Help Prove This Math Trick!!!?

can someone please help prove this math trick with inequalities





(a/b) %26lt; [(a+c)/(b+d)] %26lt; (c/d)





if you don't believe me that this trick works try it out for yourself.


like for instance:





(2/3) %26lt; x %26lt; (5/6)





just add the numerators and the denominators across


(2+5)/(3+6) = 7/9





and indeed


(2/3) %26lt; (7/9) %26lt; (5/6).

Help Prove This Math Trick!!!?
Well, first off, this can only be true if:





(a/b) %26lt; (c/d),





for fairly obvious reasons. :) But, if we assume that that's true, and that neither b nor d equals 0, then we can do this:





(a/b)*b*d %26lt; (c/d)*b*d, which yields


ad %26lt; cb





From here, we'll go two different ways. First:





ad %26lt; cb


ab + ad %26lt; ab + cb


a(b+d) %26lt; (a+c)b


a(b+d)/(b(b+d)) %26lt; (a+c)b/(b(b+d))


a/b %26lt; (a+c)/(b+d).





Which makes the left-hand inequality true. Then,





ad %26lt; cb


ad + cd %26lt; cb + cd


(a+c)d %26lt; c(b+d)


(a+c)d/(d(b+d)) %26lt; c(b+d)/(d(b+d)


(a+c)/(b+d) %26lt; c/d.





Which makes the right-hand inequality true.





This is actually a pretty neat set of inequalities; I didn't believe myself until I proved it. :)
Reply:Given:


a/b %26lt; c/d


then


(a + c)/b %26lt; c/b + c/d


(a + c)/b %26lt; c(1/b + 1/d)


(a + c)/b %26lt; c(b + d)/bd


(a + c) %26lt; (b + d)c/d


(a + c)/(b + d) %26lt; c/d





a/b + a/d %26lt; a/d + c/d


a(1/b + 1/d) %26lt; (a + c)/d


a(d/bd + b/bd) %26lt; (a + c)/d


a(b + d)/bd %26lt; (a + c)/d


(b + d)a/b %26lt; (a + c)


a/b %26lt; (a + c)/(b + d)


and


a/b %26lt; (a + c)/(b + d) %26lt; c/d
Reply:This may help.....don't think of them as fractions.....think of them as division operations. That's all a fraction really is.
Reply:first we assume that a and b are relatively prime....(just to set that a/b is the simplest form)


then there exist c/d = ax+k/bx --------- where x,k %26gt;= 0 ; to show that a/b %26lt; c/d


so we need to show that


a/b %26lt; [a(x+1)+k] / b(x+1) %26lt; (ax+k)/bx








first a/b %26lt; [a(x+1)+k] / b(x+1)


%26lt; [a(x+1)/b(x+1)] +k/b(x+1)


%26lt; (a/b) + k/b(x+1)........true.





second [a(x+1) + k]/b(x+1) %26lt; (ax+k)/bx


[a(x+1)/b(x+1)] + [k/b(x+1)] %26lt; (ax/bx)+k/bx


a/b + [k/b(x+1)] %26lt; a/b + k/bx


[k/b(x+1)] %26lt; k/bx


since b(x+1) %26gt; bx





then [a(x+1) + k]/b(x+1) %26lt; (ax+k)/bx ......true.











(a/b) %26lt; [(a+c)/(b+d)] %26lt; (c/d) ......true!
Reply:So you want to prove that, for real numbers,





(a/b) %26lt; [(a + c) / (b + d)] %26lt; (c/d)





It's false.


Choose a = 5, d = 1, b = 1, c = 1. Then





(a/b) = 5/1 = 5





[(a + c) / (b + d)] = [(5 + 1) / (1 + 1)] = 6/2 = 3





5 %26lt; 3 is already false, for half of the inequality.





What I suspect is that you're missing some additional condition. Possibly something like 0 %26lt; a %26lt; b %26lt; c %26lt; d
Reply:one example, does not a proof make.





That's like saying x^2 = 2x all the time because 2*2 = 2+2.





Tackle it from the perspective of you're dividing each peice more times. So even if you have more peices, by adding the denominators, you're making them all smaller. I don't have the time to put it into nifty generic letters to prove atm, but I'd go from:





2/9 and 5/9, where 2/9 would be %26lt;1/2 of before, and 5/9 would be %26lt;1/2 of 5/6. so less than 1/2 + less than half is less than whole, and then something similar for the greater than to sandwich the two.





btw, a=5, b=6, c=2, d=3 isn't a counter because 5/6 isn't less than 2/3.
Reply:um...looks like you just proved it yourself....
Reply:a/b %26lt; c/d


ad %26lt; bc (after multiplying both by bd)


ad + ab %26lt; bc + ab


a(b+d) %26lt; b(a+c) .... then divide by b and (b+d)


a/b %26lt; (a+c)/(b+d)





going back to step 2, add cd instead to get


ad + cd %26lt; bc + cd


d(a+c) %26lt; c(b+c) .... then divide by b and b+d


(a+c)/(b+d) %26lt; c/d





Finally put the 2 inequalities together.
Reply:Counterexample:





a=5, b=6, c=2, d=3
Reply:How simple your problem is. Why dont ........................................... got it?