Course:MATH110/Archive/2010-2011/003/Math Forum/Arithmetic Skills Test

Arithmetic Skills Test

Here is the original pdf file of the Arithmetic Skills Test.

Problem 1 - Solution

Here's a solution for this problem. First, we rewrite the expression as:

$-{\frac {7}{51}}\div {\frac {3}{12}}$

Now, we recall that dividing by a fraction is equivalent to multiply by its inverse (have a look here if you want more details). This gives us:

$=-{\frac {7}{51}}\times {\frac {12}{3}}=-{\frac {7\times 12}{51\times 3}}$

Before multiplying out, we simplify (it makes computations more simple, I guess that's why we call it simplification):

$=-{\frac {7\times 4}{51\times 1}}=-{\frac {28}{51}}$

Which is the correct answer.

Problem 3 - Solution

There are a few different ways to solve this problem.

An important fact about fractions

In all cases, it is important to remember the following rule:

$x^{-n}={\frac {1}{x^{n}}}$

This means that when you see a negative exponent, it is a short notation to mean one divided by the rest of the expression. Now, what can be a bit confusing, is what about the expression:

${\frac {1}{x^{-n}}}$

Well, let's do this slowly. First, we see a division, let us rewrite it with a nice division symbol for a second:

$=1\div x^{-n}$

Now, we rewrite the exponent without the negative sign:

$=1\div {\frac {1}{x^{n}}}$

Well, since we're dividing by a fraction, we remember from problem 1 that it means we need to actually multiply by its inverse, which gives:

$=1\times {\frac {x^{n}}{1}}=x^{n}$

So, putting everything together, we just computed that:

${\frac {1}{x^{-n}}}=x^{n}$

Actually, there's a slightly shorter way to do this, using this time the notation instead of getting rid of it. Think like this, the negative exponent notation means something like one divided by... so actually, we have that

${\frac {1}{x^{-n}}}=(x^{-n})^{-1}=x^{(-n)\times (-1)}=x^{n}$

since ${\frac {1}{A}}=A^{-1}$ for any term $A$.

First solution

So, taking all this together, we can finally solve this problem. We first start by simplifying the expression inside of the parenthesis:

${\frac {x^{-4}}{x^{-7}}}=x^{-4}\div x^{-7}$

Then we remove the negative exponents:

$={\frac {1}{x^{4}}}\div {\frac {1}{x^{7}}}={\frac {1}{x^{4}}}\times {\frac {x^{7}}{1}}={\frac {x^{7}}{x^{4}}}=x^{3}$

So the problem can be now rewritten as:

$\left({\frac {x^{-4}}{x^{-7}}}\right)^{-2}=\left(x^{3}\right)^{-2}=x^{3\times (-2)}=x^{-6}$

Which is the answer.

Second solution

Another solution for example, consists of using the fact that

$\left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}$

This allows to solve this problem as follow:

$\left({\frac {x^{-4}}{x^{-7}}}\right)^{-2}={\frac {\left(x^{-4}\right)^{-2}}{\left(x^{-7}\right)^{-2}}}={\frac {x^{(-4)\times (-2)}}{x^{(-7)\times (-2)}}}={\frac {x^{8}}{x^{14}}}={\frac {1}{x^{6}}}=x^{-6}$

Problem 5

A question

Here's a question I got by email, maybe you guys can help out:

I can't figure out why you are to put a "negative" in front of the infinity symbol, when you're representing sets of numbers using the interval notation. For example, if we know that x is smaller than or equal to b, I'd have thought that their interval notation would be $(\infty ,b)$. However, the answer should be $(-\infty ,b)$. We don't know for sure if b=0, which warrants anything smaller than it a negative value. I hope that makes sense.

So, here's an answer: the set $(-\infty ,b)$ is all the numbers on the left of $b$, so all numbers smaller than that, all the way to negative infinity, hence the notation. On the other hand, the set $(b,\infty )$ is the set of all numbers on the right of $b$, so all numbers greater than $b$, that is all the way to positive infinity and hence the notation. This also means that the set $(\infty ,b)$ doesn't make much sense since there is no number greater than infinity (we read the interval notation from left to right as if we were looking at the axis of real numbers).

Is it more clear? Is it also clear to you that we write simply $\infty$ for $+\infty$. Let me know!

--DavidKohler 04:31, 15 September 2010 (UTC)