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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: 3,646
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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
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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: 3,646
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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
Posts: 1,504
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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: 3,646
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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: 507
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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.
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: 507
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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.
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)
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?