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Author Topic: Ratio
cyber_junkie
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Icon 5 posted June 14, 2005 08:49      Profile for cyber_junkie     Send New Private Message       Edit/Delete Post   Reply With Quote 
I got this new book- Higher Algebra by H.S. Hall and S.R. Knight and i didn't understand this-

1. A ratio of greater inequality is diminished , and a ratio of less inequality is increased, by adding the same quantity to both terms. I understood this and i tried substituting numbers and it obviously worked but i didn't understand the proof given.

PROOF:-

Let a/b be the ratio and let (a+x)/(b+x) be the new ratio formed by adding X to both terms.

Now, a a+x ax-bx x(a-b)
- - --- = ----- = ------
b b+x b(b+x) b(b+z)

a-b is positive or negative according to as "a" is greater than or less than "b"


Got this to but i dint make anything of what followed this-

Hence if a > b
a a+x
- > ---
b b+x

and if a < b
a a+x
- < ---
b b+x

Why is it so?

2. If an equation is homogeneous with respect to certain quantities, we may for these quantities substitute in the equation any others proportional to them.

What does homogeneous with respect to certain quantities mean?
I got what this implied if such an equation was lx^3y + mxy^2z + ny^2z^2 = 0
But i was still doubtful what it meant by homogeneous with respect to certain quantities.

3. Heres a lame one-
If an equation is homogeneous with respect to certain quantities, we may for these quantities substitute in the equation any others proportional to them.

I know what a ratio is but abstract in what sence??

--------------------
"My crime is that of curiosity." -Mentor

Posts: 42 | From: India | Registered: Apr 2005  |  IP: Logged
Stereo

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Icon 1 posted June 14, 2005 10:15      Profile for Stereo     Send New Private Message       Edit/Delete Post   Reply With Quote 
1 : from the last term of the equality, you will see that (a-b) was "isolated" (not quite sure that's the correct word in English... sorry!). If you examine the rest of the term, you will see that the (a-b) part is the only one that can change the sign. if (a-b) is negative, so is [x(a-b)]/[b(b+x)] (taken that x is constant and positive, of course).

Now, see:

code:
a   a+x   x(a-b)    a   x(a-b)   a+x
- - --- = ------ -> - = ------ + ---
b b+x b(b+x) b b(b+x) b+x

Now, for simplification, let's state that y = |[x(a-b)]/[b(b+x)]|. Then:

code:
a   a+x
- = --- + y (if a-b is positive)
b b+x

a a+x
- = --- - y (if a-b is negative)
b b+x

Take the y out of the two equations to get the inequalities, and there you are.

2 I can't answer, either I've learned it with different wording, or I never learned it (or forgot it... [blush] ) [Confused]

3 looks quite similar to 2; a little cut-and-paste error, maybe? [Wink]

--------------------
Eppur, si muove!

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drunkennewfiemidget
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Icon 1 posted June 14, 2005 10:35      Profile for drunkennewfiemidget     Send New Private Message       Edit/Delete Post   Reply With Quote 
quote:
Originally posted by Stereo:
1 : from the last term of the equality, you will see that (a-b) was "isolated"

It is. [Smile]
Posts: 4897 | From: Cambridge, ON, Canada | Registered: Jun 2004  |  IP: Logged
cyber_junkie
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Icon 1 posted June 15, 2005 05:34      Profile for cyber_junkie     Send New Private Message       Edit/Delete Post   Reply With Quote 
Thanksalot now i can move on to variation with peace. [thumbsup]
Better learn BBCode the alignment got screwd.

--------------------
"My crime is that of curiosity." -Mentor

Posts: 42 | From: India | Registered: Apr 2005  |  IP: Logged
cyber_junkie
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Icon 1 posted June 15, 2005 05:35      Profile for cyber_junkie     Send New Private Message       Edit/Delete Post   Reply With Quote 
Thanksalot now i can move on to variation with peace. [thumbsup]
Better learn BBCode the alignment got screwd.
Be ready for a quiz on that. [Geek]

--------------------
"My crime is that of curiosity." -Mentor

Posts: 42 | From: India | Registered: Apr 2005  |  IP: Logged
Alien Investor
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Icon 10 posted June 15, 2005 06:33      Profile for Alien Investor     Send New Private Message       Edit/Delete Post   Reply With Quote 
I love a little math in the morning.

Start with a simple equation in two variables, such as: y = 2x. Pick a solution to this equation, such as x=3, y=6. Now scale up the solution by a factor of 10: x=30, y=60. The scaled-up values are also a solution of y = 2x.

In general, for any solution (x,y) of the equation y = 2x, then the values (10x, 10y) are also a solution. So are the values (kx, ky), for any scaling factor ("scalar") k.

That means "y = 2x" is homogenous.

Now consider the equation y = x + 5. A solution to this equation is: x=3, y=8. Try scaling that up: x=30, y=80. Is that a solution of y = x + 5? Nope! So that equation, y = x + 5, is not homogenous.

Here's a quick theorem. If a homogenous equation has any solution at all, let that solution be (x, y). Then let the scaling factor k = 0. So (0x, 0y) = (0,0) is also a solution. So if a homogenous equation has any solutions at all, then (0, 0) is also a solution. Conversely, if an equation does not have (0, 0) as a solution, and it has at least one solution anywhere, then that equation is not homogenous.

"Homogenous" equations can have any number of variables; I just started with 2 for the first example. The homogenous equations with a single variable are pretty boring. Either they have no solutions at all (that's always homogenous); or the only solution is x=0; or every value of x is a solution. Because if x=a is a solution, and a != 0, then x=ka is a solution for every value of k.

The equation x^2 + y^2 = r is not homogenous because x=3 y=4 r=25 is a solution, but x=30 y=40 z=250 is not. The equation x^2 + y^2 = r^2 is homogenous: x=3 y=4 r=5 is a solution, and x=30 y=40 z=50 is also a solution (try it).

If you want to prove an equation is not homogenous, you just have to find 1 solution that does not scale. But if you want to prove an equation is homogenous, you have to prove that every solution scales.

Generally, homogenous equations have the same order or power on every term. For example, x^4 + x^2y^2 + y^4 = z^4 is a homegous equation of order 4.

Why does it matter if an equation is homogenous? Consider the solution point as a point in N-dimensional space (where N is the number of variables in the original equation). Take any solution point that is not all 0's and draw a line that includes the origin and that solution point. For a homogenous equation, *every* point on that line is also a solution point.

... Or, in other words, it matters because it will be important in next year's math and four years in the future in physics.

Many of the great equations of physics are homogenous: F = m a , P V = u R T, nu = h /f. However, the equation for kinetic energy, E = 1/2 m v^2 is not homogenous! If you scale up v by 10, then E does not scale up by 10; it scales up by 100.

--------------------
"love without fear"

Posts: 627 | From: New York City | Registered: Jan 2000  |  IP: Logged
cyber_junkie
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Icon 1 posted June 18, 2005 06:52      Profile for cyber_junkie     Send New Private Message       Edit/Delete Post   Reply With Quote 
So do I and thanks for that insight on homogeneous equations.
Posts: 42 | From: India | Registered: Apr 2005  |  IP: Logged


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