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Topic: Four Integers And Inequality Problem (Read 547 times) |
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K Sengupta
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Four Integers And Inequality Problem
« on: Jun 23rd, 2007, 1:15am » |
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u, v, w and x are four positive whole numbers with u> v> w> x> 0, satisfying :
ux = vw.
Analytically determine, whether or not the inequality :
(u-x)2> = 4x+8 is always true for all u and x satisfying the conditions of the problem.
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| « Last Edit: Jun 23rd, 2007, 1:22am by K Sengupta » |
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Eigenray
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Re: Four Integers And Inequality Problem
« Reply #1 on: Jun 23rd, 2007, 2:43pm » |
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I believe this can be strengthened to
(u-x)2 > 4x + 4 x + 1,
or equivalently,
u > (x + 1)2,
which is sharp, in the sense that the difference between the two sides may be made arbitrarily small.
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| « Last Edit: Jun 23rd, 2007, 2:46pm by Eigenray » |
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FiBsTeR
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Re: Four Integers And Inequality Problem
« Reply #2 on: Jun 23rd, 2007, 4:48pm » |
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I haven't written proofs since freshman year, and even then I've never been too rigorous, so here I go:
To minimize the value for u given some value for x, we should define w = x+2 and v = x+1. This would give ux = (x+1)(x+2), u = (x2+2x+2)/x.
Thus:
u >= (x2+2x+2)/x. Multiplying by x (which is positive, so the verb remains the same):
ux >= (x2+2x+2). Subtracting by x2:
ux - x2 >= 2x+2. Factoring:
x(u-x) >= 2x+2. Squaring both sides:
x2(u-x)2 >= 4x2+8x+4. Dividing by x2:
(u-x)2 >= (4x2+8x+4)/(x2),
(u-x)2 >= [(4x+8)/x] + 4/(x2).
From here, it can be seen that the right-hand expression is greater than 4x+8 for all x>2, since the first term is always less than 4x+8 for all x>2. Also, the second term will always be less than 1 for all x>2. Since I'm too lazy to find a way to finish my proof for x=1 and x=2, I simply showed that for x=1 and x=2, u is still greater than 4x+8 (yes, I am lazy).
If x=1, the minimal value for u comes when w=2 and v=3, and thus u=6. (u-x)2 = (5)2 = 25 >= 4(1) + 8; 25>=12, so this is true for x=1.
If x=2, the minimal value for u comes when w=3 and v=4, and thus u=6. (u-x)2 = (4)2 = 16 >= 4(2) + 8; 16>=16, so this is true for x=2.
Therefore (u-x)2 >= 4x+8 for all x.
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| « Last Edit: Jun 23rd, 2007, 4:55pm by FiBsTeR » |
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FiBsTeR
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Re: Four Integers And Inequality Problem
« Reply #3 on: Jun 23rd, 2007, 4:56pm » |
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BTW:
on Jun 23rd, 2007, 1:15am, K Sengupta wrote:| u, v, w and x are four positive whole numbers with u> v> w> x> 0 |
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You could also say that u, v, w and x are natural numbers.  
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Obob
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Re: Four Integers And Inequality Problem
« Reply #4 on: Jun 23rd, 2007, 6:36pm » |
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The definition of natural number is not as universal as you might think. I for one consider zero to be a natural number. A number theorist, however, is likely to exclude zero from the natural numbers.
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Eigenray
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Re: Four Integers And Inequality Problem
« Reply #5 on: Jun 23rd, 2007, 8:34pm » |
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on Jun 23rd, 2007, 4:48pm, FiBsTeR wrote:| To minimize the value for u given some value for x, we should define w = x+2 and v = x+1. This would give ux = (x+1)(x+2), u = (x2+2x+2)/x |
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But x divides x2+2x+2 iff x divides 2, so this only works for x=1,2.
If you take v=x+1, then the smallest w could be, with x | vw, is w=2x, giving u=2x+2. But in general you can do better. For example, if x=12, we can take u as low as 20.
Quote:(u-x)2 >= [(4x+8)/x] + 4/(x2).
From here, it can be seen that the right-hand expression is greater than 4x+8 for all x>2, |
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Don't you mean less than? Which is the opposite of what you want.
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FiBsTeR
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Re: Four Integers And Inequality Problem
« Reply #6 on: Jun 24th, 2007, 12:47pm » |
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on Jun 23rd, 2007, 6:36pm, Obob wrote:| The definition of natural number is not as universal as you might think. I for one consider zero to be a natural number. A number theorist, however, is likely to exclude zero from the natural numbers. |
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Fine, then I would still prefer: u, v, w, and x are four positive whole numbers integers
And completely disregard my first attempt, not sure what I was thinking yesterday.
on Jun 23rd, 2007, 8:34pm, Eigenray wrote:
But x divides x2+2x+2 iff x divides 2, so this only works for x=1,2. |
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What I meant was that (x2+2x+2)/x is a lower bound for u for all values of x. For example, if x=3, u>=17/3, u>5.
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| « Last Edit: Jun 24th, 2007, 1:29pm by FiBsTeR » |
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