1+1=?

[beenjammin2lp]

9 rounds up you poo-poo eaters!..im the only person to get that....jeez....so when repeating finally gets to google(the highest number). it rounds up

 

[TheHeretic]

to invert nexus, theorm 3=4 is false a+b-c=0 you dont follow the order of operations.
parentheses, exponents multiply dived add subract
I think i learned that in 4th grade. THat also disproves the following one, and probally the rest of them.

 

[TheHeretic]

i dont now if this helps but in calc 1/infinity =0 because the number is getting infinatley smaller so i guess that an infinatly smaller number as a finite point of zero. SO i guess a infinitely number of .9999999999 would eventually reach a finite point.

 

[el-half]

SO i guess a infinitely number of .9999999999 would eventually reach a finite point.

That is a paradox. If something infinite reaches a finite point it is not infinite.

I don't get how you can think that 0.9r = 1
0.9r + 0.0r1 = 1

 

[TheHeretic]

Well maybe infinity does not exist. Its possible that infinity is just used to represent and exremely large finite point. Could some one give me an example of something that is actually infinate.

 

[el-half]

Infinity as a concept exists. Thus 0.9 recurring infinitely is possible.
You are correct though in questioning the existance of something infinite.

 

[Nasor]

I don't get how you can think that 0.9r = 1
0.9r + 0.0r1 = 1

No, 0.9r + 0.01r = 1.011111111....

Adding any number to 0.9r give you a number greater than 1. The difference between 0.999... and 1 is infinitely small; since there is no difference, they equal each other.

woops, I didn't read your post very carefully you wanted to add 0.0r1 to 0.9r, not 0.01r. Actually, 0.0r1 is impossible - writing 0.xr implies that x is repeated out to infinity, so you can't have anything else after it.

 

[el-half]

That is correct, but 0.9r + 0.0r1 equals 1 for r is a very large finite value reaching to infinity. Well, that doesn't really make sense since it beats the concept of infinity but what I am trying to say is that you will never reach 1 unless you add a cetain value to 0.9r

 

[TheHeretic]

So if .9r=1 this bascially proves that infinity really does not exist. The only thing that could be infinte would be an infinte large number. But really this is restricted by time. For example lets say that every second we add a new number for infinity. OK every second a number is added. Then time stops ceasing the numbers bieng added.

 

[dristam]

i heard ... how 1+1 does not equal 2?

One + one is never =2 if the addends are in units of speed

 

[Nasor]

That is correct, but 0.9r + 0.0r1 equals 1 for r is a very large finite value reaching to infinity.

In "0.0r1" you have an infinite number of zeros before the 1, which means that the 1 has an infinitely small value and therefore equals zero. By saying ".9r + .0r1 = 1" you are basically saying that .9r + 0 = 1.

 

[LifeIsPeachy]

the proof that 1=.999999 is,


1/3=.3333333333333333
.: .33333333333333 * 3 = .999999999999999999
.: 1=.99999999999999999999

 

[LifeIsPeachy]

"So if .9r=1 this bascially proves that infinity really does not exist. The only thing that could be infinte would be an infinte large number. But really this is restricted by time. For example lets say that every second we add a new number for infinity. OK every second a number is added. Then time stops ceasing the numbers bieng added. "




not true. on a number scale, how many numbers are there greater than 1?

 

[Yuriy]

LifeIsPeachy,
3*1/3 = 0.9999...
It seems pretty easy and clear... Well done...
But actuall sense of such expression is
0.333333... → 1/3
and
0.999999... → 3*1/3 = 1
Math says in such cases:
"1/3 is a accumulation point of the sequence (0.3, 0.33, 0.333, ... , 0.3333..., ...) = {0.3333...}"
So is 1 for {0.9999...}

 

[AndersHermansson]

I have heard of (1+1 !== 2).
But it was a trick question. It was in base 2, not base 10. In the binary (base 2) system, 1 + 1 = 10. I think. I am a bit rusty on my binary calculations. Anyway, it doesn't equal 2.

Actually (10)_2 is the same number as (2)_10 =) ( _x denotes base).

 

[el-half]

and therefore equals zero

0.0r1 does not equal zero. If you claim that 0.0r1 equals zero then certain functions that reach to zero for infinity are actually 0 for that value. That is incorrect.

 

[Nasor]

0.0r1 does not equal zero. If you claim that 0.0r1 equals zero then certain functions that reach to zero for infinity are actually 0 for that value. That is incorrect.

As the number of zeros before the 1 in 0.0r1 goes to infinity, the total value of the number goes to zero. At infinity, it equals zero.

 

[el-half]

At infinity, it equals zero.

This statement constradicts the concept infinity. If you say "at infinity" infinity is a marked point and thus not infinite anymore.

 

[Silas]

0.0r1 does not equal zero. If you claim that 0.0r1 equals zero then certain functions that reach to zero for infinity are actually 0 for that value. That is incorrect.

Yes, but in that case, your forumulation 0.0r1 is not mathematically valid, as was pointed out above. You cannot have aleph-null zeros and then add a one at the end as if that means something. 0.999r is equal to 1.0.

In one way it isn't even that those algebraic manipulations involve dividing by zero. It is the fact that you generally state a = b, but then algebraically you are treating them as different. At some point you can always add stuff that only works because a - b = 0. This could be regarded as a proof that algebra has to assume that every element in an equation is a free variable, which is not the case if one is always tied to the other through absolute equality.

 

[el-half]

I still can't agree with you, but you are correct in saying that my 0.0r1 does not make sense.