QuanticaAlgebra IOpen in the courseWelcome to algebra, the study of what numbers can do. Before putting numbers to work it helps to know what they are, so this first lesson sorts them into families. The sorting itself is easy once you know the value of each number. The catch is that one number can be written in many different ways, so how a number looks is a poor guide to what it is.
Problem
Here are six numbers. $$\sqrt{81},\qquad \frac{35}{7},\qquad 4.0,\qquad \frac{-12}{4},\qquad 0.75,\qquad \sqrt{10}$$ How many of them are integers, the numbers \(\ldots,-2,-1,0,1,2,\ldots\)?
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- Evaluate each number before you judge it. Some of these are not what they look like.
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\(\sqrt{81}=9\), \(\frac{35}{7}=5\), \(4.0=4\), and \(\frac{-12}{4}=-3\), so those four are integers. Only \(0.75\) and \(\sqrt{10}\) miss a whole value, giving \(\boxed{4}\). A number's family comes from its value, not from how it happens to be written.
Problem
Call a number nonnegative when it is not negative, and nonpositive when it is not positive. Exactly one integer manages to be both at once. Which integer?
Show a hint
- Nonnegative rules out one sign. Nonpositive rules out the other. What is left?
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Nonnegative rules out being negative, and nonpositive rules out being positive, so a number that is both must be neither, and every other number falls on one side or the other. The only one left is \(\boxed{0}\).
Problem
A ratio is one integer divided by a nonzero integer. The number \(8\) has no fraction bar written on it, yet it is a ratio in many different ways. Write \(8\) as a ratio with denominator \(25\). What numerator do you need?
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- The fraction must collapse back to \(8\) when you divide.
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\(\frac{200}{25}=8\), so the numerator is \(\boxed{200}\). Nothing about \(25\) was special. The same \(8\) is also \(\frac{8}{1}\) and \(\frac{16}{2}\), so an integer passes the ratio test in infinitely many ways.
Problem
The decimal \(0.272727\ldots\) repeats the block \(27\) forever. It never ends, yet it is still a ratio of integers with denominator \(11\). Find the numerator.
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- A two-digit block repeating forever is that block over \(99\). Then reduce.
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A repeating two-digit block sits over \(99\), so \(0.272727\ldots=\frac{27}{99}\). Dividing top and bottom by \(9\) leaves \(\frac{3}{11}\), so the numerator is \(\boxed{3}\). A decimal can run forever and still be a ratio of integers, as long as it repeats.
Problem
The decimal of any ratio of integers either stops or falls into a block that repeats forever. Now build $$0.31311311131111\ldots$$ where each \(3\) is followed by one more \(1\) than the last. Does this decimal ever settle into a block that repeats forever, yes or no?
Show a hint
- A repeating block has a fixed length. Watch what the growing runs of 1s do to any fixed length.
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A repeating block has some fixed length, but the runs of \(1\)s grow past any fixed length eventually. So the block would have to be all \(1\)s, and the \(3\)s that keep arriving rule that out. The answer is \(\boxed{\text{no}}\). This decimal neither stops nor repeats, so no ratio of integers equals it, and it falls outside every family met so far.
Problem
How many of these four numbers are irrational? $$\pi \qquad \frac{22}{7} \qquad \sqrt{5} \qquad \sqrt{169}$$
Show a hint
- \(\frac{22}{7}\) is famous for approximating something irrational. What is it, itself?
- Evaluate \(\sqrt{169}\) before you judge it.
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\(\frac{22}{7}\) approximates \(\pi\), but it is an integer over a nonzero integer, rational no matter what it approximates, and \(\sqrt{169}=13\), an integer. That leaves \(\pi\) and \(\sqrt{5}\) as the irrationals, and the count is \(\boxed{2}\).
Problem
The positive integers have a smallest member, the number \(1\). Is there a smallest positive rational number, yes or no?
Show a hint
- Whatever positive rational a challenger names, cut it in half. What do you get?
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Half of any positive rational is smaller, still positive, and still a ratio of integers, so every candidate for smallest is beaten by its own half. No candidate can be the smallest, and the answer is \(\boxed{\text{no}}\).
Practice these ideas
Practice
Count the nonpositive integers that are strictly greater than \(-5.5\), every single one of them. How many do you find?
Show the solution
The nonpositive integers greater than \(-5.5\) are \(-5,-4,-3,-2,-1,0\), so there are \(\boxed{6}\). Zero counts, since nonpositive means zero or negative.
Practice
The number \(\frac{-56}{14}\) fits at least one of the three family names integer, rational, and real. Which name is the most exclusive one that fits?
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\(\frac{-56}{14}=-4\), an integer. Every integer is automatically rational and real, so the tightest label is \(\boxed{\text{integer}}\).
Practice
The decimal \(0.58\) stops after two digits, so it is really a ratio of integers. Written in lowest terms, that ratio has what denominator?
Show the solution
\(0.58=\frac{58}{100}\), and dividing top and bottom by \(2\) gives \(\frac{29}{50}\), which no longer reduces. The denominator is \(\boxed{50}\).
Practice
The decimal \(0.\overline{2}=0.222\ldots\) repeats a single digit forever. Write it as a fraction in lowest terms.
Show the solution
Since \(0.111\ldots=\frac{1}{9}\), doubling both sides gives \(0.222\ldots=\frac{2}{9}\), already in lowest terms. The fraction is \(\boxed{\frac{2}{9}}\).
Practice
The number \(\sqrt{225}\) is written with a radical sign. Evaluate it and classify it, rational or irrational?
Show the solution
\(15^2=225\), so \(\sqrt{225}=15\), an integer, and every integer is rational. The answer is \(\boxed{\text{rational}}\). A radical sign does not make a number irrational, so evaluate a root before classifying it.
Practice
Every integer is rational, that street runs one way for sure. Does it run both ways, is every rational number an integer, yes or no?
Show the solution
\(\frac{3}{8}\) is rational and sits strictly between \(0\) and \(1\), so it is no integer. The nesting works inward to outward only, and the answer is \(\boxed{\text{no}}\).
Practice
Find the number sitting exactly halfway between \(\frac{2}{5}\) and \(\frac{3}{5}\) on the number line. Answer as a fraction in lowest terms.
Show the solution
Add the two and halve. \(\frac{2}{5}+\frac{3}{5}=1\), and half of \(1\) is \(\boxed{\frac{1}{2}}\). Between any two rationals you can always find another one, the same crowding that made a smallest positive rational impossible.
Practice
The number \(\sqrt{7}\) is irrational, its decimal never stops and never repeats. Is the product \(\sqrt{7}\times\sqrt{7}\) rational, yes or no?
Show the solution
By definition \(\sqrt{7}\) squares to exactly \(7\), and \(7\) is an integer, so the product is rational, \(\boxed{\text{yes}}\). Two irrational numbers can multiply to a rational one.