Problem of the Day


Let a, b, and c be real numbers. Is ab>0?

(1) \displaystyle \frac{1}{a}<1

(2) b^3c^2>0

 

Reveal Answer

Answer

E. Statements (1) and (2) together are not sufficient.

See the Solution

Solution

[latexpage]

We want to determine whether $ab$ is positive. $ab$ will be positive if $a$ and $b$ have the same sign and $ab$ will be negative if $a$ and $b$ have opposite signs. Let’s examine the statements.

Statement 1: INSUFFICIENT. $\displaystyle \frac{1}{a}$ will be less than 1 if $a<0$ or $a>1$. This statement tells us nothing about the sign of $a$ or $b$.

Statement 2: INSUFFICIENT. This statement implies that $b$ is positive. $c^2$ is always positive (as long as $c\neq ¬†0$, which is impossible in this case), so $b$ MUST be positive also if $b^3c^2$ is positive. This gives us definitive information about $b$, but we can’t conclude anything about $ab$ unless we know something about $a$

Statements 1&2 together: INSUFFICIENT. We know $b$ is positive, but $a$ could be positive or negative, so we cannot determine the sign of $ab$.

This problem is pretty easy as long as you handle the inequality in Statement 1 correctly. The only potential pitfall is if you assume that you can solve $\displaystyle \frac{1}{a}<1$ by multiplying both sides of the inequality by $a$ to get $1<a$. If you do this, you may think that $a$ has to be positive. However, this is not a legitimate way to solve this inequality. Remember that any time you multiply both sides of an inequality by a negative number, you have to flip the inequality sign. The problem here is that we have no idea if $a$ is positive OR negative, because it could represent any real number. So, you can’t perform an operation that is only legal for positive numbers if the variable may represent a negative number. Here is a better way to solve an inequality like this if you need a method:

$\displaystyle \frac{1}{a}<1$       Given.

$\displaystyle \frac{1}{a}-1<0$     Subtract everything to one side, leaving zero on the other side.

$\displaystyle \frac{1-a}{a}<0$      Combine the left side into one fraction using a common denominator.

Note that the numerator of the fraction on the left is equal to zero when $a=1$, and the denominator is equal to zero when $a=0$. Now, think of these values as dividing the real number line up into three zones: Numbers less than 0, numbers between 0 and 1, and numbers greater than 1. Or, mathematically, $a<0, 0<a<1, a>1$. Now, plug in any number from each of these zones as a test value to see if it makes the expression $\displaystyle \frac{1-a}{a}<0$¬†positive or negative. Let’s choose $a=-1, a=\frac{1}{2}$, and $a=2$.

1. $a=-1 \rightarrow \displaystyle \frac{2}{-1}=-2$. This tells us that numbers less than zero make this expression negative, which is what we want, so $a<0$ is part of our solution set.

2. $\displaystyle a=\frac{1}{2} \rightarrow \frac{\frac{1}{2}}{\frac{1}{2}}=1$. This test value gives a positive result, which would not solve the inequality, so $0<a<1$ is NOT part of our solution set.

3. $\displaystyle a=2 \rightarrow \frac{-1}{2}$. This test value gives a negative result, so we also want to include a>1 in our solution set.

Overall, the solution set is a<0 or a>1.


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