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3.10 — iNumbers
You know several kinds of numbers, from integers to complex numbers. Today we ask what happens when we operate on two numbers from the same group. For example, adding two integers still gives an integer—but dividing them might not. This simple idea links the arithmetic of numbers with the arithmetic of polynomials.
1. In class: A Practice Understanding Task
Number systems and closure
A set is a collection of objects. Its elements are members of that collection. A set is closed under an operation when operating on any two members (where the operation is defined) produces another member of that same set.
| Set | Meaning and examples |
|---|---|
| Natural numbers \(\mathbb N\) | Counting numbers \(1,2,3,\ldots\); the packet uses this convention. |
| Integers \(\mathbb Z\) | \(\ldots,-2,-1,0,1,2,\ldots\) |
| Rational numbers \(\mathbb Q\) | Fractions \(p/q\) where \(p,q\in\mathbb Z\), \(q\ne0\). |
| Irrational real numbers | Real numbers not in \(\mathbb Q\), such as \(\sqrt2\) and \(\pi\). |
| Real numbers \(\mathbb R\) | Rational and irrational numbers together. |
| Complex numbers \(\mathbb C\) | \(a+bi\), with \(a,b\) real and \(i^2=-1\). |
Always: explain why every permitted pair works. Sometimes: provide a pair that works and a pair that does not. Never: explain why no permitted pair can work. A few successful examples alone cannot prove an “always” claim.
For any integers \(m,n\), \(m+n\) is an integer; this is a basic property of integer arithmetic. Instances: \(3+8=11\), \(-4+9=5\), and \(-6+(-2)=-8\). These examples illustrate the general fact, rather than establish it by themselves.
Addition and multiplication across the number systems
Multiplication and addition preserve several important sets. When two rational numbers are added, for instance, their denominators can be combined into a new integer denominator:
\[\frac ab+\frac cd=\frac{ad+bc}{bd},\qquad b\ne0,\ d\ne0.\]Because \(ad+bc\) and \(bd\) are integers and \(bd\ne0\), the answer is rational.
Add \(\frac23\) and \(-\frac54\):
\[\frac23-\frac54=\frac{8-15}{12}=-\frac7{12}.\]Multiply them:
\[\frac23\left(-\frac54\right)=-\frac{10}{12}=-\frac56.\]In general, multiplying \(a/b\) and \(c/d\) gives \(ac/bd\), also rational. Therefore both conjectures are always.
Both \(\sqrt2\) and \(-\sqrt2\) are irrational, but their sum is \(0\), which is rational. On the other hand, \(\sqrt2+\sqrt2=2\sqrt2\) is irrational. Therefore the correct answer is sometimes.
Why? If \(2\sqrt2\) were rational, dividing by 2 would make \(\sqrt2\) rational—a contradiction.
Thus the claim is sometimes, not always. Notice how different this is from multiplying two rational numbers.
Write complex numbers as \(a+bi\) and \(c+di\). Then
\[(a+bi)+(c+di)=(a+c)+(b+d)i.\] \[(a+bi)(c+di)=(ac-bd)+(ad+bc)i.\]Both results are still of the form “real part + imaginary part times \(i\).” So complex numbers are always closed under both operations. As a numerical example,
\[(2+3i)(4-i)=8-2i+12i-3i^2=11+10i.\]| Set | Sum of two members | Product of two members |
|---|---|---|
| Integers | Always an integer | Always an integer |
| Rational numbers | Always rational | Always rational |
| Irrational numbers | Sometimes irrational | Sometimes irrational |
| Real numbers | Always real | Always real |
| Complex numbers | Always complex | Always complex |
These statements answer main-task conjectures 1–6, 8, 10–12 (with the separate quotient discussion below).
Why quotients require more care
The integer quotient \(8\div4=2\) is an integer, but \(8\div3=8/3\) is not. This proves that integer division is sometimes an integer—not always.
Successful pairs: \(12\div3=4\), \(-15\div5=-3\), \(0\div7=0\). A counterexample: \(5\div2=2.5\), not an integer. Thus sometimes.
Provided the divisor is not zero,
\[\frac{a/b}{c/d}=\frac{ad}{bc},\qquad b,c,d\ne0.\]This is rational. For example, \((2/3)\div(5/7)=14/15\). So it is always rational when the quotient is defined. Division by zero has no value; it is not a counterexample that produces an irrational number.
For a circle, \(C=\pi d\), with diameter \(d>0\). If both \(C\) and \(d\) were rational, their quotient \(C/d\) would be rational, but it equals \(\pi\), which is irrational. Therefore a circle cannot have both a positive rational diameter and a rational circumference.
Polynomials behave like integers under three operations
A polynomial in \(x\) is a sum of terms with nonnegative whole-number exponents: \(a_nx^n+\cdots+a_1x+a_0\). Constant terms correspond to exponent zero. Unlike \(x^{-1}\) or \(\sqrt x\), each allowed power is a nonnegative integer.
Combining like terms leaves only allowed powers of \(x\). So sums of polynomials are always polynomials.
Distribute the minus sign across every term of the second polynomial; the result is still a polynomial. Thus always.
The exponents produced by multiplication add: \(x^mx^n=x^{m+n}\). Adding two nonnegative integers cannot create a negative or fractional exponent, so products are always polynomials.
P1 — Addition: Always a polynomial.
P2 — Subtraction: Always a polynomial.
P3 — Multiplication: Always a polynomial.
Polynomial division need not produce a polynomial: \(1\div x=1/x\). This last example helps explain why the task specifically asks about addition, subtraction, and multiplication.
2. Ready — Compare linear, quadratic, and exponential functions
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Linear functions have constant first differences for equally spaced inputs; quadratics have constant second differences; and exponentials have a constant multiplicative ratio across equally spaced inputs, when written \(ab^x\) with \(b>0\).
| Function family | Example | Values at \(x=0,1,2,3\) | Recognizable feature |
|---|---|---|---|
| Linear | \(y=2x+1\) | 1, 3, 5, 7 | First difference \(+2\) |
| Quadratic | \(y=x^2\) | 0, 1, 4, 9 | First differences 1,3,5; second differences 2 |
| Exponential | \(y=2^x\) | 1, 2, 4, 8 | Consecutive ratio \(\times2\) |
Consider \(y=x^2-4x+3=(x-1)(x-3)=(x-2)^2-1\). Its x-intercepts are \((1,0)\), \((3,0)\); its y-intercept is \((0,3)\); the vertex is \((2,-1)\); and the axis of symmetry is \(x=2\). Since the leading coefficient is positive, the parabola opens upward.
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(y\) | 3 | 0 | -1 | 0 | 3 |
These features, its table, and the graph below form a complete model for Ready question 1.
For \(y=2x+1\), moving one unit right always raises the output by 2. For \(y=x^2-4x+3\), the change depends on position and the graph turns at \(x=2\). For \(y=2^x\), moving one unit right doubles the positive output. A linear graph is a straight line, a quadratic is a parabola, and the exponential curve bends upward without a vertex.
Check your work — Ready 1–2 answer key
- Example summary: a quadratic has form \(f(x)=ax^2+bx+c\), \(a e0\), and a parabolic graph. Vertex form \(a(x-h)^2+k\) shows vertex \((h,k)\) and axis \(x=h\). It opens up for \(a>0\) and down for \(a<0\). Factored form displays real roots; the discriminant determines whether roots are distinct real, repeated real, or nonreal complex. Example: \(f(x)=x^2-4x+3=(x-2)^2-1=(x-1)(x-3)\) has vertex \((2,-1)\), x-intercepts \((1,0),(3,0)\), and y-intercept \((0,3)\). Sample points for \(x=0,1,2,3,4\) have outputs 3, 0, −1, 0, 3. Plot not shown.
- All assign one output to each input and can be represented by equations, points, and graphs. With equally spaced inputs, linear functions have constant first differences, quadratics constant second differences, and basic exponentials constant output ratios. Lines have constant slope; parabolas have an axis and vertex; basic exponentials have a horizontal asymptote and no x-intercept. A nonconstant line has one real root; a quadratic may have zero, one, or two distinct real roots. Examples \(y=2x+1,y=x^2,y=2^x\) give outputs 1, 3, 5; 0, 1, 4; and 1, 2, 4 at inputs 0, 1, 2.
3. Set — Which operations require a larger number system?
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The number sets fit into a chain \(\mathbb N\subset\mathbb Z\subset\mathbb Q\subset\mathbb R\subset\mathbb C\). Subtracting \(5-8=-3\) forces us beyond the counting numbers. Dividing \(2\div3=2/3\) forces us beyond integers. Solving \(x^2=2\) leads to an irrational real number, and \(x^2=-1\) leads to a complex number.
Two integers added: never irrational, because integer addition always produces an integer. Two naturals subtracted: always an integer, although not always natural; \(2-9=-7\). Two integers divided: sometimes an integer (\(8/2=4\)) and sometimes not (\(8/3\)).
Two irrationals subtracted: sometimes irrational: \(\sqrt5-\sqrt5=0\) is rational, whereas \(\sqrt5-(-\sqrt5)=2\sqrt5\) is irrational. Two complex numbers subtracted: always complex, since \((a+bi)-(c+di)=(a-c)+(b-d)i\).
The packet labels “Irrational numbers” using \(\mathbb Q\), but \(\mathbb Q\) is the conventional symbol for the rational numbers. Interpret these prompts as referring to irrational real numbers, which are \(\mathbb R\setminus\mathbb Q\). The numerical conclusion still follows from the words and examples in the problems.
Check your work — Set 3–8 answer key
- Subtraction requires zero and negative integers, e.g. \(2-2=0,2-5=-3\). Division requires rational numbers, e.g. \(1\div2=\frac12\). Roots require irrational real numbers, e.g. \(\sqrt2\); solving \(x^2=-1\) requires complex numbers \(\pm i\).
- Never. Adding integers gives an integer, hence a rational number; for example \(-3+5=2\).
- Always. Subtracting two counting numbers gives an integer: \(2-5=-3,5-2=3,2-2=0\).
- Sometimes. \(6\div3=2\) is an integer, whereas \(1\div2=\frac12\) is not. The divisor must be nonzero; division by zero is undefined.
- Sometimes. \(2\sqrt2-\sqrt2=\sqrt2\) is irrational, but \(\sqrt2-\sqrt2=0\) is rational.
- Always. \((a+bi)-(c+di)=(a-c)+(b-d)i\), which is complex, including when the imaginary part is zero.
4. Go — Quadratic equations and radical notation
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Predict the kind of roots before solving (problems 9–12)
For \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Its sign tells whether real roots exist. A positive perfect-square discriminant with integer coefficients gives rational roots; a positive nonsquare discriminant gives irrational real roots (assuming rational coefficients); and a negative discriminant gives nonreal complex roots.
\(D>0\): two distinct real roots.
\(D=0\): one repeated real root.
\(D<0\): two nonreal complex conjugate roots.
Solve \(3x^2-5x-2=0\). Factor:
\[(3x+1)(x-2)=0\quad\Longrightarrow\quad x=-\frac13\ \text{or}\ x=2.\]Both roots are rational; one is an integer and the other is not. It is important to classify each root accurately, not to assume that all rational roots are integers.
Solve \(x^2-2x-7=0\). Here \(D=(-2)^2-4(1)(-7)=32\).
\[x=\frac{2\pm\sqrt{32}}2=1\pm2\sqrt2.\]Both roots are real and irrational because \(D\) is positive but not a perfect square.
Solve \(x^2+2x+6=0\):
\[D=4-24=-20,\qquad x=\frac{-2\pm\sqrt{-20}}2=-1\pm i\sqrt5.\]There are no real x-intercepts, but both complex solutions exist.
Simplify radicals (problems 13–18)
For even roots, the principal root is nonnegative; extracting an even power of a real variable may require absolute-value signs. For odd roots, negatives stay negative and absolute values are not needed.
Assuming \(x,y\) are real,
\[\sqrt[4]{81x^8y^{12}}=3x^2|y|^3.\]Here \(81=3^4\), \(x^8=(x^2)^4\), and \(y^{12}=(y^3)^4\). Meanwhile, for a fifth root,
\[\sqrt[5]{32a^{10}b^{15}}=2a^2b^3.\]No absolute values are necessary for the fifth root.
For real \(a,b\), with \(a\ne0\),
\[\sqrt{\frac{a^7b^{10}}{a^3}}=\sqrt{a^4b^{10}}=a^2|b|^5.\]The absolute value is needed because the square root cannot be negative.
Both radicands are nonnegative, so combining the two square roots is legitimate. Also \(\sqrt[3]{-27}=-3\): odd roots of negatives are real and negative.
Convert radicals to rational exponents (problems 19–24)
For even \(n\), use the usual real-number domain restrictions; formulas with variables must respect the principal-root convention.
Both factors share the fourth-root index. Another conversion:
\[16^{3/2}\cdot4^{1/2}=\sqrt{16^3}\cdot\sqrt4.\]provided the real tenth-root expression is defined (in particular, for positive \(x,y\)). If signs are unknown, retain the combined radicand and check the even-root domain rather than applying real rational powers indiscriminately.
Additional fully worked Go problems
Predict and solve \(x^2-7x+12=0\).
\(D=49-48=1\), so expect two rational roots. Factor:
\[(x-3)(x-4)=0\quad\Longrightarrow\quad\boxed{x=3,\;4}.\]Both are integers; the prediction was correct.
Predict and solve \(2x^2+4x+7=0\).
\(D=16-56=-40\), so expect nonreal complex conjugates:
\[x=\frac{-4\pm\sqrt{-40}}4=\frac{-4\pm2i\sqrt{10}}4=\boxed{-1\pm\frac{i\sqrt{10}}2}.\]Simplify \(\sqrt[4]{16p^{12}q^8}\), with real \(p,q\).
\[\sqrt[4]{16p^{12}q^8}=2|p|^3q^2.\]Although \(q^2\) is already nonnegative, \(p^3\) may be negative, so we keep \(|p|^3\).
Write \(\sqrt[5]{32u^8v^{11}}\) in rational-exponent form and simplify the fifth-root power where possible.
\[\sqrt[5]{32u^8v^{11}}=2u^{8/5}v^{11/5}=2uv^2\sqrt[5]{u^3v}.\]The fifth root is odd, so this works for any real \(u,v\).
Check your work — Go 9–24 answer key
- Prediction: two rational real roots (one integer). Factoring \(-(5x+2)(x-1)=0\) gives \(x=-\frac25,1\), confirming the prediction.
- Prediction: two integer real roots. Factoring \((x+1)(x+2)=0\) gives \(x=-1,-2\), confirming the prediction.
- Prediction: two irrational real roots, since \(D=57\). Quadratic formula gives \(x=\frac{-3\pm\sqrt{57}}2\), confirming the prediction.
- Prediction: two rational real roots (one integer). Factoring \((4x+1)(x-5)=0\) gives \(x=-\frac14,5\), confirming the prediction.
- \(3x^2|y|^3\).
- \(a^2|b|^5\) (original denominator requires \(a e0\)).
- \(x^2\sqrt[5]{625x^2}\).
- \(n^2\sqrt n=n^{\frac52}\).
- \(-3\).
- \(12\sqrt2\).
- \(8^{\frac34}\).
- \(\sqrt[4]{256^3}=64\).
- \(2^{\frac74}4^{\frac54}=2^{\frac{17}{4}}\).
- \(\sqrt{16^3}\sqrt4=128\).
- \((x^{23}y^{31})^{\frac1{10}}\); the factors stay grouped to preserve the real domain. For nonnegative \(x,y\) this is also \(x^{\frac{23}{10}}y^{\frac{31}{10}}\).
- \(64^{\frac15}a^{\frac95}b^{\frac{18}{5}}\).
5. Packet clarifications
Interpret quotient claims only when the divisor is nonzero. Division by zero is undefined, so it does not belong to the set of possible quotient results.
The description states that the exponents are positive integers. A complete polynomial definition also permits exponent \(0\), which is how constant terms are included. The task itself has a constant coefficient in its displayed expression; use nonnegative integer exponents for the formal definition.
6. Additional learning resources
| Resource and direct link | Why use it | Scope / What to skip |
|---|---|---|
| Paul’s Online Math Notes — Complex Numbers | Careful algebraic examples of complex addition, subtraction, and multiplication; reinforces the main task. | Some extra material. Focus on standard form and three operations; division by complex conjugates is optional for today. |
| Paul’s Online Math Notes — Polynomials | Explains the polynomial definition and shows worked addition, subtraction, and multiplication. | Close match for the polynomial conjectures. Omit any later polynomial procedures not needed for the three operations. |
| Paul’s Online Math Notes — Polynomial practice | Extra practice on the three polynomial operations, with worked solutions available through the site. | Close match. Choose addition, subtraction, and multiplication problems only. |
| Khan Academy — Algebra 2, Unit 1 and Unit 2 | Short videos and interactive exercises on polynomial operations and complex arithmetic. | Some extra material. Use “Add & subtract polynomials,” “Multiply binomials by polynomials,” and complex addition/multiplication; skip polynomial division and complex-plane extension topics. |
| Professor Leonard — Intermediate Algebra Lecture 10.7: Operations with Complex Numbers | Optional extended classroom-style explanation of the arithmetic of complex numbers. | Some extra material. Focus on addition, subtraction and multiplication; division extends beyond today’s main task. Timestamps were not independently verified. |
Teacher’s review — For teachers and parents
Serious issue 1 — Set problems 4 and 7: the printed worksheet calls irrational numbers \(\mathbb Q\), contradicting the standard meaning of that symbol (rational numbers). Replace this symbol with “irrational real numbers” or \(\mathbb R\setminus\mathbb Q\).
Serious issue 2 — Polynomial definition in the main task: “positive integer” exponents exclude a constant term although the polynomial form includes it. Say “nonnegative integer” exponents instead.
Division qualification: Main-task conjecture 9 should specify a nonzero divisor. The ordinary mathematical convention is clear, but explicitly noting it avoids treating an undefined quotient as a member of a number set.
All other exercises can be understood using the standard conventions addressed in this lesson.
7. Coverage of Section 3.10
| Packet | What students need to do | Where taught |
|---|---|---|
| Main task, conjectures 1–13 | Closure of integer, rational, irrational, real, and complex numbers; quotients; \(\pi\) and circles | In class |
| Main task P1–P3 | Polynomial sums, differences, products | In class |
| Ready 1–2 | Describe and compare quadratic, linear, exponential functions using equations, tables, and graphs | Section 2 |
| Set 3–8 | Explain extensions of the number system and classify operation outcomes | Section 3 |
| Go 9–12 | Predict and solve different types of quadratic roots | Section 4, examples 15–17, new A–B |
| Go 13–18 | Simplify even and odd radicals; variable signs | Section 4, examples 18–20, new C |
| Go 19–24 | Convert between radical notation and rational exponents | Section 4, examples 21–22, new D |
End of Section 3.10. This material is independent supplementary instruction, not an official MVP publication. Equations are rendered using MathJax, which requires internet access.
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