end of nose

So I said that you look pretty with your glasses on...

 

And yes, I am well versed in both Integral Calculus and differential equations.

 

 

Therefore we do not have a common vocabulary.

No common vocabulary because I think you're pretty with glasses on the end of your nose???

 

f(x)=0 is a function expression and ummm...without a lot of ground work here, I would be at a disadvantage to explain it to you because you are looking at my equation in an algebraically frame of mind, when in fact, it is Calculus. Ummm...I know...Google Search, Newton's Method. I'm sure you will find a zillion hits with better explinations than I can offer!

This is why I asked what you meant, I know exactly the meaning of the two functions.

 

"Roots" are values where the function of f(x) are ZERO.

You are correct about the vocabulary because I would have used the term "value" or "values of x" and those "roots" you gave are valid (though they have nothing to do with your proof)

 

In the attached pic, look at the dots on the X asis.

What pic???

 

They are "roots" of the function 1tan(x)=0

true, but irrelevant.

 

f(x) and f '(x) are Calculus terms and are used in calculations regarding the plotting of functions.

f'(x) is the derivative of f(x)

 

Suffice it to say that even with the substitution of .999 Ad Infinitum the resulting differences in plotting points of the function of both f(x) and f '(x) would still be clearly evident, thereby proving that .999 Ad Infinitum does not equal 1. At least in Calculus. Apologies to Sir Issac Newton for my feeble attempt at understanding his great math, but I try!

No, it is not evident.. see the following

 

1 tan(x) = 0

.999(repeating) tan(x) = 0

 

like you said, plot the points for every possible "root" or value of x for f(x), you get (in radians of course):

3(Pi)

2(Pi)

1(Pi)

0(Pi)

-1(Pi)

-2(Pi)

-3(Pi)

 

and for f'(x) you get

1/3(Pi)

1/2(Pi)

1/Pi

1/-Pi

1/-2(Pi)

1/-3(Pi)

 

take any of these points and multiply thru by 1 or (.999(repeating)) you may or may not get the same result. I don't believe that addidtion or multiplication is actually defined for repeating decimals. It is only defined for fractions that represent rational numbers, so your line of thinking is skewed.

 

First you would have to prove that .999(repeating) is a rational number, and then show its coresponding fractional equivolent.

 

the sumation series 1/a + 1/[a (10 ^ 2)] + 1/[a (10 ^ 3)] + ... 1/[a (10 ^ n)] is equal to = [a / (10 - 1)] where a is a repeating decimal, when a is 9 (.999(repeating)) the formula is 9/(10 - 1) = 9/9 = 1

 

I guess in a type of mathematics of which I know nothing, where one was "theorizing", perhaps Steven Hawkins could provide "theoretical" argument in defense of the claim, but under strict scrutiny in terms of functions within Calculus, I believe that I am correct.

I don't believe you are correct, I will agree to disagree (but mathematics do not lie) Algabra is the foundation of calculus (both integral and differential) in fact Algaebra is essential to calculus. There is no way that two would ever be in discord.

[snafu]

Total brain warp!

 

Wizzz!!!!!!

 

I see I missed a whole lot in school!

I love how you guys throw those big phrases around like,

[Mohammed_Rots_In_Hell]

Total brain warp!

 

Wizzz!!!!!!

 

I see I missed a whole lot in school!

I love how you guys throw those big phrases around like,

[Mohammed_Rots_In_Hell]

And you've just demonstrated the difference between practical reality, and possessive ideology.

 

I use maths for profit. You abuse it for perverse entertainment.

I hope they don't pay you what you're worth ... you'd starve to death!

[builder]

I hope they don't pay you what you're worth ... you'd starve to death!

 

 

Yep, I'm wasting away. Buuuuuuuurp.