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I am preparing for GRE. These two problems are from Manhattan 5lb book. I am finding absolute value equations very daunting. Sometimes it is time consuming, sometimes it seems very difficult to me. Following two are comparison problems. You have to compare quantity A with quantity B (i.e. if Quantity A> Quantity B, Quantity A< Quantity B, Quantity A= Quantity B, or it cannot be determined from the given information )

Prolblem 1:

If $p+|k|>|p|+k$, compare

Quantity A: $p$

Quantity B: $k$

Problem 2:

If $|x|+|y|>|x+z|$, compare

Quantity A: $y$

Quantity B: $z$

My solution approach

a) p $(+ve)$ and k $(+ve)$ --> not possible

b) p $(+ve)$ and k $(-ve)$ --> possible (In this case, Quantity A> Quantity B)

c) p $(-ve)$ and k $(+ve)$ --> not possible

d) p $(-ve)$ and k $(-ve)$ --> possible if $|k|>|p$| (In this case, Quantity A> Quantity B)

e) p =0 and k $(+ve)$ --> not possible

f) p =0 and k $(-ve)$ --> possible (In this case, Quantity A> Quantity B)

g) p $(+ve)$ and k =0, --> not possible

h) p $(-ve)$ and k =0, --> not possible

In all three possible cases, Quantity A> Quantity B. so, Quantity A> Quantity B is the answer.

I also solve problem 2 in a similar way. Can you show me the easier approach or the right approach to deal with these problems?

Mahmudul Hasan
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    Not clear what a) says. Is there a notational convention from the book you are using?

    For Problem 1, I would first rewrite the inequality in the more symmetric form: $$ |k|-k > |p| - p. $$ Thus, at least one of $p, k$ must be negative; otherwise, we would get $0 > 0$. That leads to two cases: (i) only one of $p, k$ is negative, and (ii) both are negative. Case (i) can be only $k < 0$, which makes the inequality $-2k > 0$ and gives no information about $p$. Case (ii) implies that $-2k > -2p$ and allows the desired comparison.

     – avs Aug 25 '16 at 05:43

2 Answers2

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Add symmetry to the first equality by rewriting as: $$ p -|p| > k - |k| $$ Now define the function: $$ f(x) = x - |x| = \begin{cases} 2x & \text{if } x < 0 \\ 0 &\text{otherwise} \end{cases} $$ and graph it. Then the question reduces to:

Does $f$ ever go both down and up, or does it never go down, or does it never go up?

Since $f$ is an increasing function (never goes down, just up or flat), we observe that $f(p) > f(k) \implies p > k$.

I want to compare $y$ with $z$, so my first instinct was to eliminate the other variable by setting $x = 0$ and seeing what happens. This yields $|y| > |z|$, which immediately leads to at least two conflicting counterexamples: we could have $y = 5$ and $z = 2$ (leading to $y > z$) or $y = -5$ and $z = 2$ (leading to $y < z$).

Adriano
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If $|p| + k < p + |k|$, then $|k| - k > |p| - p$. Now, for this to happen, note that $k$ must be negative, so that $|k|-k > 0$, and $p$ must be non-negative, so that $|p|=p $ and hence $|p|-p=0$, implying $|p| - p < |k|-k$. Thus, $p > k$ anyway.

For the second question, have a hunch, and see the result. Put $x=0$, which results in $|y| > |z|$. This is of course not decidable in general : if $y = 2, z=-1$ then it is true, but if $y=-1,z=2$ then it is not true, for example. Hence the second question is not decidable.

As a general tip, since you are giving the GRE which is a heavily time constrained exam, when possible try to partially substitute obvious values into your expression, as we did above, to get a clue of the answer. In the first problem, we used a clever trick, and that unfortunately comes with practice, but at least you have been acquainted with it here.

Sarvesh Ravichandran Iyer
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