fraction properties - If p and q are positive integers, and

[Brent@GMATPrepNow]

If p and q are positive integers, and p < q, then which of the following MUST be true?

I) p/q < (p+1)/(q+1)
II) (p-1)/q < (p+1)/q
III) (p-1)/q < (p-1)/(q+1)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

Answer: B

Source: www.gmatprepnow.com
Difficulty level: 600-650

[elias.latour.apex]

The tendency of many students on questions such as these is to pick some numbers and run a scenario. If, for example, we pick the numbers p=3 and q=5. If we run some scenarios based on these numbers, we may well conclude that answer choice (E) is the best.

More careful students may run two scenarios: p=3 and q=5 vs. p=5 and q=3. These students will also conclude that (E) is the answer.

What we must realize is that if p=1 then the third answer choice will end up 0<0, which is false.

Accordingly, we must also think of the extremes of the answer. What if p is very large or very small? That mental exercise will lead us in the right direction.

[Matt@VeritasPrep]

We could also take each fraction and try to show that it reduces to p < q, or to anything else that would work with positive integers p and q. Let me do this in three separate posts, since each case will be hard to read.

Case I:

The goal: transform (p + 1)/(q + 1) > p/q into q > p.

(p + 1)/(q + 1) > p/q

q*(p + 1) > p*(q + 1)

pq + q > pq + p

q > p

Success!

[Matt@VeritasPrep]

Case II:

(p + 1)/q > (p - 1)/q

(p + 1) > (p - 1)

Well ... duh. So this case will hold for any positive integers p and q, whether or not q > p.

[Matt@VeritasPrep]

Case III:

(p - 1)/(q + 1) > (p - 1)/q

q * (p - 1) > (p - 1) * (q + 1)

pq - q > pq - q + p - 1

1 > p

Oof! This won't work for any positive integer p, since it forces p to be less than 1.

[Matt@VeritasPrep]

Would add that the algebra here is friendly enough -- thank you, Brent, for making these positive integers, greatly simplifying the inequality multiplication -- that I think it's quicker to do that than to play around with numbers.

[Brent@GMATPrepNow]

thank you, Brent, for making these positive integers, greatly simplifying the inequality multiplication

Just trying to make things easier for you, Matt

[Brent@GMATPrepNow]

If p and q are positive integers, and p < q, then which of the following MUST be true?

I) p/q < (p+1)/(q+1)
II) (p-1)/q < (p+1)/q
III) (p-1)/q < (p-1)/(q+1)

A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

Answer: B

Source: www.gmatprepnow.com
Difficulty level: 600-650

Statement I: There's a nice rule that says "If we add the same positive value to the numerator and denominator of a positive fraction, the resulting fraction is closer to one than the original fraction was.
For example (23+8)/(50+8) is closer to 1 than is 23/50

Since p < q, we know that p/q is less than 1
By the above rule, we know that (p+1)/(q+1) is closer to 1 than is p/q, which means p/q < (p+1)/(q+1) < 1
Statement I is TRUE

Statement II: The positive denominators are the same, but the numerator p+1 is greater than p-1
So, it must be the case that (p-1)/q < (p+1)/q
Statement II is TRUE

Statement III: This time the numerators are the same, but the denominators are different (q and p+1)
We can right away that if p = 1, then the two sides are EQUAL
In other words, it is NOT the case that (p-1)/q < (p-1)/(p+1)
Statement III need NOT be true

Answer: B

[Matt@VeritasPrep]

thank you, Brent, for making these positive integers, greatly simplifying the inequality multiplication

Just trying to make things easier for you, Matt

Lord knows I need it! ⊙�⊙

[Matt@VeritasPrep]

Statement I: There's a nice rule that says "If we add the same positive value to the numerator and denominator of a positive fraction, the resulting fraction is closer to one than the original fraction was.

Here's a neat proof:

Let's say m > n > 0.

If we can reduce m/n - 1 > (m + 1)/(n + 1) - 1 to m > n, we can say that adding 1 to the numerator and the denominator pushes the fraction closer to 0. (From there, we can say that adding 1 again will push it even closer, etc.)

m/n - 1 > (m + 1)/(n + 1) - 1

m/n > (m + 1)/(n + 1)

m * (n + 1) > n * (m + 1)

mn + m > mn + n

m > n

Success!

m > n

[Matt@VeritasPrep]

We also need to do the other case, n > m > 0.

This time, the inequality looks different:

(m + 1)/(n + 1) - 1 > m/n - 1

(It's backwards this time because each side of the inequality will be negative.)

From there:

(m + 1)/(n + 1) > m/n

n * (m + 1) > m * (n + 1)

mn + n > mn + m

n > m

Touchdown!

[Matt@VeritasPrep]

If we wanted a friendly introduction to get a feel for this, we could look at some fractions.

Suppose I start with 1/2. Adding to the num and denom, I get 2/3, 4/5, 5/6, 6/7, 7/8, ...

Hmm! So they keep getting bigger. What if I pick a more random fraction to start with?

1/5, 3/7, 5/9, 7/11, ...

So it's happening again! Seems reasonable enough to suppose that this will continue, and that our fractions will get closer and closer to 1 - in fact, as arbitrarily close to 1 as we'd like, provided that we never actually reach 1.