Help Prove This Math Trick!!!?

Question

can someone please help prove this math trick with inequalities

(a/b) < [(a+c)/(b+d)] < (c/d)

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

(2/3) < x < (5/6)

just add the numerators and the denominators across
(2+5)/(3+6) = 7/9

and indeed
(2/3) < (7/9) < (5/6).

prove algebraicly, just because it works for an example doesnt mean its proven, also, adding it across isnt a valid mathematical operation, i.e you normally cant add numerators and denomators together

Answer 1

Well, first off, this can only be true if:

(a/b) < (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 < (c/d)*b*d, which yields
ad < cb

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

ad < cb
ab + ad < ab + cb
a(b+d) < (a+c)b
a(b+d)/(b(b+d)) < (a+c)b/(b(b+d))
a/b < (a+c)/(b+d).

Which makes the left-hand inequality true. Then,

ad < cb
ad + cd < cb + cd
(a+c)d < c(b+d)
(a+c)d/(d(b+d)) < c(b+d)/(d(b+d)
(a+c)/(b+d) < 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. :)

Answer 2

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 <1/2 of before, and 5/9 would be <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.

Answer 3

So you want to prove that, for real numbers,

(a/b) < [(a + c) / (b + d)] < (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 < 3 is already false, for half of the inequality.

What I suspect is that you're missing some additional condition. Possibly something like 0 < a < b < c < d

Answer 4

Counterexample:

a=5, b=6, c=2, d=3

Answer 5

a/b < c/d
ad < bc (after multiplying both by bd)
ad + ab < bc + ab
a(b+d) < b(a+c) .... then divide by b and (b+d)
a/b < (a+c)/(b+d)

going back to step 2, add cd instead to get
ad + cd < bc + cd
d(a+c) < c(b+c) .... then divide by b and b+d
(a+c)/(b+d) < c/d

Finally put the 2 inequalities together.

Answer 6

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 >= 0 ; to show that a/b < c/d
so we need to show that
a/b < [a(x+1)+k] / b(x+1) < (ax+k)/bx


first a/b < [a(x+1)+k] / b(x+1)
< [a(x+1)/b(x+1)] +k/b(x+1)
< (a/b) + k/b(x+1)........true.

second [a(x+1) + k]/b(x+1) < (ax+k)/bx
[a(x+1)/b(x+1)] + [k/b(x+1)] < (ax/bx)+k/bx
a/b + [k/b(x+1)] < a/b + k/bx
[k/b(x+1)] < k/bx
since b(x+1) > bx

then [a(x+1) + k]/b(x+1) < (ax+k)/bx ......true.



(a/b) < [(a+c)/(b+d)] < (c/d) ......true!

Answer 7

Given:
a/b < c/d
then
(a + c)/b < c/b + c/d
(a + c)/b < c(1/b + 1/d)
(a + c)/b < c(b + d)/bd
(a + c) < (b + d)c/d
(a + c)/(b + d) < c/d

a/b + a/d < a/d + c/d
a(1/b + 1/d) < (a + c)/d
a(d/bd + b/bd) < (a + c)/d
a(b + d)/bd < (a + c)/d
(b + d)a/b < (a + c)
a/b < (a + c)/(b + d)
and
a/b < (a + c)/(b + d) < c/d

Answer 8

This may help.....don't think of them as fractions.....think of them as division operations. That's all a fraction really is.

Answer 9

um...looks like you just proved it yourself....

Answer 10

How simple your problem is. Why dont .................................................... got it?