1=4

[Shaden]

1-1=4-4
1(1-1)=4(1-1)
1=4

You can't argue against that logic :P

[Jill]

You can't argue against that logic

Huh?

Sure you can. Your answer is simply wrong as shown below:

1(1-1) = 1*0 = 0

4(1-1) = 4*0 = 0

0=0

[Shaden]

Lol I know I was just being sarcastic.

In algebra the general concensus is that you don't split whole numbers - in other words if you have a series of whole numbers in an equation you make them into one whole number by adding them together. In that sense the original equation would have just amounted to 0=0 with no need to expand it any further.

This is just the kind of thing my friend constantly goes on about though and it just sort of gets drilled into my head

Woot, century!

[Stephen]

1 = 0.9999999999999999999999999999999


3
- = 1
3

1
- = 0.3333333333333333333333333333333
3



3
- = 0.3333333333333333333333333333333 x 3
3



= 0.9999999999999999999999999999999999


3
- = 1
3

[Erki]

1-0.(9)=0.(0)=0
1=0.(9)

[andrew]

Not another 0.9 repeating equals 1 thread again...

Sigh.

[Baltar]

Try and prove this wrong.

667 = 6.67e2 = 1010011011 = 01233 = 0x29b

[Matthew Shea]

Originally Posted by Baltar

Try and prove this wrong.

667 = 6.67e2 = 1010011011 = 01233 = 0x29b

Impossible. You've written the same number in decimal, exponential, binary, octal and hexadecimal notation.

[icesar]

Ok, don't take this the wrong way, but as long as we're playing with equations

Women = time x money

But time = money, so
women = money x money = money^2

However, we know that money is the root of all evil, so

Women = (√evil)^2

Therefore,
Women = evil

QED

[skinnyninja]

This proved that 2 is equal to 1 using basic Algebra. My teacher told me that it held up and no one could figure it out for about a hundred years, but then a long time ago someone saw the flaw in it. See if you can find the flaw:


A=B

and neither A or B is equal to zero

so the equation starts

A=B

Multiply both sides by A

A squared = AB

Subtract B squared from both sides of the equation

A squared – B squared = AB – B squared

Now factor, which gives you

(A-B)(A+B) = B(A-B)

Now, divide both sides by (A-B)….this will cancel out the (A-B) on both sides of the equation

So you are left with

(A+B) = B

Substitute in any number for A (like 1), and you get

2 = 1

[Shaden]

My teacher told me that it held up and no one could figure it out for about a hundred years, but then a long time ago someone saw the flaw in it. See if you can find the flaw:

Give me a hundred years and I'll get back to you on it.

Seriously though I'll have a think about it. Probably something really basic.

Edit: Actually I think I figured it out. If (A+B)=B then A has to equal 0, otherwise B could not logically equal B.

[gberardi]

Originally Posted by skinnyninja

Now factor, which gives you

(A-B)(A+B) = B(A-B)

Now, divide both sides by (A-B)….this will cancel out the (A-B) on both sides of the equation

You can't divide by zero: does not compute.

A = B, so A-B is 0, and you can't divide by 0.

[Matthew Shea]

Originally Posted by Stephen

1 = 0.9999999999999999999999999999999


3
- = 1
3

1
- = 0.3333333333333333333333333333333
3



3
- = 0.3333333333333333333333333333333 x 3
3



= 0.9999999999999999999999999999999999


3
- = 1
3

The problem with this is that .333333333333333333333 and .9999999999999999999999999 are only approximations of the value of the fractions 1/3 and 3 x 1/3. It is impossible to write the true value of 1/3 in decimal form since you'd need an infinite number of digits. The logic above fails to account for the rounding errors you get when dealing with approximations. The fact is 3/3 = 1 by definition. .99999999999999999999999 is what you get when you multiply the decimal approximation of 1/3 by 3.

[DiscoDan]

Originally Posted by Matthew Shea

The problem with this is that .333333333333333333333 and .9999999999999999999999999 are only approximations of the value of the fractions 1/3 and 3 x 1/3. It is impossible to write the true value of 1/3 in decimal form since you'd need an infinite number of digits. The logic above fails to account for the rounding errors you get when dealing with approximations. The fact is 3/3 = 1 by definition. .99999999999999999999999 is what you get when you multiply the decimal approximation of 1/3 by 3.

I don't think Stephen was talking about the decimal approximation, and was in fact talking about the actual representation of 1/3 in decimal notation (ie. 0.3, with the 3 repeating infinitely many times). Here's another 'proof':

x = 0.9... (where ... is defined to mean the previous digit repeating infinitely many times)
10x = 9.9...
10x - x = 9.9... - 0.9...
9x = 9
x = 1

[natopoto11]

This is a theory that I came up with in highschool about repeating decimals:

There exist different densities for each number system, resulting in an infinite amount of numbers contained within the set, but there exist holes in the number line, meaning that certain numbers are not contained within the set.

My theory basically states that for a number to exist in a number set (binary, decimal, etc), it must follow the form of an interger divided by a combination of the factors of the base

for example, the factors of 10 (decimal) are 1,2,5 (also 10, but it's not prime).

so the numbers contained within the decimal system are :
x/(1^a*2^b*5^c), where x,a,b, and c are intergers.

since 1/3 does not follow this form, it simply does not exist in base 10. This is not to say that it does not exist at all, but it is one of the holes in the decimal system.

Some may argue that this theory is really describing terminating decimals, but my argument is that only numbers that can be expressed using a number system (i.e. terminating decimals) are the only numbers to exist in that particular system. The missing numbers do exists, but they are not contained within that particular set.

My explaination for this is taken from my observation of line segments. Consider a line segment that spans from 0 to 1 compared and a line segment that spans from 0 to 2. Both line segments contain an infinite amount of points, yet the second segment is twice as long. In other words, the second segment is twice as dense and contains points (numbers) that are not contained within the first line segment. The points that are not contained within the first line segment (1 to 2) do exist, but just not in that segment.

Since non-terminating decimals do not exist in a particular number system, using abstractions of those numbers in a proof is invalid.

What you are essentially saying when you say that 0.999...999 = 1 is that 0.000...001 = 0 or that a single point is nothing. Saying that something is nothing is obviously flawed. The problem is in rounding to cope with our finite minds. We think that 0.999...999 / 3 = 0.333...333. This too is incorrect, because 0.999...999 is equal to 1-0.000...001 or one point less than 1. So 0.999...999 / 3 would be the same as (1-000...001) / 3 or 1/3 + (000...001)/3. A single point is the smallest numerical unit possible, so it cannot be divided by 3 into a smaller unit. To cope with this paradox, we simply round 0.000...001 / 3 to it's closet equivalent, which is 0.000...000 (1/3 is closer to 0 than to 1).

It may seem strange, since we are rounding to the infinith degree, but let's look at an example using the interger number set (i.e. no partial numbers exist within the set).

The closest representation for 1/2 that we can get is 1 (since 1/2 is not contained within the set).
Using the interger number system, (1/2) + (1/2) = 2, which is obviously wrong.

In the same way, we are rounding off non-terminating decimals to conform to it's closet representation in our current number set.

[DiscoDan]

Originally Posted by natopoto11

My theory basically states that for a number to exist in a number set (binary, decimal, etc), it must follow the form of an interger divided by a combination of the factors of the base

for example, the factors of 10 (decimal) are 1,2,5 (also 10, but it's not prime).

I don't understand. Surely the set of all reals contains values that do not fit this framework. Take for example, the square root of three, or pi. There is simply no way we can write these numbers as an integer divided by other integers.

so the numbers contained within the decimal system are :
x/(1^a*2^b*5^c), where x,a,b, and c are integers.

since 1/3 does not follow this form, it simply does not exist in base 10.

I think you are confused. Base 10 is not a number system, it is merely a representation of one. For example, the set of all rational numbers would be the exact same set regardless of whether we are working in base 10 or base 3 (since we can simply set up a one-to-one mapping system). The only thing that will change is the symbols we use to represent each unique value of the set.

My explaination for this is taken from my observation of line segments. Consider a line segment that spans from 0 to 1 compared and a line segment that spans from 0 to 2. Both line segments contain an infinite amount of points, yet the second segment is twice as long. In other words, the second segment is twice as dense and contains points (numbers) that are not contained within the first line segment. The points that are not contained within the first line segment (1 to 2) do exist, but just not in that segment.

I don't understand your explanation as to how the line segment from one to two is half as dense as the one from zero to two. Surely they have the same number of points per unit length?

What you are essentially saying when you say that 0.999...999 = 1 is that 0.000...001 = 0 or that a single point is nothing.

I don't quite understand your concept of 0.000...001. How would we define such a number? For example, what is the "length" between the number 0.000...001 and 0.000...002? Then what is the length between 0.000...001 and 0.000...999? At what point do our infinitesimal lengths converge to finite lengths?

[natopoto11]

Hey DiscoDan. Thanks for replying to my theory. Let me try to answer some of your questions.

Originally Posted by DiscoDan

I don't understand. Surely the set of all reals contains values that do not fit this framework. Take for example, the square root of three, or pi. There is simply no way we can write these numbers as an integer divided by other integers.

It is true that there is no way to represent the square root of three or pi as a terminating decimal. Although they cannot be represented, I did not say that they do not exist. We do math of Real Numbers, but use Base 10 in our calculations. My argument is that there exist numbers in the Real Number set that do not exist in Base 10, so when you try to calculate certain problems (like 1/3), you get an error. The closest approximation that you can get is 0.333...333. The same problem occurs when trying to perform math according to Base 10, using binary in computers (like 1/10).

Originally Posted by DiscoDan

I think you are confused. Base 10 is not a number system, it is merely a representation of one. For example, the set of all rational numbers would be the exact same set regardless of whether we are working in base 10 or base 3 (since we can simply set up a one-to-one mapping system). The only thing that will change is the symbols we use to represent each unique value of the set.

You're right. I confused number system with the medium we use to represent the system. A number is a number regardless on how you represent it. The only problem is that there is meaning lost through the medium. It's like trying to write Chinese using the English alphabet. It's not that the Chinese word does not exist, but they simply do not exist in the English alphabet. So we must approximate with what we have.

Originally Posted by DiscoDan

I don't understand your explanation as to how the line segment from one to two is half as dense as the one from zero to two. Surely they have the same number of points per unit length?

What I was trying to say here was that each segment contains an infinite number of point, yet one segment is twice as long as another. Since they both contain an infinite amount of points, it would seem as if they were equal, but we know that they are not because they are different lengths. In the same way, we think that using binary and decimal are the same because they both contain an infinite amount of points, but they may in fact be different.

Originally Posted by DiscoDan

I don't quite understand your concept of 0.000...001. How would we define such a number? For example, what is the "length" between the number 0.000...001 and 0.000...002? Then what is the length between 0.000...001 and 0.000...999? At what point do our infinitesimal lengths converge to finite lengths?

My concept of 0.000...001 is the number that is directly after 0. If 0 and .5 have a distinct position on a number line as points, then why can't the point directly next to those points? There is no "length" between the number 0.000...001 and 0.000...002, but there is a 998 point distance between 0.000...001 and 0.000...999.

The infinitesimally small size converges to a finite length at infinity. Now we get into the problem of Zeno's Arrow. In order for an arrow to hit it's target, it must first pass through the halfway point. Before it reaches the halfway point, it also must reach the halfway point from the origin to the halfway point. If there are an infinite halfway points, then it seems that it is impossible for the arrow to reach it's target, but somehow it does. As we get smaller and smaller, we reach the limit of the single point.

Infinity is not a number, nor is it constant. Saying that there are an infinite amount of numbers using binary is not the same as saying that there are an infinite amount of numbers using decimal.

I know that you may still not agree with me, but hopefully I cleared some confusion (rather than made more confusion).

[DiscoDan]

Originally Posted by natopoto11

Hey DiscoDan. Thanks for replying to my theory. Let me try to answer some of your questions.

Anytime you wanna chat math, I'd happily oblige.

It is true that there is no way to represent the square root of three or pi as a terminating decimal. Although they cannot be represented, I did not say that they do not exist. We do math of Real Numbers, but use Base 10 in our calculations. My argument is that there exist numbers in the Real Number set that do not exist in Base 10, so when you try to calculate certain problems (like 1/3), you get an error. The closest approximation that you can get is 0.333...333. The same problem occurs when trying to perform math according to Base 10, using binary in computers (like 1/10).

OK. Let's try a new definition for base 10 then. Let's say that all base 10 numbers are represented by:

(integer in base 10) divided by (some product of integers).

However, in our new base 10 system, we never use decimals.

In this case, by definition, one third is in our system (as it follows the integer divided by integer definition), while 0.3333... is not.

This definition is sufficient to prove that there can be no such point that is "one step bigger" than any number in our system. For example, take the base 10 number 1/7. We can look at the number (1/7) plus one.

Now, divide 1 in half to get the new base 10 number (1/7 plus (1/2) ).

By definition, this number is still base 10 (since it follows the given rule of integer divided by some product of integers), and it is also closer to one-seventh than one-seventh plus one.

Now divide the second number (ie. 1/2) by 2 again. Notice we still have a base 10 number, and we are closer to one-seventh yet again.

How many times can we continue this process?

The logical answer would seem to be as many times as we want. In other words, if you say, "I bet you can't get closer to one-seventh than (insert value here) in base 10," I can always give you a value closer than the one you gave, simply by taking your number, subtracting one-seventh, then dividing that number by two, and then adding that back to one-seventh. As a result, there can be no "least upper bound" to one-seventh that is not one-seventh itself (in other words, we cannot have a "next greatest value" above one-seventh in our new base 10 system).

Notice that this definition of "base 10 numbers" should also be sufficient to prove that the sum of (1/9)^n, where n goes from one to infinity is in fact, equal to one. I say "should" because I've pretty much completely forgotten my calc series material, although, if you really want me to, I can look it up.

[natopoto11]

Originally Posted by DiscoDan

Notice that this definition of "base 10 numbers" should also be sufficient to prove that the sum of (1/9)^n, where n goes from one to infinity is in fact, equal to one. I say "should" because I've pretty much completely forgotten my calc series material, although, if you really want me to, I can look it up.

Yeah, it's been a while since I took calc as well. We are such geeks.

[DiscoDan]

I don't know if you were expecting a proof or not for that formula, but check out this wikipedia page:

Geometric progression - Wikipedia, the free encyclopedia

Then scroll down to the "Infinite geometric series" section. Use a=(9/10) and r=(1/10), both values in our new base 10 system, and you will yield the result of 1. Unfortunately, wikipedia does not provide a proof of the formula, but I will keep searching.