## Where Algebra Fails

### Where Algebra Fails

1) Let n = 3/2 + 9/4 + 27/8 + ...

Multiply each side of 1) by 2/3

2) (2/3)n = 1 + 3/2 + 9/4 + 27/8 + ...

Subtract n or its equivalent from each side of 2)

3) (-1/3)n = 1

Multiply each side of 3) by -3

4) n = -3

Therefore 3/2 + 9/4 + 27/8 + ... = -3

Where is the error?

Multiply each side of 1) by 2/3

2) (2/3)n = 1 + 3/2 + 9/4 + 27/8 + ...

Subtract n or its equivalent from each side of 2)

3) (-1/3)n = 1

Multiply each side of 3) by -3

4) n = -3

Therefore 3/2 + 9/4 + 27/8 + ... = -3

Where is the error?

*Paidion**Man judges a person by his past deeds, and administers penalties for his wrongdoing. God judges a person by his present character, and disciplines him that he may become righteous.*

Avatar shows me at 75 years old. I am now 82.

- willowtree
**Posts:**100**Joined:**Sun Nov 02, 2014 1:56 pm**Location:**Sooke BC Canada

### Re: Where Algebra Fails

Paidion wrote:1) Let n = 3/2 + 9/4 + 27/8 + ...

Multiply each side of 1) by 2/3

2) (2/3)n = 1 + 3/2 + 9/4 + 27/8 + ...

Subtract n or its equivalent from each side of 2)

3) (-1/3)n = 1

Multiply each side of 3) by -3

4) n = -3

Therefore 3/2 + 9/4 + 27/8 + ... = -3

Where is the error?

I don't think the rules of algebra are the problem.

It seems to me that n is an ongoing progression, the next one in line 1) would be 81/16, etc.

In order to keep the equation balanced you must determine how many progressions you want to solve for the value of n and not keep adding more as you go along.

Let n = 3/2 + 9/4 + 27/8 + ... (That is, after three progressions, n = 1.50 +2.15 +3.375 = 7.125)

So for the location of the error we must stop after 27/8 ( the third progression: you cannot just keep adding to one side of the equation and not the other).

Multiply each side of 1) by 2/3

2) (2/3)n = 1 + 3/2 + 9/4 + 27/8 + ... WRONG. (2/3)n = (only) 1 + 3/2 + 9/4

(limit to 3 progressions +27/8 adds another progression on the right and upsets the equation. )

Wish all theological problems were this easy.

Graeme

If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

### Re: Where Algebra Fails

Willow, nothing has been added to one side and not the other.

Here is a valid proof based on the same algebraic principles:

Proof that 1/2 + 1/4 +1/8 + 1/16 + ... = 1

1) Let n = 1/2 + 1/4 +1/8 + 1/16 + ...

Multiply each side of 1) by 2

2) 2n = 1 + 1/2 + 1/4 +1/8 + 1/16 + ...

Subtract n or its equivalent from each side of 2)

3) n=1

Therefore 1/2 + 1/4 +1/8 + 1/16 + ... = 1

The series is an infinite sum, but it is still exactly equal to 1 and not a little bit less as our intuition tells us.

Any competent mathematician will tell you that this proof is valid.

Similarly, we shall prove that .999... =1 (and not a little bit less as our intuition tells us)

1) Let n = .999..

Multiply each side of 1) by 10

2) 10n = 9.999...

Subtract n or its equivalent from each side of 2)

3) 9n = 9

Divide each side of 3) by 9

4) n=1

Therefore .999... = 1

This is another valid proof. So why is the original "proof" that I offered NOT valid?

Here is a valid proof based on the same algebraic principles:

Proof that 1/2 + 1/4 +1/8 + 1/16 + ... = 1

1) Let n = 1/2 + 1/4 +1/8 + 1/16 + ...

Multiply each side of 1) by 2

2) 2n = 1 + 1/2 + 1/4 +1/8 + 1/16 + ...

Subtract n or its equivalent from each side of 2)

3) n=1

Therefore 1/2 + 1/4 +1/8 + 1/16 + ... = 1

The series is an infinite sum, but it is still exactly equal to 1 and not a little bit less as our intuition tells us.

Any competent mathematician will tell you that this proof is valid.

Similarly, we shall prove that .999... =1 (and not a little bit less as our intuition tells us)

1) Let n = .999..

Multiply each side of 1) by 10

2) 10n = 9.999...

Subtract n or its equivalent from each side of 2)

3) 9n = 9

Divide each side of 3) by 9

4) n=1

Therefore .999... = 1

This is another valid proof. So why is the original "proof" that I offered NOT valid?

*Paidion**Man judges a person by his past deeds, and administers penalties for his wrongdoing. God judges a person by his present character, and disciplines him that he may become righteous.*

Avatar shows me at 75 years old. I am now 82.

- willowtree
**Posts:**100**Joined:**Sun Nov 02, 2014 1:56 pm**Location:**Sooke BC Canada

### Re: Where Algebra Fails

Paidion wrote:Willow, nothing has been added to one side and not the other.

Here is a valid proof based on the same algebraic principles:

Proof that 1/2 + 1/4 +1/8 + 1/16 + ... = 1

1) Let n = 1/2 + 1/4 +1/8 + 1/16 + ...

Multiply each side of 1) by 2

2) 2n = 1 + 1/2 + 1/4 +1/8 + 1/16 + ...

Subtract n or its equivalent from each side of 2)

3) n=1

Therefore 1/2 + 1/4 +1/8 + 1/16 + ... = 1

The series is an infinite sum, but it is still exactly equal to 1 and not a little bit less as our intuition tells us.

Any competent mathematician will tell you that this proof is valid.

So why is the original "proof" that I offered NOT valid?

I give in, but just to give me some comfort, please show me how it works with the following:

Since it is is well established in mathematics that 1 +2 + 3 + 4 is equal to 4 + 3 + 2 + 1 would you please, using the same logic as above, convince me that n = 1 in this series.

let n = 1/16 + 1/8 + 1/4 + 1/2 + ...

Graeme

If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

### Re: Where Algebra Fails

You haven't actually reversed the series, Graeme. The original series had 1/2 as the first element, and there is no element in the series larger than 1/2

In the series you offered, there are an infinite number of elements which are larger than 1/2. Also, the smallest element in your series is 1/16, whereas there are an infinite number of elements in the original series which are less than 1/16. However, you could reverse the series by putting the three dots at the beginning, if that will give you any comfort:

n= ...1/16 + 1/8 + 1/4 + 1/2

I think the algebra would still work fine with that one.

But I will point the following out to any who doubt that .999... =1

We have all learned in elementary school that 1/3 = .333... and 2/3 = .666...

Add 1/3 and 2/3 and you have 3/3 or 1.

Add

.333...

.666...

.999...

Therefore .999... = 1

By the way, this has some interesting implications 8 = 7.999..., 13 = 12.999..., etc.

If no one figures out why algebra doesn't work (fails) in the original post, I'll give a reason in my next post which was given to me by some expert mathematicians.

In the series you offered, there are an infinite number of elements which are larger than 1/2. Also, the smallest element in your series is 1/16, whereas there are an infinite number of elements in the original series which are less than 1/16. However, you could reverse the series by putting the three dots at the beginning, if that will give you any comfort:

n= ...1/16 + 1/8 + 1/4 + 1/2

I think the algebra would still work fine with that one.

But I will point the following out to any who doubt that .999... =1

We have all learned in elementary school that 1/3 = .333... and 2/3 = .666...

Add 1/3 and 2/3 and you have 3/3 or 1.

Add

.333...

.666...

.999...

Therefore .999... = 1

By the way, this has some interesting implications 8 = 7.999..., 13 = 12.999..., etc.

If no one figures out why algebra doesn't work (fails) in the original post, I'll give a reason in my next post which was given to me by some expert mathematicians.

*Paidion**Man judges a person by his past deeds, and administers penalties for his wrongdoing. God judges a person by his present character, and disciplines him that he may become righteous.*

Avatar shows me at 75 years old. I am now 82.

### Re: Where Algebra Fails

Math can assume an actual infinity for mathematical purposes, but don't endless sequences require special handling? What makes me suspicious is multiplying an endless sequence by a number.

~ Classical Arminian Christian Mystic ~

- willowtree
**Posts:**100**Joined:**Sun Nov 02, 2014 1:56 pm**Location:**Sooke BC Canada

### Re: Where Algebra Fails

[quote="Paidion"

In the series you offered, there are an infinite number of elements which are larger than 1/2.[/quote]

My point exactly - your first series, with errors, includes 3/2, 9/4, 27/8 ? These are all greater than 1/2 and begin an infinite series which increase in value.

Graeme

In the series you offered, there are an infinite number of elements which are larger than 1/2.[/quote]

My point exactly - your first series, with errors, includes 3/2, 9/4, 27/8 ? These are all greater than 1/2 and begin an infinite series which increase in value.

Graeme

If you find yourself between a rock and a hard place, always head for the rock. Ps 62..

### Re: Where Algebra Fails

Hi Dizerner,

It is mathematically correct to multiply each term of and endless sequence by a number, as was done in the proof that 1/2 + 1/4 + 1/8 + ... =1Math can assume an actual infinity for mathematical purposes, but don't endless sequences require special handling? What makes me suspicious is multiplying an endless sequence by a number.

*Paidion*Avatar shows me at 75 years old. I am now 82.

### Re: Where Algebra Fails

Hi Graeme,

I gave reason WHY it is not the reverse, one being that it contains elements that are larger than 1/2. But my series 1/2 + 1/4 + 1/8 +... do NOT contain elements larger than 1/2.

The reverse of a series must contain the same elements as is the case with 4+3+2+1. It contains identical elements to the series 1+2+3+4.

I understood what you were trying to do by making the elements increase as in the original series. And you were on the right track!

So now I must do what I promised. The experts said that while algebra works with infinite series that converge, such as 1/2 + 1/4 + 1/8 +..., it does not work with series that diverge such as 3/2 + 9/4 + 27/8 +... They did not state WHY algebra doesn't work with divergent series. So I still do not know why. If you should find out, please post the answer. I would very much like to know!

That is true that the original series which I gave increases in value. But you seemed to state that you were going to give the series 1/2 + 1/4 + 1/8 +... in reverse order, your having given 1+2+3+4 as equivalent to 4+3+2+1 as an example. But the series you gave, was 1/16 + 1/8 + 1/4 + 1/2 + ... and this is NOT the reverse of 1/2 + 1/4 + 1/8 +...You wrote:My point exactly - your first series, with errors, includes 3/2, 9/4, 27/8 ? These are all greater than 1/2 and begin an infinite series which increase in value.Paidion wrote:In the series you offered, there are an infinite number of elements which are larger than 1/2.

I gave reason WHY it is not the reverse, one being that it contains elements that are larger than 1/2. But my series 1/2 + 1/4 + 1/8 +... do NOT contain elements larger than 1/2.

The reverse of a series must contain the same elements as is the case with 4+3+2+1. It contains identical elements to the series 1+2+3+4.

I understood what you were trying to do by making the elements increase as in the original series. And you were on the right track!

So now I must do what I promised. The experts said that while algebra works with infinite series that converge, such as 1/2 + 1/4 + 1/8 +..., it does not work with series that diverge such as 3/2 + 9/4 + 27/8 +... They did not state WHY algebra doesn't work with divergent series. So I still do not know why. If you should find out, please post the answer. I would very much like to know!

*Paidion*Avatar shows me at 75 years old. I am now 82.