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#1 2007-10-20 13:46:37

mikau
Member
Registered: 2005-08-22
Posts: 1,504

whoah, is this right? Proof help

I was working on some recreational maths, and I was faced with a problem, a problem which required a proof. I tried several times to disprove it without luck so.. i then decided to try to prove it, and...  I think i did it. But i'm not certain. Hopefully one of you can verify if its valid and if not, where the error in my logic is.

show that there is some solution, y = y1, to the inequality

(1): (m + b)y <= (q + d)

such that (y = y1, x = x1) is a solution to the inequalities

a(x) + b(y) <= d AND
-a(x) + m(y) <= q

for some value x1 (which must be shown to exist) and for any non zero real number a.

suppose now, that  m + b != 0, then it just so happens that the system of equations

a(x) + b(y) = d and
-a(x) + m(y) = q has a solution, which also happens to give you y = (q + d)/(b+m) which satisfies (1), and x = (d - by)/a. (recall a != 0)

thus the three inequalities

(m + b)y <= (q+d)
a(x1) + b(y1) <= d
-a(x1) + m(y1) <= q

have a solution where m + b != 0, namely, the solution to the two equations listed above.

Now suppose that (m+b) = 0, we have then that (m+b)y <= q + d,  and it must be so that q + d >= 0. Moreover, y may be anything at all and will still satisfy (1).

we must now show that

a(x) + b(y) <= d AND
-a(x) + m(y) <= q

has a solution, and since m = -b, we need only show that

a(x) + b(y) <= d AND
-a(x) -b(y) <= q     

for some two values (x,y) with no restrictions placed on their values. Equivilently, this can be written as

-q <= a(x1) + b(y1) <= d,

note that 0 <= q + d so -q <= d, in otherwords, there must lie some number s between -q and d, if we therefore chose any values of x and y such that a(x)+ b(y) = s (trivial) all inequalities are thereby satisfied and the proof is complete. ∈

that worked out a little too nicely, especially with the q + p >= 0 part. Is this proof legit? or do i need more sleep?

Last edited by mikau (2007-10-20 14:14:16)


A logarithm is just a misspelled algorithm.

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