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Fraction Debate [Problem #5260]
Ms. Nomial's class is studying fractions, and she asks them to try the following experiment:
Pick any proper positive fraction where the numerator and denominator are both positive integers. Add 1 to the numerator and 1 to the denominator.
Compare the original fraction and the new fraction. Which is greater?
After the students work for a while, Jacinta says, "The new fraction is always greater than the original fraction."
Lurah says, "The original fraction is always greater than the new fraction."
Yanik says, "It depends on what the original fraction was - sometimes it's greater and sometimes the new one is greater."
Question: Which student is right? How do you know? Use algebra to support your answer.
Extra: Now suppose that the numerator and denominator are not required to be positive integers but the original fraction is still proper and positive.
If the three students still make the same statements, which one do you agree with? Why? Use algebra to support your answer.
Comments and Sample Solutions
A lot of people approached this problem by checking several actual fractions to see what happened when one was added to the numerator and
denominator. When the new fraction turned out to be larger in all the trials, they concluded that was always true and Jacinta was right.
The problem with that approach is that no matter how many fractions you try, there are always lots you haven't tried yet, and you can't be entirely sure that
it will work for every possible fraction. Even if you worked 100 examples and they all turned out that the new fraction was larger, you still can't be
completely certain that the 101st fraction you try will work, too.
But if you use algebra and represent the original fraction with variables, such as n and d for the numerator and denominator, then that fraction represents
every possible fraction and if you can show that the new fraction is greater in that case, you can be certain that it will always be greater. This is a hard idea
to understand at first, but it's behind all manner of mathematical proof - you need to work in a general sense and show that your result will hold no matter
what actual values you substitute in for the variables.
Let's look at the student solutions shown below. While they all use variables to represent the original and new fractions, the overall approaches are slightly
different. A key step in all of them was knowing that since the original fraction was proper, its numerator is less than its denominator, making it have a value
between 0 and 1.
Young Hoon Kang and Sally Mai each started by writing an inequality showing that the original fraction would be greater than the new fraction. Simplifying
the inequality led to a false statement, meaning that their original inequality could never be true. That in turn meant that the original fraction could not ever
be greater than the new one.
Using a similar approach, Helen Kirkby guessed that the new fraction would always be greater, so she set up her inequality showing that. When she
simplified it she wound up with a statement which is always true, confirming her guess that the original fraction will always be less than the new one.
Jim Lee and Melody Chen each started working with the two fractions without first making a prediction about which would be greater. Their work is similar
to the others except that they don't make a decision on which way the inequality sign will go until after they've compared the two fractions. They picked
variables for the original fraction, wrote a new fraction with one added to the numerator and denominator, and then compared those two fractions
algebraically to determine which was larger.
Note that some students compared the two fractions by renaming them so they had a common denominator, in which case they just had to compare the
numerators to see which was greater. Others cross-multiplied the two fractions, while others multiplied through by an expression that would cancel out both
denominators. Not surprisingly, the results of all three approaches are very similar.
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Timothy Lee used a slightly different method of comparing the fractions. He started with an inequality, similar to others above, but moved it all to one side,
leaving 0 on the other side. Timothy then used the conditions of the problem to decide if his left side was positive or negative, with positive being greater
than 0 to show the original inequality was true.
Finally, Bryson Thornburgh started by reasoning that if you add 1 to a number and then divide by the original number, the result is greater for smaller original
numbers than for larger ones. He used that to generate an inequality, then manipulated his inequality to produce a comparison of the two fractions in the
problem.
It was fun to see so many different ways to think about and approach this problem. It's often the case that math problems can be solved in a variety of
ways, and the student work on this one clearly shows that to be the case.
Congratulations to all the students shown here, and to anyone else who used algebraic techniques to compare the two fractions and determine that the new
fraction is greater than the original.
- Riz
From: Young Hoon K , Rancho San Joaquin Middle School
Jacinta is right. Extra: Yanik is right.
In the fraction x/y, x>0, y>0, and x(x+1)/(y+1)?
x(y+1)>y(x+1)
xy+x>xy+y
x>y
However, x
From: Sally M , Johnson Middle School
Jacinta's statement was correct to Mr. Nomial's question.
State the Problem
Ms. Nomial's class is studying fractions, and she asks them to try the
following experiment:
Pick any proper positive fraction where the numerator and denominator
are both positive integers.
Add 1 to the numerator and 1 to the denominator. Compare the original
fraction and the new fraction. Which is greater?
Discuss your approach:
Since I don’t know what the fraction was I chose variable A as my
Numerator and B as my denominator or a/b. Both numerator and
denominator are positive integers. A proper positive fraction is when
the absolute value of A is less than the absolute value of B or a/b<
a/b. Because 1 is added to both the numerator and denominator, the
fractions become a/b < a+1/b+1. Then I find the least common
denominator which is b(b+1). I multiply b(b+1) to both sides of the
inequality. B(B+1)a/b < B(+B+1)a+1/B+1, then I simplify or distribute
to ab+a < ab+b. I subtract ab from both sides because they are common
to both sides which becomes a
From: Helen K , Rancho San Joaquin Middle School
Jacinata was right - the new fraction is always greater than the original fraction.
I used algebra to solve this problem. N represented my positive whole
numerator, and d represented my positive whole denominator, so my
original fraction is n/d. Adding 1 to both n and d, I get n+1/d+1.
My guess was that the new was always greater than the original. To
try and prove this, I wrote an inequality:
n
n+1
- < -d
d+1
(n/d representing the original fraction, and n+1/d+1 representing the
new fraction)
I cross-multiplied to get:
n(d+1) < d(n+1)
Simplify:
nd+n < dn+d
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Subtract nd (or dn) from both sides:
n < d
Remembering that n was the numerator and d was the denominator, and
the original fraction had to be proper, n would always be less than d.
Therefore, The new fraction was always greater than the original fraction.
EXTRA:
If it is possible to have negatives, the
right - it sometimes gets greater and it
First I used -2/3 as an example.
-2 + 1= -1
3 + 1 = 4
-2/3 < -1/4
Since the new fraction was greater, that
the new fraction is always less than the
Then I used the example -2/-3.
-2 + 1 = -1
-3 + 1 = -2
new answer is that Yanik was
sometimes gets smaller.
eliminates the option that
original.
Since -2/-3 = 2/3, and -1/-2 = equals 1/2, I simplified both my
fractions to eliminate the negatives.
2/3 > 1/2
Since 2/3 (original) is greater than 1/2 (new), but -2/3 (original) is
smaller than -1/4 (new), sometimes the new fraction can be smaller,
and sometimes it can be greater. Therefore, Yanik was right.
REFLECTION:
At first I was just trying to use guess-and-check to find my answer.
Though in all of the situations I tried the new fraction was greater,
I had no way to prove that. Besides, I must have made a careless
mistake because I found myself thinking that in the case of 1/4 and
2/5 that 1/4 was greater. Confused, I asked my mom and she pointed
out my mistake and explained I needed to use algebra to solve this
problem. This was a simpler way, and it proved that Jacinta would
always be correct. When I looked at it in the right way,
(algebraically), the problem seemed simple. But I had a little bit of
difficulty and confusion until I got there.
From: Jim L , Rancho San Joaquin Middle School
Jacinta is the one who is correct
first i took a fraction and add both the numerator and the
denomenator by one
1/2 --> 2/3
1/2 < 2/3
so now i know that Lurah is not right. then i took the fraction the
problem gave me and see which one is bigger
3/4 vs. 4/5
i tried to make the denomenator both LCM, now it equals
15/20 vs. 16/20
3/4 < 4/5
then it hit me that i can solve this algerbricaly.
it will be
x/y vs. x+1/y+1 and x
From: Melody C , Rancho San Joaquin Middle School
Jacinta's statement is correct.
Explanation:
To do this problem, I first wrote down a couple of equations.
The problem said "any proper positive fraction". Therefore, I knew
that the numerator and denominator both would have to be greater
than zero.
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a = numerator
b = denominator
0 < a < b
a/b vs a+1/b+1
I cross multiplied the two fractions to compare them...
a(b+1) vs b(a+1)
Distributive Property...
ab+a vs ab+b
Cancel "ab" from both sides...
a vs b
From my original equation, "0 < a < b", I know that "a" is less
than "b", so therefore I know that the new fraction is always
greater than the original fraction.
Extra:
For the Extra, I agree with Lurah.
To do this, I did pretty much the same thing as the main problem.
I wrote some equations:
b > a > 0
However, the problem said to suppose the numerator and denominator
were not positive integers.
-b < -a < 0
I used the same technique:
-a/-b vs -a+1/-b+1
Multiply by (-b)(-b+1) to cancel out denominators:
-a(-b+1) vs -b(-a+1)
Distributive Property:
ab-a vs ab-b
-a vs -b
Since I said that "-b < -a < 0", the original equation is larger.
Reflection:
This POW was fairly simple. At first, I didn't know how to explain
my answer by algebra, since I had figured it out at first with
logic. Then, I finally realized that I had to set a different
variable for numerator and denominator. For the Extra, I originally
was a little confused by the problem. If it wanted the numerator and
denominator to be negative, wouldn't it just be a positive fraction?
Then, my mom helped me understand what it wanted me to do.
From: Bryson T , Rancho San Joaquin Middle School
Jacinta is correct. Extra: Lurah would be correct in this case.
A proper positive fraction (N/D) is one where the numerator is
always smaller than the denominator.
N < D
If you add 1 to both the numerator and the denominator is it smaller
or larger than the original fraction?
Considering the proportional of the increase of the Numerator is
always greater than the proportional increase of the denominator…or
the knowing that the N is always less than D, adding 1 to N
increases the numerator more than adding 1 to the denominator.
Proportion of numerator (N +1) / N
Proportion of denominator ( D + 1) /D
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(N+1) / N > (D+1) / D
We then will solve for N / D
Multiply both sides by N
(N +1) > ((D+1) x N)/D
Divide both sides by (D+1)
(N+1) / (D+1) > N / D
This proves that the new proper positive fraction is always larger
than the original fraction. Jacinta is correct.
Extra:
If the original fraction is always proper and positive, then both
the numerator and denominator could be negative.
Original fraction could be –N / -D
The proportional change for the numerator would still be bigger than
the proportional change of the denominator.
(-N +1) / - N > (-D +1) / -D
We then will solve for -N / -D
Multiply both sides by – N. Since you are multiplying both sides by
a negative number, the greater than (>) sign converts to a less than
sign (<).
(-N +1) < ((-D+1) x –N)/-D
Divide both sides by (-D+1)
(-N+1) / (-D+1) < -N / -D
In the case of a negative numerator and denominator, the new number
will be less than the original number. Lurah would be correct in
this case.
Thought about what if the denominator was -1…adding 1 would make it
zero in the denominator…and an indefinite number. Then, I
remembered that -1 would not make the original fraction a proper
fraction but a whole number.
Reflection: This Problem of the week was all about explaining
logical thinking with algebraic expressions. I knew the answers all
along but it was tough. I got a little help from my father and that
was it. The main objective that I had was making the algebraic
equations. These equations were the whole problem. I would say this
is a tough POW that I barely managed to answer.
From: Timothy L
Jacinta is right.
, Rancho San Joaquin Middle School
Let a equal the numerator
Let b equal the denominator
then b > a and
a > 0 and
b > 0.
(smaller) (positive)
(larger) (positive)
This is what Jacinta said was always true:
a/b < a+1/b+1
which, when simplified:
(a+1/b+1)-(a/b) > 0
further simplified:
b-a/b(b+1) > 0
from the given condition above
b-a>0, b>0 and b+1>0, therefore b-a/b(b+1) > 0
We know that b and a are positive and b is larger than a, therefore
the difference of b-a is positive.
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And because b is positive, then b(b+1) is positive as well, the
fraction then becomes positive or greater than 0.
Extra:
I agree with Yanik, because:
Conditions:
1) to satisfy 'a/b' is a proper fraction, then |b|>|a| and |b|>1
2) to satisfy a/b>0, if a<0, then b<0
if a>0, then b>0
If b<0 and a<0, b+1<0, b-a<0, thus b(b+1)>0, then b-a/b(b+1)<0
If b>0 and a>0, b+1>0, b-a>0, thus b(b+1)>0, then b-a/b(b+1)>0
Then the new results of a and b after adding 1 can either be less
than or greater than the original fraction.
Reflection:
I thought that this problem was difficult. My Dad helped me solve
this problem. He showed me where I had to go after I algebraically
simplified the original, and then he helped me with the last
simplification of the inequality. Later, the extra was checked by my
dad and I deleted unnecessary stuff. After that I added some more
information.
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