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LatineSummary
This lesson opens the unit on number systems. It presents the path of the unit, which starts from the natural numbers and arrives, by successive extensions, at the integers, the rational numbers, the real numbers and the complex numbers; it reconstructs the history of those extensions, from Egyptian fractions and the Greek crisis of the incommensurables to the rigorous foundations of the nineteenth century; it formulates precisely the problem that gave rise to them, which is always an equation without a solution in the available set, and examines the case of Bombelli, in which an equation with real solutions requires passing through the complex numbers. It ends by pointing out where these sets are used in the rest of the course, in other courses on the site and in science and technology.
Learning objectives
- Describe the path of the unit and explain what each lesson contributes to the sequence of constructions that leads from the natural to the complex numbers.
- Formulate, for each extension, the equation that motivates it, and point out which property is gained and which is lost at each stage.
- Place in their historical context the main stages in the extension of the concept of number and identify the problem of measurement, of computation or of foundations that drove each one.
- Point out where number systems are involved in the rest of the course, in other courses and in scientific and technical applications.
Entry assessment
Before working through the unit, it is advisable to measure the starting point. This self-assessment is not graded: it serves to show where to dwell. The first ten questions cover the eight content lessons of the unit: the sixth and the seventh both correspond to powers with rational exponents (the second of them, to rationalization), and the tenth returns to the decimal expansion to set up a problem in context and report its result in scientific notation, with the appropriate rounding. The same skills are measured again at the end, in the synthesis lesson, with other data; the last three check what this unit takes as known from the previous one, on the natural numbers: division with remainder, the greatest common divisor and Euclid’s lemma.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1Which of these equations has a solution in but not in ?
Its solution, , is already rational: this equation is the one that forces the passage from the integers to the rational numbers, not from the rational numbers to the real numbers.Its solution, , is an integer: this equation motivates the passage from the natural numbers to the integers.It has no real solution, because the square of a real number is never negative; its solutions, and , are complex.Its solutions are and , which are real; and is not rational, because is not the square of a natural number. It is the same argument that proves the irrationality of .Review: Number systems: from the natural to the complex numbers
Question 2In the construction of the integers, which of these pairs of natural numbers represents the same integer as the pair ?
It represents the opposite, : swapping the components reflects the pair in the diagonal. The condition does not hold.Both components have been multiplied by , as when expanding a fraction. Pairs of natural numbers are identified by the difference, not by the quotient: .has been added to only one component, and that changes the difference: represents , because . To pass to another representative, the same number must be added to both components.Two pairs and are equivalent when ; here . Both represent the integer .Question 3Calculate and give the result as a fraction in lowest terms.
Correct answer:
The least common multiple of and is , and . With the denominator one obtains , which reduces to the same result. Whoever obtains has subtracted numerators and denominators separately.Review: The rational numbers: fractions and their operations
Question 4Which of these fractions has a terminating decimal expansion?
The denominator of the fraction in lowest terms has the prime factor , and so the expansion is repeating: It is not enough for the denominator to be a multiple of .That the denominator ends in is not enough: has the factor , and the expansion is repeating.That the denominator is even is not enough: has the factor , and the expansion is repeating.It is in lowest terms and its denominator is , with no other prime factors: the expansion is terminating, .Question 5What is the set of real numbers that satisfy ?
The sign of has been changed: from it follows that , not .The endpoints do not belong to the set, because the inequality is strict: at one has , which is not less than .Only has been solved. The absolute value also requires , and for example gives .The condition is equivalent to ; with , adding and dividing by gives .Question 6Calculate .
Correct answer:
The power is the cube root of , and since , one has . Whoever obtains has forgotten that the negative exponent gives the inverse; whoever obtains has confused the negative exponent with a sign of the result.Question 7What is the result of rationalizing the denominator of ?
The squares have been added, : the product of the sum and the difference of two numbers is the difference of their squares, not their sum.The binomial of the denominator has been kept instead of its conjugate. If this were the result, would equal , but it equals .It has been written that : the root of a difference is not the difference of the roots.The numerator and the denominator are multiplied by the conjugate of the denominator, ; the denominator becomes the difference of the squares, , which cancels with the numerator.Question 8For which value of does the system , have infinitely many solutions?
Correct answer:
The determinant of the system is , and the second equation has the coefficients of the first multiplied by . If , both equations are the same and there are infinitely many solutions; if , they are inconsistent and there are none. That the determinant is zero is not enough for there to be infinitely many solutions.Review: First-degree equations and inequalities, and linear systems
Question 9What are the solutions in of the equation ?
has been taken instead of in the formula: the sum of the solutions must be , not .The division by is missing. Substituting into the equation does not give .The root of has been taken instead of that of . These numbers are not solutions: their product is , not .The discriminant is , whose square root in is ; the solutions are , conjugate, as befits an equation with real coefficients.Question 10A leaky faucet lets a drop fall every seconds, and each drop has a volume of milliliters. How many liters of water does it lose in a year of days? Express the result in scientific notation, rounded to two significant figures.
The surplus digits have been dropped instead of rounding: is closer to than to . Truncating can almost double the rounding error.That number counts milliliters, not liters: the division by is missing.One drop per second has been assumed; half as many fall, one every two seconds.A year has seconds; half that number of drops falls, , which add up to milliliters, that is, liters. Rounded to two significant figures, , with a relative error less than .Question 11From earlier unitsWhat are the quotient and the remainder of the division with remainder of by ?
The equality is true, but the remainder must satisfy . This is the quotient truncated toward zero, which coincides with that of the division with remainder only when the division is exact or the dividend is not negative.has been divided by and only the sign of the quotient has been changed: , not .One has with , and the quotient and the remainder satisfying those conditions are unique.Question 12From earlier unitsCalculate the greatest common divisor of and .
Correct answer:
With the Euclidean algorithm, , and ; the last nonzero remainder is . Also, and share . Whoever obtains has calculated the least common multiple, and whoever obtains has forgotten the common factor .Question 13From earlier unitsIf a prime divides , with a natural number, what can be asserted?
Euclid’s lemma holds for any product of natural numbers; what is required is that the divisor be prime, not the factors.It is the other way round: since divides , the square divides .The argument does not depend on parity: Euclid’s lemma holds for every prime.By Euclid’s lemma, a prime that divides a product divides one of the factors; here both factors are . It is the key point in the proof that is not rational.Review: Prime numbers and the fundamental theorem of arithmetic
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