Total Posts:28|Showing Posts:1-28

# Mathematics Problem: Need Help/Confirmation

 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/25/2014 1:35:53 PMPosted: 3 years agoGiven that x + 1/x = 1Show 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)
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 1:58:20 PMPosted: 3 years agoAt 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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=qq 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 = 12) q + 1/q = 1This is exactly the same formulation as 3) x + 1/x = 1. So IF 3) is true, then 2 must be true.
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 2:00:46 PMPosted: 3 years agoAt 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/25/2014 2:01:13 PMPosted: 3 years agoAt 11/25/2014 1:58:20 PM, Envisage wrote:At 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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=qq 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 = 12) q + 1/q = 1This 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.
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 2:02:33 PMPosted: 3 years agoAt 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 = 1Show 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=qq 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 = 12) q + 1/q = 1This 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"
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 2:22:05 PMPosted: 3 years agoAt 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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...
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 11/25/2014 4:00:35 PMPosted: 3 years agoThe 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/25/2014 9:12:40 PMPosted: 3 years agoAt 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
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/25/2014 9:13:15 PMPosted: 3 years agoAt 11/25/2014 2:22:05 PM, Envisage wrote:At 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 9:16:59 PMPosted: 3 years agoAt 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 = 1Show 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. :PSo about that debate....
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/25/2014 9:18:42 PMPosted: 3 years agoAt 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 = 1Show 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. :PSo about that debate....By tonight promise. :P
 Posts: 3,881 Add as FriendChallenge to a DebateSend a Message 11/25/2014 9:27:46 PMPosted: 3 years agoAt 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 = 1Show 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. :PSo about that debate....By tonight promise. :PI 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
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 11/27/2014 1:48:40 AMPosted: 3 years agoAt 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. :PDo 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 5:57:57 AMPosted: 3 years agoAll 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 = 1w^7 + 1/x^7 = 1w^6*w + 1/w*6*w = 1(w^3)^2*w + 1/(w^3)^2*w =1Substitute w^3 as 11*w + 1/1*w = 1w + 1/w = 1Substitute w as xx + 1/x = 1We 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
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 11/27/2014 6:44:58 AMPosted: 3 years agoAt 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 7:32:37 AMPosted: 3 years agoAt 11/27/2014 6:44:58 AM, chui wrote:At 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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.
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 11/27/2014 8:50:42 AMPosted: 3 years agoAt 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 = 1Show 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 10:41:12 AMPosted: 3 years agoAt 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 = 1Show 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.
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 11/27/2014 11:00:35 AMPosted: 3 years agoAt 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 = 1w^7 + 1/x^7 = 1w^6*w + 1/w*6*w = 1(w^3)^2*w + 1/(w^3)^2*w =1Substitute w^3 as 11*w + 1/1*w = 1w + 1/w = 1Substitute w as xx + 1/x = 1We 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. :PI 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 11:07:09 AMPosted: 3 years agoAt 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 = 1w^7 + 1/x^7 = 1w^6*w + 1/w*6*w = 1(w^3)^2*w + 1/(w^3)^2*w =1Substitute w^3 as 11*w + 1/1*w = 1w + 1/w = 1Substitute w as xx + 1/x = 1We 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. :PI am happy that everything after the beginning is logical (There are a few typos but your meaning is clear)
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 11:08:33 AMPosted: 3 years agoSince 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.
 Posts: 4,250 Add as FriendChallenge to a DebateSend a Message 11/27/2014 11:24:50 AMPosted: 3 years agoAt 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.Equestrian election http://www.debate.org... This House would impose democracy http://www.debate.org... Reign of Terror is unjustified http://www.debate.org... Raise min. wage to \$10.10 http://www.debate.org...
 Posts: 14,720 Add as FriendChallenge to a DebateSend a Message 11/27/2014 1:34:04 PMPosted: 3 years agoAt 11/25/2014 1:35:53 PM, Ajabi wrote:Given that x + 1/x = 1Show 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!Solonkr~ I don't care about whether an ideology is "necessary" or not, I care about how to solve problems, which is what everyone else should also care about. Ken~ In essence, the world is fucked up and you can either ignore it, become cynical or bitter about it. Deep down, we're all dumbassses who act like shittheads Me~ "BAILEY + SOLON = SAILEY MY SHIP SAILEY MUST SAIL" SCREW THAT SHIZ #BANNIE = BAILEY & ANNIE P.S. Shipped Sailey before it was cannon bitches.
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 11/27/2014 3:25:33 PMPosted: 3 years agoAt 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?
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/27/2014 8:42:57 PMPosted: 3 years agoAt 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.
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 11/27/2014 10:21:21 PMPosted: 3 years agoAt 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.
 Posts: 1,510 Add as FriendChallenge to a DebateSend a Message 11/28/2014 12:06:49 AMPosted: 3 years agoAt 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.
 Posts: 561 Add as FriendChallenge to a DebateSend a Message 11/28/2014 4:25:23 AMPosted: 3 years agoAt 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=1Then 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.