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Question 1190710: Suppose that a, b, c and d are positive integers and c is not a square.
Given that a/(b + sqrt(c)) + d/sqrt(c) is rational, prove that b^2 * d = c (a + d).
--> I tried rationalising the denominator of a/(b + sqrt(c)) + d/sqrt(c) to remove sqrt(c) but I got stuck. Thanks for your help.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52800)   (Show Source):
You can put this solution on YOUR website! .
Suppose that a, b, c and d are positive integers and c is not a square.
Given that a/(b + sqrt(c)) + d/sqrt(c) is rational, prove that b^2 * d = c (a + d).
--> I tried rationalising the denominator of a/(b + sqrt(c)) + d/sqrt(c) to remove sqrt(c) but I got stuck.
Thanks for your help.
~~~~~~~~~~~~~~~~~~~~~~~~~
As I see from your post, you are trying to solve the problem on your own and, probably, you have fun of it.
I like such people; therefore, I will help you to solve the problem ON YOUR OWN,
giving you the directions; but I will not make your job instead of you.
You need make these steps, one after another.
(1) Write the sum of these two fractions with the common denominator.
You will get something in the form  .
(2) At this point, rationalize this fraction by multiplying by  , as you usually do rationalizing.
(3) You will get long expression. Its denominator will be integer number  .
Its precise form DOES NOT MATTER, because it is integer number in the denominator.
Only the numerator does matter for you now - - - so work with it separately.
Make FOIL in the numerator and collect all the terms with  in one group.
The other terms in the numerator do not matter for you now - - - only those that go with .
(4) Collect all this terms and write them in the form (...)*sqrt(c).
What you will get in parentheses (...) is (b^2d - ac - dc).
So, the necessary and sufficient condition for the fraction to be a rational number is equality to zero of this combination
b^2*d - ac - dc = 0.
(5) It is exactly this equality, which you need to prove.
(6) I did all this stuff. It requires a lot of writing, but the final result is as I described.
So, you have all necessary instructions from me.
Good luck (!)
You can report me about your progress . . .
Come again to this forum soon to learn something new (!)
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Here's the original expression  in which we are told is rational.
Any rational number is of the form p/q where p,q are integers and q is nonzero.
Let's rationalize each denominator.
For the first denominator, we'll have to multiply and bottom of the first fraction by 
Next, we'll rationalize the denominator of the second fraction
At this point, both fractions have rational denominators.
Unfortunately the denominators are different, so we need to get each fraction to the LCD. In this case, the LCD is c(b^2-c)
Now we can combine the fractions
We have one big fraction of the form M/N
---------------------------------------------------------
Let's focus solely on the numerator for now. I'll come back to the denominator later.
Let's rearrange terms like so
---------------------------------------------------------
The conclusion of the previous section means this messy fraction
turns into
The denominator is an integer (I'm skipping steps a bit but it's probably fairly obvious c(b^2-c) is an integer), so we must have the numerator be an integer as well.
Otherwise, the entire object isn't a rational number.
To ensure that the numerator is an integer, we need to have the two pieces  and  be integers themselves.
Clearly abc is an integer, so there's no need to worry about that too much.
is only an integer when the coefficient (-ac+b^2*d-cd) is zero.
If it were nonzero, then the sqrt(c) part will mean we have something irrational (since c is a non-perfect square) and thereby breaking the entire object from being rational.
Set that coefficient mentioned equal to zero and rearrange the terms to get what we want like so
This concludes the proof.
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