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The Magic Cafe Forum Index » » Puzzle me this... » » Oy, fractions! (0 Likes) Printer Friendly Version

landmark
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Given fractions a/b and c/d not equal to each other or zero. Give a convincing argument that (a+b)/(c+d) lies somewhere between the two given fractions.


Jack Shalom
LobowolfXXX
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I'm assuming you mean (a+c)/(b+d).
"Torture doesn't work" lol
Guess they forgot to tell Bill Buckley.

"...as we reason and love, we are able to hope. And hope enables us to resist those things that would enslave us."
stanalger
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Assume a,b,c, and d are positive.
and a/b < c/d.


Multiply both sides of this inequality by bd to get:
ad < cb.

Add ab to both sides to get
ad + ab < cb + ab.
Factor:
a(d+b) < b(c+a).
Divide both sides by b*(d+b) to get
a/b < (a+c)/(b+d).

Going back to ad < cb,
add cd to both sides to get
ad + cd < cb +cd.
Factor:
d(a+c) < c(b+d).
Divide both sides by d*(b+d) to get
(a+c)/(b+d) < c/d.

Hence a/b < (a+c)/(b+d) < c/d.
stanalger
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The argument in the previous post applies whenever b and d are nonzero
and have the same sign. (I.e. whenever bd>0.)
But if b and d don't have the same sign, the result may or may not hold.
(Let a=1, b=-2,c=1, d=3 to see a counterexample.)
stanalger
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Quote:
On 2006-02-22 12:37, stanalger wrote:
But if b and d don't have the same sign, the result may or may not hold.
(Let a=1, b=-2,c=1, d=3 to see a counterexample.)


Correction: If bd<0, the result WILL NOT hold.
Either (a/b) > (a+c)/(b+d)
or (a+c)/(b+d) > (c/d)
or (a+c)/(b+d) is undefined.
landmark
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Quote:
On 2006-02-22 11:54, LobowolfXXX wrote:
I'm assuming you mean (a+c)/(b+d).


Quite right.
landmark
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Nice, stan. Especially like the bd<0 analysis.

Okay, now how about a nice non-algebraic argument that requires no more than fifth grade math? (Assume a, b, c, d all > 0)

Jack
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They all have to be whole numbers, greater than 0.

a/b and c/d

where (a+c)/(b+d) is somewhere is the middle.

c/d < (a+c)/(b+d) > a/b

Which is greater a/b or c/d
landmark
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Not sure I understand your post. Are you clarifying/re-stating the conditions, or offering a solution?

If clarifying, then a,b,c,d must all be >0. Not necessarily whole numbers. For convenience's sake, let's say that a/b < c/d.

The problem now is to give a convincing non-algebraic argument that a/b < (a+c)/(b+d) <c/d using no more than fifth grade mathematics.
landmark
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Okay, due to underwhelming demand, here is my non-algebraic solution:

Think of each fraction as a rate, specifically apples per dollar.

So suppose one store sells 3 apples for $8, and another store sells 5 apples for $10. If I buy from both stores I'll have 8 apples for $18. (Notice this buying from both stores is equivalent to (a+c)/(b+d))

Now let's think about this for a minute. If the first store has a lousy deal, then buying from the second store as well will make my lousy deal somewhat better. But how much better? No matter how good the second store's deal is, my combined deal must be worse than the second store alone, because the bad deal at the first store weakens the strength of the good deal.

Hence, the combined deal will be better than the first store, but worse than the second store.

Baseball fans might also look at a batting average analogy: if I'm 3 for 10 on one day, and then do better the second day, my combined average is better than 3 for 10, but less than what I did on the second day.

Jack
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