Mathematical-Reasoning · Unit 7 · Video 4 · Interactive Practice

Constructive and Nonconstructive Proofs of Existence

IKey Formulas

Statement	Name	Role
$(\exists x)(P(x))$	Existence statement	The goal of an existence theorem
$ax + b = c$ , $a \neq 0$ $\Rightarrow$ $x = \frac{c-b}{a}$	Linear equation in one variable	Constructive: solve (any solution must be this), then check (it is one)
$f$ continuous on $[a, b]$ , $q$ strictly between $f(a)$ and $f(b)$ $\Rightarrow$ $(\exists c \in (a, b))(f(c) = q)$	Intermediate Value Theorem	Nonconstructive: proves $c$ exists, never names it
$f(-2) = -5 < 0 < 1 = f(0)$	$f(x) = x^3 - x + 1$	Some $c \in (-2, 0)$ has $c^3 - c + 1 = 0$

Key Insight: A constructive proof names the object and checks that it works; a nonconstructive proof shows that the object must exist and gives no value for it. Both are complete proofs of $(\exists x)(P(x))$ .

IISolve, Then Check

For $ax + b = c$ with $a \neq 0$ , solving shows any solution must be $\frac{c-b}{a}$ ; checking shows it is one.

a 3 b 8 c 23

Solve

Check

IIIThe Intermediate Value Theorem

A continuous graph that starts below the level $q$ and ends above it must meet the line $y = q$ .

Level q 1 Function

IVClosing In on c

Halving the interval and keeping the sign change traps the solution of $x^3 - x + 1 = 0$ without ever naming it.

Step 0

💡 The existence proof is already complete at step 0: $f(-2) < 0 < f(0)$ and the Intermediate Value Theorem prove that a solution exists. The halving only approximates it.

VQuiz Questions

Problem 1 · Name the Solution

Given: the linear equation $4x + 9 = 1$ , where $x$ is a real number. Which number does the constructive proof name as its only real solution?

✅ Correct!

x = \frac{c-b}{a} = \frac{1-9}{4} = -2

, and the check

4(-2) + 9 = -8 + 9 = 1

confirms it.

❌ One step is missing.

1 - 9 = -8

4x

, not

x

. Divide by

a = 4

, which is allowed because

4 \neq 0

❌ Check the first step. To remove

+9

, add

-9

to both sides:

4x = 1 - 9

, not

1 + 9

. The check fails too:

4 \cdot \frac{5}{2} + 9 = 19 \neq 1

❌ The subtraction is reversed. The formula is

\frac{c-b}{a} = \frac{1-9}{4}

, not

\frac{9-1}{4}

. The check fails:

4(2) + 9 = 17 \neq 1

Show solution

Here $a = 4$ , $b = 9$ , $c = 1$ , and $a \neq 0$ .

Solve. Adding $-9$ to both sides gives $4x = -8$ . Since $4 \neq 0$ , dividing both sides by $4$ gives $x = -2$ . So if there is a solution, it must be $-2$ .

Check. Substituting $-2$ for $x$ in the left side:

4(-2) + 9 = -8 + 9 = 1.

So $-2$ is a solution, and $-2$ is the only real solution of $4x + 9 = 1$ . This agrees with the formula $x = \frac{c-b}{a} = \frac{1-9}{4} = -2$ .

Problem 2 · What the Solve Step Proves

Given: real numbers $a, b, c$ with $a \neq 0$ . In the proof for $ax + b = c$ , the solve step adds $-b$ to both sides and then divides by $a$ . Which statement does the solve step prove?

✅ Correct! Solving starts from a solution and derives its value: any solution must be

\frac{c-b}{a}

, so there is at most one.

❌ That is the converse. It is proved by the check step, which substitutes

\frac{c-b}{a}

into the left side and gets

c

❌ Not from the solve step alone. Solving assumes a solution and shows what it must be; existence comes from the check, which shows

\frac{c-b}{a}

actually works.

❌ That is false in general. For

3x + 8 = 23

, the value

x = 0

gives

8 \neq 23

. Only one real number solves the equation.

Show solution

The solve step begins with "assume $ax + b = c$ " and ends with " $x = \frac{c-b}{a}$ ". That is a direct proof of the conditional

ax + b = c \;\Rightarrow\; x = \frac{c-b}{a},

which says that any solution must equal $\frac{c-b}{a}$ . It does not say that $\frac{c-b}{a}$ is a solution. That is the converse,

x = \frac{c-b}{a} \;\Rightarrow\; ax + b = c,

proved by the check: $a \cdot \frac{c-b}{a} + b = (c - b) + b = c$ . Together the two conditionals give the biconditional $ax + b = c \Leftrightarrow x = \frac{c-b}{a}$ : exactly one real solution.

Problem 3 · Shrink the Interval

Given: $f(x) = x^3 - x + 1$ with $f(-2) = -5$ , $f(-1.5) = -0.875$ , $f(-1.25) = 0.296875$ and $f(-1) = 1$ . Which is the shortest of these open intervals that the Intermediate Value Theorem (with $q = 0$ ) guarantees contains a solution of $x^3 - x + 1 = 0$ ?

✅ Correct!

f(-1.5) < 0 < f(-1.25)

, so some

c \in (-1.5, -1.25)

has

f(c) = 0

, and this interval has length

0.25

❌ Guaranteed, but not the shortest.

f(-2) < 0 < f(-1)

does give a solution in

(-2, -1)

, an interval of length

1

. The given values allow a shorter one.

❌ No sign change there.

f(-1.25) = 0.296875

and

f(-1) = 1

are both positive, so

q = 0

is not strictly between them and the theorem promises nothing.

❌ Guaranteed, but not the shortest.

f(-1.5) < 0 < f(-1)

gives length

0.5

. The value at

-1.25

halves it again.

Show solution

$f$ is a polynomial, so it is continuous on every closed interval. The theorem with $q = 0$ applies to $[s, t]$ exactly when $0$ is strictly between $f(s)$ and $f(t)$ , that is, when $f(s)$ and $f(t)$ have opposite signs.

The signs are $f(-2) < 0$ , $f(-1.5) < 0$ , $f(-1.25) > 0$ , $f(-1) > 0$ . A sign change occurs between $-2$ and $-1$ , between $-1.5$ and $-1$ , and between $-1.5$ and $-1.25$ , but not between $-1.25$ and $-1$ .

The shortest of the guaranteed intervals is $(-1.5, -1.25)$ , of length $0.25$ . For example, $f(-1.25) = (-1.25)^3 - (-1.25) + 1 = -1.953125 + 1.25 + 1 = 0.296875$ .

Problem 4 · Set Up the Theorem

Given: to prove that $x^3 + 2x = 5$ has a real number solution, apply the Intermediate Value Theorem to $f(x) = x^3 + 2x$ , which is continuous on every closed interval.

Which value of $q$ ?

Which interval $[a, b]$ makes the theorem apply?

✅ Correct!

f(1) = 3 < 5 < 12 = f(2)

, so some

c \in (1, 2)

has

c^3 + 2c = 5

. The proof is nonconstructive: it names no value of

c

❌ That level answers a different equation.

f(c) = 0

means

c^3 + 2c = 0

. The equation asks for

f(c) = 5

, so

q = 5

❌ Wrong sign.

f(c) = -5

means

c^3 + 2c = -5

. The right side of the equation is

+5

, so

q = 5

❌ That is an endpoint value, not the level.

f(1) = 3

, but the equation asks for

f(c) = 5

, so

q = 5

❌ Check $q$ . The theorem's conclusion is

f(c) = q

; match it to

c^3 + 2c = 5

❌ Not quite.

f(0) = 0

and

f(1) = 3

; the level

5

is not between them.

❌ Not quite.

f(2) = 12

and

f(3) = 33

; both are above

5

❌ Not quite.

f(-1) = -3

and

f(0) = 0

; both are below

5

❌ Check the interval. Compute

f

at both endpoints and look for

5

strictly between the two values.

Show solution

Proof. Let $f(x) = x^3 + 2x$ . Since $f$ is a polynomial, it is continuous on $[1, 2]$ . We compute

f(1) = 1 + 2 = 3, \qquad f(2) = 8 + 4 = 12.

Since $3 < 5 < 12$ , the number $q = 5$ is strictly between $f(1)$ and $f(2)$ . By the Intermediate Value Theorem there exists a real number $c \in (1, 2)$ with $f(c) = 5$ , that is, $c^3 + 2c = 5$ . Hence the equation $x^3 + 2x = 5$ has a real number solution. $\blacksquare$

The other intervals fail: $f(-1) = -3$ and $f(0) = 0$ ; $f(0) = 0$ and $f(1) = 3$ ; $f(2) = 12$ and $f(3) = 33$ . In none of them is $5$ strictly between the endpoint values.

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