SOLUTION: If b is a positive number and a < b (recall that this is another way of saying b-a>0), show that a/b < (a+1)/(b+1)

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: If b is a positive number and a < b (recall that this is another way of saying b-a>0), show that a/b < (a+1)/(b+1)      Log On


   



Question 1015790: If b is a positive number and a < b (recall that this is another
way of saying b-a>0), show that a/b < (a+1)/(b+1)

Found 2 solutions by fractalier, Edwin McCravy:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
From
a%2Fb+%3C+%28a%2B1%29%2F%28b%2B1%29
let us cross-multiply and get
a(b+1) < b(a+1)
ab + a < ab + b
Subtract off ab and get
a < b
which is what you supposed to be true to begin with...

Answer by Edwin McCravy(20062) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor's proof is wrong.  He proved the converse
of what you are to prove.  He started by assuming
the conclusion and proved what was given.  You must
start with what is given, not with what you are to
prove.   

Here is the correct proof:

If b is a positive number and a < b
show that a%2Fb%22%22%3C%22%22%28a%2B1%29%2F%28b%2B1%29
We start with what is given,
(not with what we are to prove, as he did):

a%22%22%3C%22%22b

and use the hint:

b-a%22%22%3E%22%220

We replace 0 by the equivalent ab - ab.

b-a%22%22%3E%22%22ab-ab

We add ab to both sides:
 
ab%2Bb-a%22%22%3E%22%22ab

We add a to both sides

ab%2B+b%22%22%3E%22%22ab%2Ba

Factor out common factors on each side

b%28a%2B1%29%22%22%3E%22%22a%28b%2B1%29

Divide both sides by b+1, which is positive
since b is positive. Therefore we will not 
reverse the inequality when we divide by b+1:

b%28a%2B1%29%2F%28b%2B1%29%22%22%3E%22%22a

Divide both sides by b which is positive and
also will not reverse the inequality when we 
divide by it:

%28a%2B1%29%2F%28b%2B1%29%22%22%3E%22%22a%2Fb

which is equivalent to

a%2Fb%22%22%3C%22%22%28a%2B1%29%2F%28b%2B1%29

Edwin