Lesson test: The real numbers

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The real numbers

1.[1p]
Which property of $ℝ$ fails in $ℚ$ ?

Every nonempty set bounded above has a least upper bound

Correct

Every nonzero element has a reciprocal

The answer is: Every nonempty set bounded above has a least upper bound

The order is total

The answer is: Every nonempty set bounded above has a least upper bound

$x \leq y$ implies $x + z \leq y + z$

The answer is: Every nonempty set bounded above has a least upper bound

2.[1p]
The ordered field axioms alone are enough to prove that some number has square $2$ .

True

The answer is: False

False

Correct

3.[2p]
Let $s = \sup A$ and $ε = 0.01$ . What does the characterisation of the supremum guarantee?

Some $a \in A$ satisfies $a > s - 0.01$

Correct

Some $a \in A$ satisfies $a = s - 0.01$

The answer is: Some $a \in A$ satisfies $a > s - 0.01$

Every $a \in A$ satisfies $a > s - 0.01$

The answer is: Some $a \in A$ satisfies $a > s - 0.01$

$s - 0.01$ is an upper bound of $A$

The answer is: Some $a \in A$ satisfies $a > s - 0.01$

4.[2p]
By the Archimedean property some $n \in ℕ$ has $\frac{1}{n} < 0.03$ . What is the smallest such $n$ ?

CorrectNot quite: 34

5.[2p]
What is $\sup {x \in ℝ : x^{2} + x < 6}$ ?

CorrectNot quite: 2

6.[2p]
Put the lines of the proof of the Archimedean property in order.

Since $ℕ$ is nonempty, completeness gives it a supremum $s$ .
Since $s - 1$ is not an upper bound, some $n \in ℕ$ has $n > s - 1$ .
This contradicts that $s$ is an upper bound of $ℕ$ .
Then $n + 1 > s$ and $n + 1 \in ℕ$ .
Suppose, for contradiction, that $ℕ$ is bounded above in $ℝ$ .

Show the answer

a, b, c, d, e

7.[3p]
Match each result to the step its proof turns on.

No least upper bound of ${ q \in ℚ
Archimedean property
Density of $ℚ$ in $ℝ$
Existence of $\sqrt{2}$

the least integer $m$ greater than $n a$
the supremum of ${x \in ℝ : x^{2} < 2}$ has neither $s^{2} < 2$ nor $s^{2} > 2$
q^2 < 2 }$ in $ℚ$ : the irrationality of $\sqrt{2}$ rules out $u^{2} = 2$
a supremum of $ℕ$ would be exceeded by $n + 1$

Show the answer

No least upper bound of ${q \in ℚ : q^{2} < 2}$ in $ℚ$ : the irrationality of $\sqrt{2}$ rules out $u^{2} = 2$ Archimedean property: a supremum of $ℕ$ would be exceeded by $n + 1$ Density of $ℚ$ in $ℝ$ : the least integer $m$ greater than $n a$ Existence of $\sqrt{2}$ : the supremum of ${x \in ℝ : x^{2} < 2}$ has neither $s^{2} < 2$ nor $s^{2} > 2$

8.[3p]
Which statements are true?

Select all that apply

$\inf {\frac{1}{n} : n \in ℕ} = 0$

Correct

Between any two distinct reals lies a rational

Correct

Between any two distinct reals lies an irrational

Correct

Every nonempty set of reals bounded below has an infimum

Correct

Every nonempty set of rationals bounded above has a least upper bound in $ℚ$

The answer is: $\inf {\frac{1}{n} : n \in ℕ} = 0$ , Between any two distinct reals lies a rational, Between any two distinct reals lies an irrational, Every nonempty set of reals bounded below has an infimum

$ℕ$ has a supremum in $ℝ$

9.[2p]
In the proof that $s = \sup {x \in ℝ : x^{2} < 2}$ has $s^{2} = 2$ , why can $s^{2} > 2$ not hold?

Then $s - k$ , with $k = \frac{s^{2} - 2}{2 s}$ , would be a smaller upper bound

Correct

Then $s + h$ would lie in the set and exceed $s$

The answer is: Then $s - k$ , with $k = \frac{s^{2} - 2}{2 s}$ , would be a smaller upper bound

Then $s$ would be rational, contradicting the irrationality of $\sqrt{2}$

The answer is: Then $s - k$ , with $k = \frac{s^{2} - 2}{2 s}$ , would be a smaller upper bound

Then the set would have no upper bound at all

The answer is: Then $s - k$ , with $k = \frac{s^{2} - 2}{2 s}$ , would be a smaller upper bound

10.[1p]
What is $⌊ - 2.5 ⌋$ ?

CorrectNot quite: -3

11.[2p]
What is $\sup {(- 1)^{n} + \frac{1}{n} : n \in ℕ}$ ? Give a decimal.

CorrectNot quite: 1.5

12.[2p]
Why is $0.999 \dots = 1$ ?

The decimal names the supremum of its truncations $1 - \frac{1}{10^{n}}$ , and by the Archimedean property that supremum is $1$

Correct

The truncations get closer and closer to $1$ without reaching it, so the two are equal only approximately

The answer is: The decimal names the supremum of its truncations $1 - \frac{1}{10^{n}}$ , and by the Archimedean property that supremum is $1$

The two differ by an infinitesimal, which is treated as zero

The answer is: The decimal names the supremum of its truncations $1 - \frac{1}{10^{n}}$ , and by the Archimedean property that supremum is $1$

Every truncation of $0.999 \dots$ is already equal to $1$

The answer is: The decimal names the supremum of its truncations $1 - \frac{1}{10^{n}}$ , and by the Archimedean property that supremum is $1$

13.[2p]
Which of these nested sequences of intervals in $ℝ$ must have a point common to all of them?

Select all that apply

$[0, \frac{1}{n}]$ for $n \in ℕ$

Correct

$(0, \frac{1}{n})$ for $n \in ℕ$

The answer is: $[0, \frac{1}{n}]$ for $n \in ℕ$ , $[- \frac{1}{n}, 1 + \frac{1}{n}]$ for $n \in ℕ$

$[n, \infty)$ for $n \in ℕ$

The answer is: $[0, \frac{1}{n}]$ for $n \in ℕ$ , $[- \frac{1}{n}, 1 + \frac{1}{n}]$ for $n \in ℕ$

$[- \frac{1}{n}, 1 + \frac{1}{n}]$ for $n \in ℕ$

Correct

14.[2p]
Why can the complex numbers not be made into an ordered field?

In an ordered field every square is non-negative and $- 1$ is negative, but $- 1 = i^{2}$

Correct

The complex numbers are uncountable

The answer is: In an ordered field every square is non-negative and $- 1$ is negative, but $- 1 = i^{2}$

The complex numbers do not satisfy the completeness axiom

The answer is: In an ordered field every square is non-negative and $- 1$ is negative, but $- 1 = i^{2}$

No total order on the set $ℂ$ exists

The answer is: In an ordered field every square is non-negative and $- 1$ is negative, but $- 1 = i^{2}$

1.[1p] Which property of ℝ fails in ℚ?

2.[1p] The ordered field axioms alone are enough to prove that some number has square 2.

3.[2p] Let s=supA and ε=0.01. What does the characterisation of the supremum guarantee?

4.[2p] By the Archimedean property some n∈ℕ has 1n<0.03. What is the smallest such n?

5.[2p] What is sup{x∈ℝ:x2+x<6}?

6.[2p] Put the lines of the proof of the Archimedean property in order.

7.[3p] Match each result to the step its proof turns on.

8.[3p] Which statements are true?

9.[2p] In the proof that s=sup{x∈ℝ:x2<2} has s2=2, why can s2>2 not hold?

10.[1p] What is ⌊−2.5⌋?

11.[2p] What is sup{(−1)n+1n:n∈ℕ}? Give a decimal.

12.[2p] Why is 0.999…=1?

13.[2p] Which of these nested sequences of intervals in ℝ must have a point common to all of them?

14.[2p] Why can the complex numbers not be made into an ordered field?

1.[1p]
Which property of $ℝ$ fails in $ℚ$ ?

2.[1p]
The ordered field axioms alone are enough to prove that some number has square $2$ .

3.[2p]
Let $s = \sup A$ and $ε = 0.01$ . What does the characterisation of the supremum guarantee?

4.[2p]
By the Archimedean property some $n \in ℕ$ has $\frac{1}{n} < 0.03$ . What is the smallest such $n$ ?

5.[2p]
What is $\sup {x \in ℝ : x^{2} + x < 6}$ ?

6.[2p]
Put the lines of the proof of the Archimedean property in order.

7.[3p]
Match each result to the step its proof turns on.

8.[3p]
Which statements are true?

9.[2p]
In the proof that $s = \sup {x \in ℝ : x^{2} < 2}$ has $s^{2} = 2$ , why can $s^{2} > 2$ not hold?

10.[1p]
What is $⌊ - 2.5 ⌋$ ?

11.[2p]
What is $\sup {(- 1)^{n} + \frac{1}{n} : n \in ℕ}$ ? Give a decimal.

12.[2p]
Why is $0.999 \dots = 1$ ?

13.[2p]
Which of these nested sequences of intervals in $ℝ$ must have a point common to all of them?

14.[2p]
Why can the complex numbers not be made into an ordered field?