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Mathematics Problem: Need Help/Confirmation

Ajabi
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11/25/2014 1:35:53 PM
Posted: 2 years ago
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)
Envisage
Posts: 3,648
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11/25/2014 1:58:20 PM
Posted: 2 years ago
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Just let x^7=q

q is just a label for a variable, so just let x^7 be your variable, substitute in.

Then we get:

1) x^7 + 1/x^7 = 1
2) q + 1/q = 1

This is exactly the same formulation as 3) x + 1/x = 1. So IF 3) is true, then 2 must be true.
Envisage
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11/25/2014 2:00:46 PM
Posted: 2 years ago
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

By the same logic, the formula Ax^4 + Bx^2 + C can just be rewritten as Aq^2 + Bq + c with q = x^2. Therefore you can do quadratics with this without worrying too much.
Ajabi
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11/25/2014 2:01:13 PM
Posted: 2 years ago
At 11/25/2014 1:58:20 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Just let x^7=q

q is just a label for a variable, so just let x^7 be your variable, substitute in.

Then we get:

1) x^7 + 1/x^7 = 1
2) q + 1/q = 1

This is exactly the same formulation as 3) x + 1/x = 1. So IF 3) is true, then 2 must be true.

Your proof already assumes that the second formation by implication is true. All we have is the first formation viz. 1 + 1/x = 1 and we are being asked whether this can be phrased as x^7 + 1/x^7 = 1.
Envisage
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11/25/2014 2:02:33 PM
Posted: 2 years ago
At 11/25/2014 2:01:13 PM, Ajabi wrote:
At 11/25/2014 1:58:20 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Just let x^7=q

q is just a label for a variable, so just let x^7 be your variable, substitute in.

Then we get:

1) x^7 + 1/x^7 = 1
2) q + 1/q = 1

This is exactly the same formulation as 3) x + 1/x = 1. So IF 3) is true, then 2 must be true.

Your proof already assumes that the second formation by implication is true. All we have is the first formation viz. 1 + 1/x = 1 and we are being asked whether this can be phrased as x^7 + 1/x^7 = 1.

Thats because the second formulation IS true, since you have already said:

"Given that x + 1/x = 1"
Envisage
Posts: 3,648
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11/25/2014 2:22:05 PM
Posted: 2 years ago
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Nvm.. did some more work on it. Seems my math is horribly shite nowadays...
Enji
Posts: 1,022
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11/25/2014 4:00:35 PM
Posted: 2 years ago
The induction hypthesis is x^n + 1/x^n = 1. You can assume this is true because you're given the n=1 case; now you need to prove that the n case implies the n+1 case. Something I would try would be multiplying the left side of the induction hypothesis by a fancy way to write 1 (for example, x+1/x) and see where that takes me. With a mix of factoring and rearranging terms, I think you should be able to get x^n+1 + 1/x^n+1 = 1 or equivalent.
Ajabi
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11/25/2014 9:12:40 PM
Posted: 2 years ago
At 11/25/2014 4:00:35 PM, Enji wrote:
The induction hypthesis is x^n + 1/x^n = 1. You can assume this is true because you're given the n=1 case; now you need to prove that the n case implies the n+1 case. Something I would try would be multiplying the left side of the induction hypothesis by a fancy way to write 1 (for example, x+1/x) and see where that takes me. With a mix of factoring and rearranging terms, I think you should be able to get x^n+1 + 1/x^n+1 = 1 or equivalent.

Thats how I originally went but for some reasons it does not work. :P
Ajabi
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11/25/2014 9:13:15 PM
Posted: 2 years ago
At 11/25/2014 2:22:05 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Nvm.. did some more work on it. Seems my math is horribly shite nowadays...

Yeah lol, also I went to sleep. :P
Envisage
Posts: 3,648
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11/25/2014 9:16:59 PM
Posted: 2 years ago
At 11/25/2014 9:13:15 PM, Ajabi wrote:
At 11/25/2014 2:22:05 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Nvm.. did some more work on it. Seems my math is horribly shite nowadays...

Yeah lol, also I went to sleep. :P

So about that debate....
Ajabi
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11/25/2014 9:18:42 PM
Posted: 2 years ago
At 11/25/2014 9:16:59 PM, Envisage wrote:
At 11/25/2014 9:13:15 PM, Ajabi wrote:
At 11/25/2014 2:22:05 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Nvm.. did some more work on it. Seems my math is horribly shite nowadays...

Yeah lol, also I went to sleep. :P

So about that debate....

By tonight promise. :P
Envisage
Posts: 3,648
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11/25/2014 9:27:46 PM
Posted: 2 years ago
At 11/25/2014 9:18:42 PM, Ajabi wrote:
At 11/25/2014 9:16:59 PM, Envisage wrote:
At 11/25/2014 9:13:15 PM, Ajabi wrote:
At 11/25/2014 2:22:05 PM, Envisage wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Nvm.. did some more work on it. Seems my math is horribly shite nowadays...

Yeah lol, also I went to sleep. :P

So about that debate....

By tonight promise. :P

I have already started writing arguments...

I have half written rebuttals to the ontology, and homosexuality. As well as a revamped version of teenage humping.

=D
Enji
Posts: 1,022
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11/27/2014 1:48:40 AM
Posted: 2 years ago
At 11/25/2014 9:12:40 PM, Ajabi wrote:
At 11/25/2014 4:00:35 PM, Enji wrote:
The induction hypthesis is x^n + 1/x^n = 1. You can assume this is true because you're given the n=1 case; now you need to prove that the n case implies the n+1 case. Something I would try would be multiplying the left side of the induction hypothesis by a fancy way to write 1 (for example, x+1/x) and see where that takes me. With a mix of factoring and rearranging terms, I think you should be able to get x^n+1 + 1/x^n+1 = 1 or equivalent.

Thats how I originally went but for some reasons it does not work. :P

Do you know the reasons, or did you just not get the answer you were looking for? I'm too lazy to do the algebra myself.
Ajabi
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11/27/2014 5:57:57 AM
Posted: 2 years ago
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Start with x^7 + 1/x^7 = 1

w^7 + 1/x^7 = 1

w^6*w + 1/w*6*w = 1

(w^3)^2*w + 1/(w^3)^2*w =1

Substitute w^3 as 1

1*w + 1/1*w = 1

w + 1/w = 1

Substitute w as x

x + 1/x = 1

We have successfully gone from our second equation to our first equation. I think there is enough proof that there exists a bridge between the two. Now if only we can manipulate this to go from our first to our second. :P
chui
Posts: 546
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11/27/2014 6:44:58 AM
Posted: 2 years ago
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Isn't this all about the properties of imaginary numbers?

This might be a possible solution depending how you interpret not solving for x.
The first equation has a general solution (ie not a specific solution) for x which is (2, +-(pi/3+2npi)) in argand form.
Raising this number to the seventh power would give (2,+-(7pi/3+14npi)) which is equivalent to (2,+_(pi/3+15npi)) and if you plot this point on an argand diagram it would be identical to (2,+_(pi/3+2npi)) so x is identical to x^7.

This suggests that in general x^(6n+1) gives a family of equations of the form x^(6n+1) + 1/x^(6n+1)=1 that are also implied.
Let's hope "the truth is out there" cos there is bugger all round here.
Ajabi
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11/27/2014 7:32:37 AM
Posted: 2 years ago
At 11/27/2014 6:44:58 AM, chui wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Isn't this all about the properties of imaginary numbers?

This might be a possible solution depending how you interpret not solving for x.
The first equation has a general solution (ie not a specific solution) for x which is (2, +-(pi/3+2npi)) in argand form.
Raising this number to the seventh power would give (2,+-(7pi/3+14npi)) which is equivalent to (2,+_(pi/3+15npi)) and if you plot this point on an argand diagram it would be identical to (2,+_(pi/3+2npi)) so x is identical to x^7.

This suggests that in general x^(6n+1) gives a family of equations of the form x^(6n+1) + 1/x^(6n+1)=1 that are also implied.

But see we can't solve explicitly for X something which you have done.
chui
Posts: 546
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11/27/2014 8:50:42 AM
Posted: 2 years ago
At 11/27/2014 7:32:37 AM, Ajabi wrote:
At 11/27/2014 6:44:58 AM, chui wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Isn't this all about the properties of imaginary numbers?

This might be a possible solution depending how you interpret not solving for x.
The first equation has a general solution (ie not a specific solution) for x which is (2, +-(pi/3+2npi)) in argand form.
Raising this number to the seventh power would give (2,+-(7pi/3+14npi)) which is equivalent to (2,+_(pi/3+15npi)) and if you plot this point on an argand diagram it would be identical to (2,+_(pi/3+2npi)) so x is identical to x^7.

This suggests that in general x^(6n+1) gives a family of equations of the form x^(6n+1) + 1/x^(6n+1)=1 that are also implied.

But see we can't solve explicitly for X something which you have done.

I was hoping that since I used a general solution for x not one single value it could count as a general proof but I do see your point. But it seems to me that it is pointless to not solve for x since it is defined by the first equation. Alternative solution tend to need horrible binomial expansions.
Let's hope "the truth is out there" cos there is bugger all round here.
Ajabi
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11/27/2014 10:41:12 AM
Posted: 2 years ago
At 11/27/2014 8:50:42 AM, chui wrote:
At 11/27/2014 7:32:37 AM, Ajabi wrote:
At 11/27/2014 6:44:58 AM, chui wrote:
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

Isn't this all about the properties of imaginary numbers?

This might be a possible solution depending how you interpret not solving for x.
The first equation has a general solution (ie not a specific solution) for x which is (2, +-(pi/3+2npi)) in argand form.
Raising this number to the seventh power would give (2,+-(7pi/3+14npi)) which is equivalent to (2,+_(pi/3+15npi)) and if you plot this point on an argand diagram it would be identical to (2,+_(pi/3+2npi)) so x is identical to x^7.

This suggests that in general x^(6n+1) gives a family of equations of the form x^(6n+1) + 1/x^(6n+1)=1 that are also implied.

But see we can't solve explicitly for X something which you have done.

I was hoping that since I used a general solution for x not one single value it could count as a general proof but I do see your point. But it seems to me that it is pointless to not solve for x since it is defined by the first equation. Alternative solution tend to need horrible binomial expansions.

The question was given to try and prepare the reader to "think" mathematically. I gave a pseudo-proof above, do tell me what you think.

An ideal proof would be with x^n yielding n to naturally be 7, or greater/less than equals to 7.
chui
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11/27/2014 11:00:35 AM
Posted: 2 years ago
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Sorry I don't follow the above. I am not familiar with omega. Is omega a particular function or value? Why is omega cubed always 1? This suggests omega is 1 and therefore you are substituting an explicit and inconsistent value for x. What am I missing here?

Start with x^7 + 1/x^7 = 1

w^7 + 1/x^7 = 1

w^6*w + 1/w*6*w = 1

(w^3)^2*w + 1/(w^3)^2*w =1

Substitute w^3 as 1

1*w + 1/1*w = 1

w + 1/w = 1

Substitute w as x

x + 1/x = 1

We have successfully gone from our second equation to our first equation. I think there is enough proof that there exists a bridge between the two. Now if only we can manipulate this to go from our first to our second. :P

I am happy that everything after the beginning is logical (There are a few typos but your meaning is clear)
Let's hope "the truth is out there" cos there is bugger all round here.
Ajabi
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11/27/2014 11:07:09 AM
Posted: 2 years ago
At 11/27/2014 11:00:35 AM, chui wrote:
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Sorry I don't follow the above. I am not familiar with omega. Is omega a particular function or value? Why is omega cubed always 1? This suggests omega is 1 and therefore you are substituting an explicit and inconsistent value for x. What am I missing here?

http://en.wikipedia.org...

Omega itself isn't one, but omega cubed (the imaginery root) is always 1. This means that we do not solve for x, we have x after all as omega, and as long as the constant x remains (even if it does so in another symbolic form) we did not solve for x explicityly, but used a substitution.


Start with x^7 + 1/x^7 = 1

w^7 + 1/x^7 = 1

w^6*w + 1/w*6*w = 1

(w^3)^2*w + 1/(w^3)^2*w =1

Substitute w^3 as 1

1*w + 1/1*w = 1

w + 1/w = 1

Substitute w as x

x + 1/x = 1

We have successfully gone from our second equation to our first equation. I think there is enough proof that there exists a bridge between the two. Now if only we can manipulate this to go from our first to our second. :P

I am happy that everything after the beginning is logical (There are a few typos but your meaning is clear)
Ajabi
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11/27/2014 11:08:33 AM
Posted: 2 years ago
Since omega itself does not have a value, and we supposed x as omega we have not solved anything explicitly. We use the fact that omeg^3 is 1 for our benefit.
9spaceking
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11/27/2014 11:24:50 AM
Posted: 2 years ago
At 11/27/2014 11:08:33 AM, Ajabi wrote:
Since omega itself does not have a value, and we supposed x as omega we have not solved anything explicitly. We use the fact that omeg^3 is 1 for our benefit.

wut.
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ESocialBookworm
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11/27/2014 1:34:04 PM
Posted: 2 years ago
At 11/25/2014 1:35:53 PM, Ajabi wrote:
Given that x + 1/x = 1
Show that the above statement implies x^7 + 1/x^7 = 1 without solving x (it should be a general proof).

I have had a few ideas for this, and I believe I got it finally, but I want to see how others try and solve it. A rather entertaining problem I believe. Well ddo give me your ideas.

(Please remember the principles of mathematical induction)

I got a test on this today! Lol!
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Enji
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11/27/2014 3:25:33 PM
Posted: 2 years ago
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Does x=w? It's easy to solve for x= cubedroot(-1) which is not equal to the omega constant. x^3 is -1 instead of 1. This doesn't change things too much; you get (-1)^2*x+1/[(-1)^2*x]=1 -- but how can you go backwards if you don't know x^3=-1?
Ajabi
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11/27/2014 8:42:57 PM
Posted: 2 years ago
At 11/27/2014 3:25:33 PM, Enji wrote:
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Does x=w? It's easy to solve for x= cubedroot(-1) which is not equal to the omega constant. x^3 is -1 instead of 1. This doesn't change things too much; you get (-1)^2*x+1/[(-1)^2*x]=1 -- but how can you go backwards if you don't know x^3=-1?

Thats the thing we don't need to. Since we do not have to solve x, we can substitute it as a constant, can't we? I remember reading about how in maths induction you can substitute a variable by an unknown constant, and then manipulate the constant to reach your answer.
Enji
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11/27/2014 10:21:21 PM
Posted: 2 years ago
At 11/27/2014 8:42:57 PM, Ajabi wrote:
At 11/27/2014 3:25:33 PM, Enji wrote:
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Does x=w? It's easy to solve for x= cubedroot(-1) which is not equal to the omega constant. x^3 is -1 instead of 1. This doesn't change things too much; you get (-1)^2*x+1/[(-1)^2*x]=1 -- but how can you go backwards if you don't know x^3=-1?

Thats the thing we don't need to. Since we do not have to solve x, we can substitute it as a constant, can't we? I remember reading about how in maths induction you can substitute a variable by an unknown constant, and then manipulate the constant to reach your answer.

You said you need to prove that "x + 1/x = 1" is true implies "x^7 + 1/x^7 = 1" is true. "x + 1/x = 1" is not true if x=w, so I don't know how you are going to prove "x^7 + 1/x^7 = 1" given "x + 1/x = 1" by substituting x with omega.
Ajabi
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11/28/2014 12:06:49 AM
Posted: 2 years ago
At 11/27/2014 10:21:21 PM, Enji wrote:
At 11/27/2014 8:42:57 PM, Ajabi wrote:
At 11/27/2014 3:25:33 PM, Enji wrote:
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Does x=w? It's easy to solve for x= cubedroot(-1) which is not equal to the omega constant. x^3 is -1 instead of 1. This doesn't change things too much; you get (-1)^2*x+1/[(-1)^2*x]=1 -- but how can you go backwards if you don't know x^3=-1?

Thats the thing we don't need to. Since we do not have to solve x, we can substitute it as a constant, can't we? I remember reading about how in maths induction you can substitute a variable by an unknown constant, and then manipulate the constant to reach your answer.

You said you need to prove that "x + 1/x = 1" is true implies "x^7 + 1/x^7 = 1" is true. "x + 1/x = 1" is not true if x=w, so I don't know how you are going to prove "x^7 + 1/x^7 = 1" given "x + 1/x = 1" by substituting x with omega.

Oh yeah you're right Enji. :p Well back to square one.
chui
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11/28/2014 4:25:23 AM
Posted: 2 years ago
At 11/28/2014 12:06:49 AM, Ajabi wrote:
At 11/27/2014 10:21:21 PM, Enji wrote:
At 11/27/2014 8:42:57 PM, Ajabi wrote:
At 11/27/2014 3:25:33 PM, Enji wrote:
At 11/27/2014 5:57:57 AM, Ajabi wrote:
All right I have a pseudo-proof. Since we can't solve "x" explicitly this is all I have.

Suppose "x" as "w" (omega).
We know that w^3 is 1 (omega cubed is always 1)

Does x=w? It's easy to solve for x= cubedroot(-1) which is not equal to the omega constant. x^3 is -1 instead of 1. This doesn't change things too much; you get (-1)^2*x+1/[(-1)^2*x]=1 -- but how can you go backwards if you don't know x^3=-1?

Thats the thing we don't need to. Since we do not have to solve x, we can substitute it as a constant, can't we? I remember reading about how in maths induction you can substitute a variable by an unknown constant, and then manipulate the constant to reach your answer.

You said you need to prove that "x + 1/x = 1" is true implies "x^7 + 1/x^7 = 1" is true. "x + 1/x = 1" is not true if x=w, so I don't know how you are going to prove "x^7 + 1/x^7 = 1" given "x + 1/x = 1" by substituting x with omega.

Oh yeah you're right Enji. :p Well back to square one.

How about this:-

Lets assume x is a root of the equation x^6=1
Then whatever value x has it must be true that x^6=1 and your solution above solves the rest.

Now x^6=1 can be expanded to

(x-1)(x+1)(x^2+x+1)(x^2-x+1)=0 ...........(1)

But x+1/x=1 implies that x^2-x+1=0 so this implies x satisfies (1) since the last bracket is zero. QED?
Let's hope "the truth is out there" cos there is bugger all round here.