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Proof that LHS coefficients have to = RHS coefficients 
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#1
Mar912, 01:09 AM

P: 251

1. The problem statement, all variables and given/known data
If; a*x + b*y = c*x + d*y x ≠ y a,b,c,d ≥ 0 Prove that; a=c b=d 2. The attempt at a solution I've been fiddling with this equation and have been getting nowhere. 


#2
Mar912, 01:52 AM

P: 49

Try putting that equation in the matricial form. You'll get a 1x2 matrice and a 2x2 matrice on both sides. Since the 1x2 matrices are the same, you can conclude that the 2x2 matrices are also the same, and so a=c and b=d.



#3
Mar912, 02:34 AM

P: 251

Hi, thank you for your suggestion. Are you sure that this is a proof? I'm not sure;
[x y] = [x y], hence [a b] = [c d], hence a=c, b=d I mean it makes complete sense but it's a part of a much longer proof for an assignment so it's important that it's correct 


#4
Mar912, 06:53 AM

P: 426

Proof that LHS coefficients have to = RHS coefficients
For instance: (8 x 9) + (7 x 4) = (5 x 2) + (30 x 3) 


#5
Mar912, 07:05 AM

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hi imiyakawa!
try rewriting it as (a  c)x = (d  b)y 


#6
Mar912, 09:06 AM

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On the other hand, if you really mean that a*x + b*y = c*x + d*y should be true for ALL x ≠ y, then the result is true. RGV 


#7
Mar912, 09:14 AM

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The point that Ray Vickson raises is an important one  the difference between a conditional equation (true under certain conditions) and an identity (true for all values of some variable or variables).
For example, the equation 2x + y = 0 has an infinite number of solutions  all of the ordered pairs that satisfy y = 2x. The identity 2x + y ##\equiv## 0 has only one solution, x = 0, y = 0. 


#8
Mar912, 12:30 PM

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I presume that ax+ by= cx+ dy means that the two sides are equal for all x (given any a, b, c, d, you could find an infinite number of values for x and y which will make that true).
If ax+ by= cx+ dy for all x and y, take x= 1, y= 0. Then take x= 0, y= 1. 


#9
Mar912, 01:39 PM

P: 49

ax+by=cx+dy → (1x2).(2x2)=(1x2).(2x2)=(1x2) The x and y in the 1 by 2 matrices and a,b,c, and d in the 2 by 2 diagonal matrix. If you compute it, you'll have the following: (ax,by)=(cx,dy) → ax=cx [itex]\wedge[/itex] by=dy From those equations you conclude that a=c and that b=d. 


#10
Mar912, 02:12 PM

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RGV 


#11
Mar912, 07:55 PM

P: 49

This equational problem is about the same expression but with different coeficients (alphabetically, at least). Algebraically, I've proved it (for x≠0 and y≠0). For two different expressions: ax+by=cx+dy It's obvious that it's not right as a could be a different number then c and the same for b and d. However, in algebra, such is equation is correct because we're only interested to prove it's equal and what it means to be equal (a=c and b=d). Algebraically, the meaning of the equation is fundamental. Now, I'm not going to say you're wrong (you aren't, you're only facing it in another point of view, maybe geometrically, with a independent of c and b independent of d, as two separate expressions). Let me know what you think.. 


#12
Mar912, 08:28 PM

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#13
Mar912, 09:31 PM

P: 251

Hey everyone. Thanks for so much help on this question.
I need to prove this as part of a broader question in an assignment on utility theory. The equation is this; aE[u(x)] + (1a)E[u(y)] = (b+(1b)a)E[u(x)] + (1b(1b)a)E[u(y)] (1) Both sides are a gamble, E[u(G(x,y;a))] = E[u(G(x,y;b+(1b)a))]. I have to give the conditions such that LHS = RHS. Now if you're unfamiliar of a gamble under utility theory see the constraints (below). (1) is of the form a*x + b*y = c*x + d*y in my original post. Where;  each coefficient >= 0  each coefficient <= 1  a+b = 1  c+d = 1  E[u(x)] ≠ E[u(y)] (there are no other constraints in E[u(x)] and E[u(y)] apart from this one. They can be zero or negative.) I want to set the coefficients of E[u(x)] equal to each other and the coefficients of E[u(y)] equal to each other to derive required conditions such that LHS = RHS., which is why I was asking for a proof of a*x + b*y = c*x + d*y ... If this is true then I can do what I wanted to do by setting the coefficients equal to each other. Ray Vickson found an exception for 2*x + y = x + 3*y, where x=2*y given x≠0. Unfortunately I forgot to include the constraint that each coefficient is <= 1 (because we're looking at a formal gamble), a+b=1 and c+d=1 ... I apologise for this, lost my brain when I made the OP. 


#14
Mar912, 10:07 PM

P: 251

I multiplied a (1*2) matrix with a (2*1) matrix to represent a*x + b*y. Is there something wrong with this? 


#15
Mar1012, 01:59 AM

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RGV 


#16
Mar1012, 02:30 AM

P: 251




#17
Mar1012, 04:36 AM

P: 49




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