The Real Number System and Its Extensions

Classify number systems, reason about closure, prove exact number claims, and control distance, bounds, radicals, and approximation.

Why this matters

The equation \(x^2=2\) has no rational solution but has two real solutions. The equation \(x^2=-1\) has no real solution but has two complex solutions. Dividing two integers can leave the integers, and subtracting two natural numbers can leave the natural numbers. Even the familiar simplification \(\sqrt{x^2}=x\) fails whenever \(x<0\).

These are not unrelated exceptions. Every algebraic statement lives in a number system, and that system decides which operations are available, which comparisons make sense, and which equations can be solved. Difficult problems often hide their decisive information in a phrase such as “integer,” “real,” “exact,” or “to three decimal places.”

This chapter builds a working map of the standard number systems. It also develops three habits used throughout the course:

  1. name the universe before making a claim;
  2. preserve exact structure until approximation is requested; and
  3. attach an error bound to every numerical approximation.

What you should be able to do

After completing the chapter, you should be able to:

  • classify a number in the smallest standard set that contains it;
  • use the inclusions among \(\mathbb N\), \(\mathbb Z\), \(\mathbb Q\), \(\mathbb R\), and \(\mathbb C\) without confusing membership with equality;
  • determine whether a set is closed under a stated operation and disprove closure with a valid counterexample;
  • convert terminating and repeating decimals to exact fractions;
  • prove elementary irrationality results and avoid false rules about sums and products of irrational numbers;
  • use density and the least-upper-bound property to explain the structure of the real line;
  • compare exact radicals without relying on rounded decimal output;
  • derive and apply absolute-value distance and triangle-inequality results;
  • simplify radicals with the correct sign and domain conditions;
  • report absolute and relative error and infer intervals from rounded data;
  • explain why complex numbers extend the reals but do not inherit a compatible order; and
  • use SymPy number objects and assumptions as checks rather than as proofs.

Prerequisites

You should be able to use set-builder notation, intervals, implications, and quantifiers. Review Sets, intervals, statements, and quantifiers if the difference between \(x\in\mathbb Q\) and \(\mathbb Q\subseteq\mathbb R\) is unclear. You should also be able to compute exactly with signed fractions and integer powers.

Five-minute retrieval warm-up

  1. Write the domain of \(1/(x-2)\) as a subset of \(\mathbb R\).
  2. Negate: “Every real solution is rational.”
  3. Is \(\{-1,0,1\}\subseteq\mathbb Z\)? Is it an element of \(\mathbb Z\)?
  4. Give a counterexample to “if \(x^2>4\), then \(x>2\)” over \(\mathbb R\).
  5. Explain why \(x=2\Rightarrow x^2=4\) is not reversible over \(\mathbb R\).
  1. \(\mathbb R\setminus\{2\}\).
  2. “There exists a real solution that is irrational.”
  3. It is a subset, but it is not an integer and therefore not an element of \(\mathbb Z\).
  4. \(x=-3\) satisfies the hypothesis but not the conclusion.
  5. \(x^2=4\) also allows \(x=-2\).

1. A nested map of number systems

We use the following conventions:

Set Description Examples
\(\mathbb N\) positive integers \(\{1,2,3,\ldots\}\) \(1,12,10^6\)
\(\mathbb N_0\) nonnegative integers \(\{0,1,2,\ldots\}\) \(0,8\)
\(\mathbb Z\) integers \(-9,0,14\)
\(\mathbb Q\) rationals \(p/q\), where \(p,q\in\mathbb Z\) and \(q\ne0\) \(-7/3,0,2.125\)
\(\mathbb R\) points on the real line \(\sqrt2,\pi,-4\)
\(\mathbb C\) numbers \(a+bi\), where \(a,b\in\mathbb R\) and \(i^2=-1\) \(3-2i,i,5\)

The inclusions are

\[ \mathbb N\subsetneq\mathbb N_0\subsetneq\mathbb Z \subsetneq\mathbb Q\subsetneq\mathbb R\subsetneq\mathbb C. \]

Every integer is rational because \(n=n/1\). Every rational is real, and every real is complex because \(a=a+0i\). The converses fail. For example, \(1/2\) is not an integer, \(\sqrt2\) is not rational, and \(i\) is not real.

The irrational numbers are \(\mathbb R\setminus\mathbb Q\). “Irrational” is not a synonym for “complicated decimal” or “contains a radical.” For example, \(\sqrt{49}=7\) is natural, while the number with decimal expansion \(0.101001000100001\ldots\) is irrational even though no radical appears.

Example 1: classify by the smallest standard set

Classify each number as narrowly as possible:

\[ -12,\qquad \frac{18}{6},\qquad \sqrt{18},\qquad 0.\overline{27}, \qquad \sqrt{-9}. \]

  1. \(-12\in\mathbb Z\) and is not natural.
  2. \(18/6=3\in\mathbb N\); an unreduced fraction can still be an integer.
  3. \(\sqrt{18}=3\sqrt2\) is real and irrational.
  4. \(0.\overline{27}=27/99=3/11\in\mathbb Q\).
  5. Over \(\mathbb C\), the principal square root is \(3i\in\mathbb C\setminus \mathbb R\). Over \(\mathbb R\), \(\sqrt{-9}\) is not defined.

Classification therefore requires simplification and an announced universe.

Classification checkpoint

What is the smallest listed set containing \((\sqrt5)^2\)?

Because \(\sqrt5\) denotes the nonnegative real whose square is \(5\), \((\sqrt5)^2=5\in\mathbb N\).

2. Rational numbers and decimal representations

A number is rational when it can be represented by an integer ratio. The representation is not unique: \(2/3=4/6=-10/-15\). A reduced fraction chooses \(q>0\) and \(\gcd(|p|,q)=1\).

Every terminating decimal is rational because a finite string of \(k\) decimal digits is an integer divided by \(10^k\). Every eventually repeating decimal is also rational. Conversely, long division of \(p\) by \(q\) has only finitely many possible remainders \(0,1,\ldots,q-1\). It must either reach remainder zero and terminate, or repeat a remainder and thereafter repeat the same digit cycle.

Decimal characterization of rational numbers

A real number is rational if and only if its decimal expansion terminates or is eventually periodic. The familiar ambiguity \(0.999\ldots=1.000\ldots\) represents the same real number, not two adjacent real numbers.

Example 2: convert an eventually repeating decimal exactly

Let \(x=0.12\overline{34}=0.12343434\ldots\). The nonrepeating part has two digits and the repeating block has two digits. Align two copies:

\[ 100x=12.343434\ldots, \qquad 10000x=1234.343434\ldots. \]

Subtracting cancels the infinite tail:

\[ 9900x=1222, \qquad x=\frac{1222}{9900}=\frac{611}{4950}. \]

Substitution by division reproduces the repeating decimal. The subtraction is exact; no infinite decimal was rounded.

Guided exercise: a one-digit repeat

Convert \(0.1\overline6=0.1666\ldots\) to a reduced fraction.

First remove the one nonrepeating digit by considering \(10x\); then align the one-digit repeating tails.

Let \(x=0.1666\ldots\). Then \(10x=1.666\ldots\) and \(100x=16.666\ldots\). Subtract:

\[90x=15,\]

so \(x=15/90=1/6\).

3. Closure: which operations stay inside?

A set \(S\) is closed under a binary operation \(\star\) if

\[ (\forall a,b\in S)\;a\star b\in S. \]

Closure always names both the set and the operation. The integers are closed under subtraction because \(a-b\in\mathbb Z\) whenever \(a,b\in\mathbb Z\), but they are not closed under division because \(1,2\in\mathbb Z\) while \(1/2\notin\mathbb Z\). Division by zero is not an operation value at all, so closure under division is stated only for nonzero divisors.

Set Addition Subtraction Multiplication Division by nonzero members
\(\mathbb N\) yes no yes no
\(\mathbb Z\) yes yes yes no
\(\mathbb Q\) yes yes yes yes
\(\mathbb R\) yes yes yes yes
\(\mathbb C\) yes yes yes yes

The irrational numbers are not closed under addition or multiplication:

\[ \sqrt2+(-\sqrt2)=0,\qquad \sqrt2\cdot\sqrt2=2. \]

One counterexample disproves a universal closure claim. A list of successful examples cannot prove closure; a proof must start with arbitrary admissible members.

Example 3: prove rational closure and identify the condition

Let \(a=p/q\) and \(b=r/s\), with integers \(p,q,r,s\) and \(q,s\ne0\). Then

\[ a+b=\frac{ps+rq}{qs}, \qquad ab=\frac{pr}{qs}. \]

Both are integer ratios with nonzero denominators, so both are rational. Also,

\[ \frac{a}{b}=\frac{ps}{qr} \]

is rational provided \(b\ne0\), which forces \(r\ne0\). The hypothesis is not optional: division by zero has no value in \(\mathbb Q\).

Guided exercise: rational times irrational

Prove that if \(r\in\mathbb Q\setminus\{0\}\) and \(u\) is irrational, then \(ru\) is irrational. Explain why \(r\ne0\) is necessary.

Suppose for contradiction that \(ru\) were rational. Because \(r\) is nonzero and rational, closure of rational division would make \(u=(ru)/r\) rational, contradicting the hypothesis. If \(r=0\), then \(ru=0\) is rational for every \(u\), so the condition is necessary.

4. Irrationality and exact proof

A decimal approximation cannot prove irrationality: every displayed finite decimal is rational. Irrationality proofs instead use definitions and structural facts such as divisibility.

Example 4: prove that \(\sqrt2\) is irrational

Assume for contradiction that \(\sqrt2=p/q\) in lowest terms, with \(p,q\in\mathbb Z\), \(q>0\). Squaring gives

\[p^2=2q^2.\]

Thus \(p^2\) is even, so \(p\) is even. Write \(p=2k\). Substitution gives

\[4k^2=2q^2\quad\Longrightarrow\quad q^2=2k^2,\]

so \(q\) is also even. Then \(p\) and \(q\) have the common factor \(2\), contradicting that \(p/q\) was reduced. Therefore \(\sqrt2\notin\mathbb Q\).

The step “\(p^2\) even implies \(p\) even” follows because an odd integer \(p=2m+1\) has odd square \(4m^2+4m+1\).

Example 5: sums of irrational numbers require analysis

No single closure rule determines the sum of two irrational numbers:

\[ \sqrt2+(-\sqrt2)=0\in\mathbb Q, \]

but \(\sqrt2+\sqrt3\) is irrational. To prove the second claim, suppose \(\sqrt2+\sqrt3=r\in\mathbb Q\). Then

\[ r^2=5+2\sqrt6, \qquad \sqrt6=\frac{r^2-5}{2}\in\mathbb Q. \]

The same reduced-fraction argument as for \(\sqrt2\) shows \(\sqrt6\) is irrational, a contradiction. The examples refute both false universal rules “irrational plus irrational is always rational” and “always irrational.”

Irrationality checkpoint

If \(r\) is rational and \(u\) is irrational, must \(r+u\) be irrational?

Yes. If \(r+u\) were rational, subtracting the rational number \(r\) would make \(u=(r+u)-r\) rational, a contradiction.

5. Density and completeness of the real line

The rational and irrational numbers are both dense in \(\mathbb R\): between any two distinct real numbers lies a rational and an irrational number. Dense does not mean that the two sets have the same size, and it does not mean every short decimal display reveals which kind a number is.

For \(a<b\), choose a natural number \(n\) with \(1/n<b-a\), and let \(m=\lfloor na\rfloor+1\). Then

\[ a<\frac mn\le a+\frac1n<b. \]

Thus \(m/n\) is a rational between \(a\) and \(b\). To obtain an irrational, first choose a rational \(q\) between \(a-\sqrt2\) and \(b-\sqrt2\); then \(q+\sqrt2\) lies between \(a\) and \(b\) and is irrational.

Example 6: place a rational beside an irrational exactly

Find a rational strictly between \(\sqrt2\) and \(1.42\). Consider \(283/200=1.415\). Both quantities are positive, so comparison by squares is reversible:

\[ \left(\frac{283}{200}\right)^2 =\frac{80089}{40000}>2. \]

Hence \(283/200>\sqrt2\), and visibly \(1.415<1.42\). Therefore

\[ \sqrt2<\frac{283}{200}<1.42. \]

The proof uses exact integers; the decimal is terminating and exact.

Density says there are no gaps around rationals or irrationals. A stronger real number property supplies endpoints for bounded sets.

Least-upper-bound property

Every nonempty subset of \(\mathbb R\) that is bounded above has a least upper bound (a supremum) in \(\mathbb R\).

This property is a form of completeness. It distinguishes \(\mathbb R\) from \(\mathbb Q\). The rational set

\[ S=\{q\in\mathbb Q:q>0\text{ and }q^2<2\} \]

is nonempty and bounded above, but its real supremum is \(\sqrt2\), which is not rational. A full construction of the real numbers belongs to analysis; here we use completeness as an explicit structural fact rather than assuming every bounded decimal search must terminate at a rational endpoint.

Example 7: compare exact positive quantities without decimals

Compare \(\sqrt7\) and \(8/3\). Both are positive, so squaring preserves order:

\[ (\sqrt7)^2=7=\frac{63}{9} <\frac{64}{9}=\left(\frac83\right)^2. \]

Therefore \(\sqrt7<8/3\). Squaring would not be automatically reversible if the signs were unknown; for example, \((-3)^2>2^2\) does not imply \(-3>2\).

Example 8: identify a missing rational endpoint

For

\[S=\{q\in\mathbb Q:q\ge0, q^2<5\},\]

\(3\) is an upper bound, but it is not least. The real number \(\sqrt5\) is an upper bound: if a nonnegative \(q\) exceeded \(\sqrt5\), then \(q^2>5\). Every number below \(\sqrt5\) fails to be an upper bound because rational density supplies a rational between it and \(\sqrt5\) whose square is still below \(5\). Thus \(\sup_{\mathbb R}S=\sqrt5\), although \(S\subseteq\mathbb Q\) and \(\sqrt5\notin\mathbb Q\).

6. Absolute value is distance

For real \(x\),

\[ |x|= \begin{cases} x,&x\ge0,\\ -x,&x<0. \end{cases} \]

Geometrically, \(|x|\) is the distance from \(x\) to \(0\), and \(|x-a|\) is the distance from \(x\) to \(a\). Distance is nonnegative, symmetric, and zero exactly when the points agree.

Two essential bounds follow from the definition:

\[ -|x|\le x\le |x|. \]

Adding the corresponding bounds for \(u\) and \(v\) gives \(-|u|-|v|\le u+v\le |u|+|v|\), so

\[ |u+v|\le |u|+|v|. \]

This is the triangle inequality. Applying it to \(u=(u-v)+v\) produces the reverse form

\[ \bigl||u|-|v|\bigr|\le |u-v|. \]

Example 9: solve a distance equation and verify both points

Solve \(|2x-5|=7\) over \(\mathbb R\).

The distance of \(2x-5\) from zero is \(7\), so

\[ 2x-5=7\quad\text{or}\quad 2x-5=-7. \]

These give \(x=6\) or \(x=-1\). Substitution checks both:

\[|2(6)-5|=7,\qquad |2(-1)-5|=7.\]

Hence the solution set is \(\{-1,6\}\). The two branches are exhaustive because \(|y|=7\) means exactly \(y=7\) or \(y=-7\).

Guided exercise: total distance

Solve \(|x-2|+|x+2|=6\).

The formulas change at \(x=-2\) and \(x=2\). Analyze the three intervals.

For \(x\le-2\), the sum is \((2-x)+(-x-2)=-2x\), so \(-2x=6\) gives \(x=-3\), which belongs to the case. For \(-2\le x\le2\), the sum is constantly \(4\), so there is no solution. For \(x\ge2\), the sum is \((x-2)+(x+2)=2x\), giving \(x=3\). The solution set is \(\{-3,3\}\).

7. Exact radicals and their conditions

For \(a\ge0\), \(\sqrt a\) means the nonnegative real number whose square is \(a\). Consequently,

\[ (\sqrt a)^2=a\quad(a\ge0), \qquad \sqrt{x^2}=|x|\quad(x\in\mathbb R). \]

The absolute value in the second identity is necessary. If \(x=-5\), then \(\sqrt{x^2}=\sqrt{25}=5\), not \(-5\).

For nonnegative \(a,b\),

\[\sqrt{ab}=\sqrt a\sqrt b.\]

The conditions matter over the reals. Applying the formula separately to two negative factors would falsely give \(\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}\), whose left side is \(1\) but whose right side is not even real.

Example 10: simplify without losing a sign

For real \(x\) and \(y\),

\[ \sqrt{72x^2y^4} =\sqrt{36\cdot2}\sqrt{x^2}\sqrt{y^4} =6\sqrt2\,|x|\,y^2. \]

We used \(\sqrt{x^2}=|x|\) and \(\sqrt{y^4}=\sqrt{(y^2)^2}=|y^2|=y^2\). The result \(6xy^2\sqrt2\) would be negative when \(x<0\), impossible for a principal square root.

Rationalizing a denominator replaces a quotient containing a radical below the fraction bar by an exact equivalent form. For example,

\[ \frac5{\sqrt3}=\frac{5\sqrt3}{3}. \]

This is not a decimal approximation, and it is valid because \(\sqrt3\ne0\). Conjugates handle two-term expressions:

\[ \frac1{\sqrt{n+1}+\sqrt n} =\sqrt{n+1}-\sqrt n \]

for \(n\ge0\), because the product of the conjugates is \(1\).

Radical checkpoint

Simplify \(\sqrt{25(t-1)^2}\) for real \(t\).

\(5|t-1|\). The answer is \(5(t-1)\) only under the additional condition \(t\ge1\).

8. Exact values, approximations, and error

An exact value identifies the number without rounding, as in \(1/3\), \(\sqrt2\), or \(2+3i\). An approximation such as \(\sqrt2\approx1.414\) must be accompanied by precision information.

If the exact value is \(x\) and the approximation is \(a\), then

\[ \text{absolute error}=|x-a|, \qquad \text{relative error}=\frac{|x-a|}{|x|}\quad(x\ne0). \]

Relative error measures error compared with the size of the quantity. A small absolute error may be large in relative terms near zero.

Reporting \(a=4.27\) “to the nearest hundredth” normally asserts

\[|x-4.27|\le0.005,\]

with one endpoint adjusted under the chosen tie-breaking convention. Reporting only 4.27 without units or a rounding convention does not determine a unique error interval.

Example 11: certify a rounded radical by exact bounds

To prove that \(\sqrt2\) rounds to \(1.414\) to three decimal places, it is enough to show

\[1.4135<\sqrt2<1.4145.\]

All quantities are positive, so square exactly:

\[ 1.4135^2=1.99798225<2, \qquad 1.4145^2=2.00081025>2. \]

The first comparison can equivalently be checked using integers: \(14135^2=199798225<2\cdot10^8\). Thus the exact number lies in the rounding interval centered at \(1.414\), and the absolute rounding error is at most \(0.0005\).

Equivalent exact expressions can behave very differently in floating-point arithmetic. When \(n\) is large, \(\sqrt{n+1}-\sqrt n\) subtracts nearly equal numbers. Its conjugate form

\[ \frac1{\sqrt{n+1}+\sqrt n} \]

usually evaluates more stably. Algebraic form is therefore relevant to numerical reliability, not only appearance.

Example 12: propagate a stated measurement interval

A measured length is \(L=12.4\) cm to the nearest \(0.1\) cm. Then

\[12.35\le L<12.45.\]

For a square with side \(L\), positivity makes squaring order-preserving:

\[ 12.35^2\le L^2<12.45^2, \]

so

\[152.5225\le A<155.0025\text{ cm}^2.\]

Simply writing \(A=12.4^2=153.76\) hides the input uncertainty. The computed central value is useful, but it is not exact evidence about the physical area.

Guided exercise: compare two close rational approximations

Which is closer to \(\sqrt2\): \(99/70\) or \(140/99\)? Do not use rounded values of \(\sqrt2\).

Show that one fraction is above \(\sqrt2\) and one below. Their errors are equal at their midpoint, so compare that midpoint with \(\sqrt2\).

\(99/70>\sqrt2\) because \(99^2>2(70^2)\), while \(140/99<\sqrt2\) because \(140^2<2(99^2)\). Their midpoint is

\[ \frac12\left(\frac{99}{70}+\frac{140}{99}\right) =\frac{19601}{13860}. \]

The remarkable exact comparison

\[19601^2-2(13860^2)=1>0\]

shows the midpoint is just above \(\sqrt2\). Therefore \(\sqrt2\) is closer to the lower fraction \(140/99\).

9. Why the complex numbers are an extension, not an order line

The real equation \(x^2=-1\) has no solution because every real square is nonnegative. The complex system introduces \(i\) with \(i^2=-1\) and contains numbers \(a+bi\). Real numbers are precisely the complex numbers with \(b=0\).

Addition and multiplication remain closed:

\[ (a+bi)+(c+di)=(a+c)+(b+d)i, \]

\[ (a+bi)(c+di)=(ac-bd)+(ad+bc)i. \]

There is no order on \(\mathbb C\) compatible with the usual ordered-field laws. If such an order existed, either \(i>0\) or \(-i>0\); in either case its square would be positive. But \(i^2=(-i)^2=-1\), contradicting \(-1<0\). Statements such as \(2+i<3+i\) have no standard ordered-number meaning.

Example 13: solve a quadratic in the announced universe

Solve \(x^2+9=0\).

  • Over \(\mathbb R\), \(x^2=-9\) is impossible, so the solution set is empty.
  • Over \(\mathbb C\), \(x=3i\) and \(x=-3i\), since both square to \(-9\).

Thus

\[ \{x\in\mathbb R:x^2+9=0\}=\varnothing, \qquad \{x\in\mathbb C:x^2+9=0\}=\{-3i,3i\}. \]

The formula did not change; the universe changed the available solutions. Chapter 5.5 will develop complex geometry, conjugates, polar form, and roots of unity in full.

10. Common mistakes and limitations

Mistake 1: treating a displayed decimal as proof of type

Seeing 1.41421356 does not prove \(\sqrt2\) is irrational. That displayed finite decimal is rational. Irrationality concerns the exact number and needs a structural proof.

Mistake 2: assuming irrational numbers are closed

The words “both inputs are irrational” do not determine whether their sum or product is irrational. Test the universal claim with counterexamples, then prove any stronger conditional statement you actually use.

Mistake 3: writing \(\sqrt{x^2}=x\)

For real \(x\), the correct identity is \(\sqrt{x^2}=|x|\). Test \(x=-1\) before accepting a proposed radical identity.

Mistake 4: squaring an inequality without checking signs

Squaring preserves order on \([0,\infty)\), not on all of \(\mathbb R\). Establish nonnegativity or split into cases before using squares to compare values.

Mistake 5: treating an approximation as equality

\(\sqrt2=1.414\) is false. Write \(\sqrt2\approx1.414\) and state the rounding or error convention. Preserve the exact form during symbolic work.

Mistake 6: ordering complex numbers as if they were real

Complex numbers have magnitude and geometry but no compatible linear order. Compare \(|z|\), real parts, or another explicitly defined quantity instead.

11. Guided practice across the chapter

Guided proof: uniqueness of rational and irrational parts

Let \(a,b\in\mathbb Q\). Prove that if \(a+b\sqrt2=0\), then \(a=b=0\).

Separate the case \(b=0\) from the case \(b\ne0\).

If \(b=0\), then the equation gives \(a=0\). If \(b\ne0\), rearrangement gives \(\sqrt2=-a/b\). Closure of \(\mathbb Q\) under division by a nonzero rational would make \(\sqrt2\) rational, contradicting Example 4. Hence \(b=0\), and then \(a=0\).

Guided construction: an irrational in a rational interval

Construct an irrational number between \(1/3\) and \(1/2\) and verify both bounds without decimals.

Take \(u=1/3+\sqrt2/100\). It is irrational because it is a rational plus a nonzero rational multiple of an irrational. It is greater than \(1/3\). Also \(\sqrt2<2\), so

\[u<\frac13+\frac2{100}=\frac{53}{150}<\frac12.\]

Thus \(1/3<u<1/2\).

12. Examination strategy clinic

A number described by several structures

Let \(x=\sqrt2+\sqrt3\).

  1. Find a polynomial with integer coefficients having \(x\) as a root.
  2. Prove that \(x\) is irrational.
  3. Locate \(x\) in a rational interval of width \(1/4\).
  4. Explain why the polynomial alone does not uniquely identify \(x\).

Inventory. The expression has two radicals. Squaring once leaves the mixed term \(\sqrt6\); squaring again removes it. Conjugate-like products can prove irrationality. Exact rational square bounds can locate the number.

Polynomial. From \(x^2=5+2\sqrt6\),

\[ (x^2-5)^2=24, \]

so

\[x^4-10x^2+1=0.\]

Every squaring step was implication-only, so this establishes that \(x\) is a root; it does not claim every root is the original number.

Irrationality. Observe

\[ (\sqrt2+\sqrt3)(\sqrt3-\sqrt2)=1, \]

so \(1/x=\sqrt3-\sqrt2\). If \(x\) were rational, then \(1/x\) would be rational and

\[x+\frac1x=2\sqrt3\]

would be rational, contradicting irrationality of \(\sqrt3\).

Bounds. Because \((7/5)^2<2\) and \((8/5)^2<3\), \(\sqrt2>7/5\) and \(\sqrt3>8/5\), giving \(x>3\). Because \((3/2)^2>2\) and \((7/4)^2>3\),

\[x<\frac32+\frac74=\frac{13}{4}.\]

Thus \(3<x<13/4\), an interval of width \(1/4\).

Completeness audit. The polynomial is even and has four real roots: \(\pm(\sqrt3+\sqrt2)\) and \(\pm(\sqrt3-\sqrt2)\). The definition and the bounds select the positive root greater than \(3\). A polynomial equation without that extra information does not uniquely recover the original radical expression.

13. Exercises by purpose and difficulty

Unless a problem says otherwise, work exactly, use no calculator, and state the smallest standard number set justified by your result.

Level A — fluency

  1. Classify \(-7\), \(0.125\), \(\sqrt{49}\), \(\sqrt5\), and \(2+3i\) as narrowly as possible.

  2. Write \(0.36\) as a reduced fraction.

  3. Convert \(0.\overline{27}\) to a reduced fraction.

  4. Is \(\mathbb Z\) closed under addition, multiplication, subtraction, and division by a nonzero integer?

  5. Evaluate \(|-9|\), \(|4-11|\), and \(|3-(-5)|\).

  6. Compare \(\sqrt{10}\) and \(19/6\) exactly.

  7. Simplify \(\sqrt{72}\).

  8. Rationalize \(5/\sqrt3\).

  9. A value is reported as \(4.27\) to the nearest hundredth. Give a standard maximum absolute rounding error.

  10. Compute the modulus \(|3-4i|\).

Level B — connected reasoning

  1. Explain why every terminating decimal is rational.

  2. Convert \(0.1\overline6\) to a fraction by aligning repeating tails.

  3. Give one pair of irrationals with rational sum and one pair with irrational sum. Justify the second classification.

  4. Prove that a nonzero rational times an irrational is irrational.

  5. Find a rational strictly between \(5/7\) and \(3/4\) without using decimals.

  6. Compare \(\sqrt{13}\) and \(18/5\) exactly.

  7. Solve \(|x+2|=5\) over \(\mathbb R\) and verify the solution set.

  8. Write \(\{x\in\mathbb R:|x-1|\le3\}\) in interval notation.

  9. Derive \(|u+v|\le|u|+|v|\) from \(-|u|\le u\le|u|\) and the corresponding bounds for \(v\).

  10. Simplify \(\sqrt{x^2y^4}\) for real \(x,y\).

  11. For the approximation \(a=1.41\) to \(x=\sqrt2\), express absolute and relative error exactly.

  12. Multiply \((2+3i)(1-4i)\) and verify real and imaginary parts separately.

Level C — synthesis

  1. Convert \(0.12\overline{34}\) to a reduced fraction.

  2. Prove that \(\sqrt3+\sqrt5\) is irrational.

  3. Construct and verify an irrational strictly between \(1/3\) and \(1/2\).

  4. Find the real supremum of \(\{q\in\mathbb Q:q\ge0, q^2<5\}\) and explain why the set has no rational least upper bound.

  5. Classify all real \(a\) for which \(a+\sqrt2\) is rational.

  6. Solve \(|x-2|+|x+2|=6\) by cases.

  7. Prove \(\bigl||u|-|v|\bigr|\le|u-v|\).

  8. Compare \(\sqrt2+\sqrt3\) with \(\sqrt{10}\) exactly.

  9. Find an integer-coefficient polynomial having \(\sqrt2+\sqrt3\) as a root and list all its real roots.

  10. Rewrite \(\sqrt{n+1}-\sqrt n\) in a numerically stable exact form for \(n\ge0\).

  11. If \(a,b\in\mathbb Q\) and \(a+b\sqrt2=0\), prove \(a=b=0\).

  12. Solve \(z^2=-16\) over \(\mathbb C\) and explain why there is no real solution.

Level D — honors and examination

  1. Prove that every eventually periodic decimal is rational, including a nonrepeating initial block.

  2. Prove that \(\sqrt n+\sqrt{n+1}\) is irrational for every positive integer \(n\).

  3. Prove \(|x-a|+|x-b|\ge|a-b|\) for real \(a,b,x\), and classify every equality case.

  4. Prove that \(0.101001000100001\ldots\) is irrational, where successive blocks of zeros have lengths \(1,2,3,\ldots\).

  5. Decide exactly which of \(99/70\) and \(140/99\) is closer to \(\sqrt2\).

  6. Suppose \(x\ne0\) and \(x+x^{-1}\in\mathbb Q\). Prove that \(x^n+x^{-n}\in\mathbb Q\) for every positive integer \(n\), and exhibit an irrational \(x\) satisfying the hypothesis.

Concise answers

  1. \(-7:\mathbb Z\); \(0.125:\mathbb Q\); \(\sqrt{49}=7:\mathbb N\); \(\sqrt5:\mathbb R\setminus\mathbb Q\); \(2+3i:\mathbb C\setminus\mathbb R\).
  2. \(9/25\).
  3. \(3/11\).
  4. Yes under \(+\), \(-\), and multiplication; no under division, since \(1/2\notin\mathbb Z\).
  5. \(9,7,8\).
  6. \(\sqrt{10}<19/6\) because \(10<361/36\).
  7. \(6\sqrt2\).
  8. \(5\sqrt3/3\).
  9. \(0.005\).
  10. \(5\).
  11. A decimal with \(k\) digits after the point is an integer divided by \(10^k\).
  12. \(1/6\).
  13. \(\sqrt2+(-\sqrt2)=0\); \(\sqrt2+\sqrt3\) is irrational as in Example 5.
  14. Use contradiction and divide by the nonzero rational.
  15. The midpoint \(41/56\).
  16. \(\sqrt{13}>18/5\) because \(13>324/25\).
  17. \(\{-7,3\}\).
  18. \([-2,4]\).
  19. Add the two pairs of bounds and use the definition of absolute value.
  20. \(|x|y^2\).
  21. Absolute error \(|\sqrt2-141/100|=\sqrt2-141/100\); relative error \(1-141\sqrt2/200\).
  22. \(14-5i\).
  23. \(611/4950\).
  24. Squaring a hypothetical rational sum makes \(\sqrt{15}\) rational, a contradiction.
  25. One answer is \(1/3+\sqrt2/100\).
  26. \(\sqrt5\); rational density gives members above every smaller proposed bound.
  27. Exactly \(a=r-\sqrt2\) for some \(r\in\mathbb Q\).
  28. \(\{-3,3\}\).
  29. Apply the triangle inequality to \(u=(u-v)+v\) and swap \(u,v\).
  30. \(\sqrt2+\sqrt3<\sqrt{10}\) because its square is \(5+2\sqrt6<10\), using \(\sqrt6<5/2\).
  31. \(t^4-10t^2+1\); roots \(\pm(\sqrt3+\sqrt2)\) and \(\pm(\sqrt3-\sqrt2)\).
  32. \(1/(\sqrt{n+1}+\sqrt n)\).
  33. Separate \(b=0\); otherwise \(\sqrt2=-a/b\) would be rational.
  34. \(z=\pm4i\); real squares are nonnegative.
  35. Align the repeating tails after powers of \(10\) and subtract; see the full solution.
  36. There are no positive-integer exceptions; see the full solution.
  37. Equality holds exactly when \(x\) lies between \(a\) and \(b\), inclusive.
  38. Its decimal is neither terminating nor eventually periodic.
  39. \(140/99\) is closer.
  40. Use the recurrence \(s_{n+1}=s_1s_n-s_{n-1}\) with \(s_n=x^n+x^{-n}\); \(x=(3+\sqrt5)/2\) is an irrational example because \(x+x^{-1}=3\).

Selected full solutions

Suppose \(r=\sqrt3+\sqrt5\) were rational. Squaring gives \(r^2=8+2\sqrt{15}\), so \(\sqrt{15}=(r^2-8)/2\) would be rational. In a reduced fraction proof, \(\sqrt{15}=p/q\) would imply \(p^2=15q^2\); prime divisibility forces both \(3\) and \(5\) into \(p\), and then into \(q\), contradicting reduction. Thus \(r\) is irrational.

From \(u=(u-v)+v\), the triangle inequality gives \(|u|\le|u-v|+|v|\), hence \(|u|-|v|\le|u-v|\). Swapping \(u\) and \(v\) gives \(|v|-|u|\le|u-v|\). Together these say \(- |u-v|\le |u|-|v|\le |u-v|\), which is equivalent to \(\bigl||u|-|v|\bigr|\le|u-v|\).

Suppose a decimal has \(k\) initial digits and then a block of \(m\) digits repeats. Multiplying the number \(x\) by \(10^k\) moves the repeating part to the decimal point. Multiplying again by \(10^m\) advances by one full period. Therefore \(10^{k+m}x-10^kx\) is an integer: the infinite repeating tails align and cancel. Thus

\[ x=\frac{N}{10^k(10^m-1)} \]

for some integer \(N\), so \(x\) is rational. Terminating decimals are the special case in which the tail is all zeros.

Assume \(r=\sqrt n+\sqrt{n+1}\) is rational for a positive integer \(n\). Then

\[r^2=2n+1+2\sqrt{n(n+1)},\]

so \(\sqrt{n(n+1)}\) is rational. The square root of an integer is rational only when that integer is a perfect square. Since \(\gcd(n,n+1)=1\), their product can be a square only if both are squares. But two positive consecutive integers cannot both be squares: consecutive positive squares differ by at least \(3\). This contradiction proves the claim. (If \(n=0\) were allowed, the sum would be \(1\), which explains the positive-integer condition.)

The triangle inequality applied to \((a-x)+(x-b)=a-b\) gives

\[|a-b|\le|a-x|+|x-b|.\]

On the real line, equality in \(|u+v|\le|u|+|v|\) occurs when \(u\) and \(v\) have the same sign or one is zero. Here that means \(a-x\) and \(x-b\) have the same sign, which is exactly that \(x\) lies between \(a\) and \(b\), including endpoints. The statement remains correct whichever of \(a,b\) is larger.

The decimal does not terminate because it contains infinitely many \(1\)s. If it were eventually periodic with period \(m\), every sufficiently late block of \(m\) consecutive digits would repeat forever. But the construction contains arbitrarily long runs of zeros between successive \(1\)s. Choose a run longer than \(m\) after the alleged periodic start. A periodic tail containing a \(1\) cannot have such an unbounded gap; a periodic all-zero tail would terminate. Both alternatives contradict the construction. By the decimal characterization theorem, the number is irrational.

The square tests place \(140/99<\sqrt2<99/70\). The midpoint of the fractions is \(19601/13860\). Since

\[19601^2-2(13860)^2=1,\]

the midpoint is greater than \(\sqrt2\). On a line, a point below the midpoint is closer to the lower endpoint, so \(140/99\) is closer to \(\sqrt2\).

Let \(s_n=x^n+x^{-n}\) and \(s_0=2\). Direct multiplication gives

\[ (x+x^{-1})(x^n+x^{-n}) =s_{n+1}+s_{n-1}, \]

so \(s_{n+1}=s_1s_n-s_{n-1}\). Because \(s_0=2\) and \(s_1=x+x^{-1}\) are rational, induction proves every \(s_n\) rational. For \(x=(3+\sqrt5)/2\), the reciprocal is \((3-\sqrt5)/2\), so \(s_1=3\) although \(x\) is irrational.

14. Cumulative retrieval

Universe and quantifier audit

Consider: “Every number satisfying \(x^2=2\) is irrational.” Write a precise version over \(\mathbb Q\), over \(\mathbb R\), and over \(\mathbb C\). Which claims are vacuously true, substantively true, or false if “irrational” means a member of \(\mathbb R\setminus\mathbb Q\)?

Over \(\mathbb Q\) there are no solutions, so the universal implication is vacuously true. Over \(\mathbb R\), the solutions \(\pm\sqrt2\) exist and are both irrational, so the statement is substantively true. Over \(\mathbb C\), the same two solutions happen to be real and irrational, so the claim remains true for this equation, but “irrational” does not classify arbitrary nonreal complex numbers. The universe must still be stated.

Negate a density statement

Negate precisely: “Between every two distinct real numbers there is an irrational number.” Then explain what a counterexample would have to provide.

One precise form is

\[ (\forall a,b\in\mathbb R)(a<b\Rightarrow (\exists u\in\mathbb R\setminus\mathbb Q)(a<u<b)). \]

Its negation is

\[ (\exists a,b\in\mathbb R)(a<b\ \text{and}\ (\forall u\in\mathbb R\setminus\mathbb Q)(u\le a\ \text{or}\ u\ge b)). \]

A counterexample would therefore be a genuine open interval \((a,b)\) containing no irrational number, not merely a failed attempt to name one.

15. Topic checkpoint

Answer from definitions and exact comparisons. The generated notebook contains the same questions, hints, and readable solutions.

16. Summary and next step

  • The standard systems form a strict chain \(\mathbb N\subsetneq\mathbb N_0\subsetneq\mathbb Z\subsetneq\mathbb Q \subsetneq\mathbb R\subsetneq\mathbb C\).
  • Rational numbers are exactly the reals with terminating or eventually periodic decimal expansions.
  • Closure must specify a set and operation; one counterexample disproves it.
  • Rational plus irrational is irrational, and nonzero rational times irrational is irrational, but two irrational inputs have no universal sum/product rule.
  • Rationals and irrationals are dense in the reals; completeness supplies least upper bounds that may be missing from \(\mathbb Q\).
  • Absolute value is distance, and the triangle inequalities control how distances combine.
  • \(\sqrt{x^2}=|x|\) over the reals; radical laws require domain and sign conditions.
  • Approximate values need error information. Equivalent exact forms can have different numerical stability.
  • Complex numbers close equations such as \(x^2=-1\) but do not carry a compatible order.

Continue to Algebraic laws and valid transformation, which turns the closure and domain habits into a line-by-line system for deciding when an algebraic transformation is reversible.

References and further study

  • OpenStax, College Algebra 2e, for additional number-system and algebra review. The book is CC BY 4.0; this lesson’s explanations and problems are original rather than adaptations.
  • Richard Hammack, Book of Proof, for contradiction, divisibility, and set arguments.
  • Serge Lang, Basic Mathematics, for exact arithmetic, inequalities, and the transition from computation to proof.
  • NIST, Digital Library of Mathematical Functions, for authoritative notation and numerical reference.
  • SymPy documentation, Numbers and Assumptions, for exact computational representations and the limits of automated inference.

Using this lesson with fcmath and SymPy

SymPy distinguishes exact rational objects from Python floating-point numbers. Construct exact values before evaluating decimals:

import sympy as sp

q = sp.Rational(611, 4950)
r = sp.sqrt(2)
q, sp.N(q, 20), r, sp.N(r, 20)
(611/4950, 0.12343434343434343434, sqrt(2), 1.4142135623730950488)

The objects Rational(611, 4950) and sqrt(2) remain exact. N(..., 20) asks for a 20-digit approximation; it does not change the underlying exact value.

Use exact comparisons or simplify their difference instead of comparing short printed decimals:

comparison = sp.simplify(sp.Rational(8, 3) ** 2 - 7)
comparison, sp.Rational(8, 3) > sp.sqrt(7)
(1/9, True)

The positive difference \(1/9\) supports the manual comparison from Example 7.

Solving a number-system equation with SymPy

For \(|2x-5|=7\), the manual reasoning was:

  1. the absolute value is a distance of \(7\);
  2. therefore \(2x-5=7\) or \(2x-5=-7\);
  3. solve the two linear branches; and
  4. substitute each candidate into the original equation.

The equivalent exact SymPy workflow is:

x = sp.symbols("x", real=True)
solution = sp.solveset(sp.Abs(2 * x - 5) - 7, x, domain=sp.S.Reals)
checks = {candidate: sp.Abs(2 * candidate - 5) == 7 for candidate in solution}
solution, checks
({-1, 6}, {-1: True, 6: True})

The expected set is \(\{-1,6\}\), and both checks are True. The explicit real domain matters: solver output is always interpreted relative to its domain.

Assumptions can prevent an invalid radical simplification:

t = sp.symbols("t", real=True)
t_nonnegative = sp.symbols("t_nonnegative", nonnegative=True)
general = sp.sqrt(t**2)
restricted = sp.sqrt(t_nonnegative**2)
general, restricted
(Abs(t), t_nonnegative)

For a general real symbol SymPy retains \(|t|\) (or an equivalent form); with a nonnegative assumption it can return the symbol itself. A software assumption is part of the mathematical statement, not a convenience to hide.

Finally, verify the examination-clinic polynomial while remembering that polynomial membership does not identify which root was intended:

value = sp.sqrt(2) + sp.sqrt(3)
polynomial = sp.expand(value**4 - 10 * value**2 + 1)
all_roots = sp.solve(sp.Symbol("y") ** 4 - 10 * sp.Symbol("y") ** 2 + 1)
polynomial, all_roots
(0,
 [-sqrt(5 - 2*sqrt(6)),
  sqrt(5 - 2*sqrt(6)),
  -sqrt(2*sqrt(6) + 5),
  sqrt(2*sqrt(6) + 5)])

The first result is exactly zero. The root list contains four values, so the original definition and the proven interval \(3<x<13/4\) are still needed to select \(\sqrt2+\sqrt3\).

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