Mathematical-Reasoning · Unit 2 · Video 4 · Interactive Practice

Which Way the Arrow Runs: Choosing Answers and Finishing Two Direct Proofs

IKey Formulas

Idea	Statement	Note
Backward answer to goal $Q$	A statement $Q_1$ with $Q_1 \Rightarrow Q$	Not $Q \Rightarrow Q_1$ : $y^2 = 4$ does not give $y = 2$ , since $(-2)^2 = 4$
Proving an integer is odd	Show it equals $2q + 1$ with $q$ an integer, or show it is a product of two odd integers	$x^2 = x \cdot x$ is odd when $x$ is odd
$m$ odd, $m = 2k + 1$	$3m^2 + 4m + 6 = 2(6k^2 + 10k + 6) + 1$	$6k^2 + 10k + 6$ is an integer by closure
Sides $m$ , $m + 1$ , $m + 2$	$m^2 + (m + 1)^2 = (m + 2)^2$ gives $(m - 3)(m + 1) = 0$	$m > 0$ rules out $m = -1$ , so $m = 3$

Key Insight: A backward answer must imply the goal. Checking that $m = 3$ satisfies the equation runs the other way: it shows the case occurs ( $3$ , $4$ , $5$ is a right triangle) but does not show that the hypothesis forces $m = 3$ .

IIWhich Way Does the Arrow Run?

A backward answer $Q_1$ works only if every real $y$ that makes $Q_1$ true also makes the goal $Q$ true.

Goal Q Backward answer Q₁

IIIFrom Six Examples to Every Odd $m$

The table suggests that odd $m$ gives an odd value; substituting $m = 2k + 1$ proves it for every odd $m$ .

m 3

Step 1 · Know — $m$ is odd, so $m = 2k + 1$ for some integer $k$ .

Step 2 · Substitute and expand

3m^2 + 4m + 6 = 3(2k + 1)^2 + 4(2k + 1) + 6

= (12k^2 + 12k + 3) + (8k + 4) + 6 = 12k^2 + 20k + 13

Step 3 · Show — split $13 = 12 + 1$ and factor a $2$ from the first three terms:

12k^2 + 20k + 13 = 2(6k^2 + 10k + 6) + 1

By closure of the integers, $q = 6k^2 + 10k + 6$ is an integer, so $3m^2 + 4m + 6 = 2q + 1$ is odd.

IVSides $m$ , $m + 1$ , $m + 2$

Legs $m$ and $m + 1$ fix the hypotenuse, and it equals $m + 2$ only where $(m - 3)(m + 1) = 0$ .

m 2

💡 Substituting $m = 3$ into $m^2 - 2m - 3 = 0$ only shows $m = 3 \Rightarrow$ the equation holds. The proof runs from the hypothesis to $m = 3$ , and it needs the fact that a length is positive to discard $m = -1$ .

VQuiz Questions

Problem 1 · Choosing a Backward Answer

Given: $y$ is a real number and the goal is $y^2 = 9$ . Which statement is an appropriate answer to the backward question “how do we prove that the square of a real number equals $9$ ?”

✅ Correct! If

y = -3

, then

y^2 = (-3)^2 = 9

, so

y = -3 \Rightarrow y^2 = 9

: the arrow runs from the answer to the goal. (

y = 3

would work too.)

❌ The arrow is reversed.

y^2 = 9

does give

y^2 \ge 9

, but

y = 4

has

y^2 = 16 \ge 9

and

y^2 \neq 9

, so proving

y^2 \ge 9

would not establish the goal.

❌ Not quite.

y = 4

satisfies

y \ge 3

while

y^2 = 16 \neq 9

, so

y \ge 3

does not imply the goal.

❌ The arrow is reversed. Every

y

with

y^2 = 9

is nonzero, but

y = 1

is nonzero with

y^2 = 1

, so

y \neq 0

does not imply

y^2 = 9

❌ Not quite. Look for the statement

Q_1

for which

Q_1 \Rightarrow y^2 = 9

Show solution

A backward answer to the goal $Q$ is a statement $Q_1$ with $Q_1 \Rightarrow Q$ : whenever $Q_1$ is true, $Q$ must be true. Test each option for a real $y$ that makes it true while $y^2 \neq 9$ :

(a) $y = 4$ : $y^2 = 16 \ge 9$ but $16 \neq 9$ . Fails. (The arrow $y^2 = 9 \Rightarrow y^2 \ge 9$ holds, which is the wrong direction.)
(b) $y = 4$ : $4 \ge 3$ but $16 \neq 9$ . Fails.
(c) The only $y$ with $y = -3$ is $-3$ itself, and $(-3)^2 = 9$ . Works.
(d) $y = 1$ : $1 \neq 0$ but $1^2 = 1 \neq 9$ . Fails.

So proving $y = -3$ (or $y = 3$ ) is an appropriate way to prove $y^2 = 9$ .

Problem 2 · The Backward Trap

Given: $t$ is a real number and the goal is $t = 5$ . Which statement $Q_1$ satisfies $Q_1 \Rightarrow t = 5$ , so that proving it would prove the goal?

✅ Correct!

t^2 = 25

leaves

t = 5

t = -5

, and

t > 0

removes

-5

, so this

Q_1

forces

t = 5

❌ That is the backward trap.

t = 5 \Rightarrow t^2 = 25

, but not the reverse:

t = -5

has

t^2 = 25

and

t \neq 5

❌ The arrow is reversed.

t = 5

gives

t \ge 5

, but

t = 6

satisfies

t \ge 5

and

t \neq 5

❌ The arrow is reversed.

t = 5

gives

t^2 = 5t

, but

t = 0

also satisfies

t^2 = 5t

(

0 = 0

) and

t \neq 5

❌ Not quite. The chosen statement must force

t = 5

Show solution

Each of (a), (b), (c) follows from $t = 5$ , so each is a consequence of the goal, not a way to reach it. A single real $t$ with $Q_1$ true and $t \neq 5$ shows $Q_1 \not\Rightarrow t = 5$ :

(a) $t = -5$ : $(-5)^2 = 25$ .
(b) $t = 6$ : $6 \ge 5$ .
(c) $t = 0$ : $0^2 = 0 = 5 \cdot 0$ .

For (d): if $t^2 = 25$ , then $t = 5$ or $t = -5$ ; since also $t > 0$ , $t = 5$ . So (d) $\Rightarrow t = 5$ , and proving (d) would prove the goal. The extra condition $t > 0$ plays the same role as “a length is positive” in the triangle proof.

Problem 3 · Another Odd Expression

Proposition: if $m$ is an odd integer, then $5m^2 + 2m + 4$ is an odd integer. The proof writes $m = 2k + 1$ for an integer $k$ and aims for $5m^2 + 2m + 4 = 2q + 1$ .

Which q works?

Why is that q an integer?

✅ Correct!

5m^2 + 2m + 4 = 20k^2 + 24k + 11 = 2(10k^2 + 12k + 5) + 1

, and closure makes

10k^2 + 12k + 5

an integer.

❌ Close, but that is $2q$ .

20k^2 + 24k + 10 = 2(10k^2 + 12k + 5)

: factor the

2

out before reading off

q

❌ Not quite. This

q

comes from dropping the

2m = 4k + 2

term:

5(2k + 1)^2 + 4 = 2(10k^2 + 10k + 4) + 1

❌ Check the constant.

2(10k^2 + 12k + 6) + 1 = 20k^2 + 24k + 13

, two more than the expression; split

11

10 + 1

❌ Not quite. Expand

5(2k + 1)^2 + 2(2k + 1) + 4

and write the result as

2(\dots) + 1

❌ Not quite. Nothing says

k

is odd (

m = 1

gives

k = 0

); the argument needs only that

k

is an integer.

❌ That is circular. Oddness of

5m^2 + 2m + 4

is what the proof is trying to show, so it cannot be a reason.

❌ Not quite. Three examples cannot cover every odd

m

; the reason must hold for every integer

k

❌ Not quite.

q

is built from

k

by adding and multiplying integers.

Show solution

Know: $m = 2k + 1$ for some integer $k$ . Show: $5m^2 + 2m + 4 = 2q + 1$ for some integer $q$ .

5m^2 + 2m + 4 = 5(2k + 1)^2 + 2(2k + 1) + 4

= (20k^2 + 20k + 5) + (4k + 2) + 4 = 20k^2 + 24k + 11

= (20k^2 + 24k + 10) + 1 = 2(10k^2 + 12k + 5) + 1.

Since $k$ is an integer and the integers are closed under multiplication and addition, $q = 10k^2 + 12k + 5$ is an integer. Hence $5m^2 + 2m + 4$ is odd.

Spot check: $m = 1$ gives $k = 0$ , $q = 5$ , and $5 + 2 + 4 = 11 = 2(5) + 1$ .

Problem 4 · Another Right Triangle

Proposition: if $m$ is a real number and $m$ , $m + 7$ and $m + 8$ are the lengths of the three sides of a right triangle, then $m = 5$ .

Which standard-form equation does the Pythagorean Theorem give?

How does the proof finish?

✅ Correct! The equation narrows

m

5

-3

, and positivity of a length removes

-3

; the sides are

5

12

13

❌ Check the hypotenuse. This uses

m + 7

as the hypotenuse, but the hypotenuse is the longest side,

m + 8

❌ Check the expansion.

(m + 7)^2 = m^2 + 14m + 49

, not

m^2 + 49

❌ Check the sign of the $m$ -term. Collecting on the left gives

14m - 16m = -2m

❌ Not quite. Expand

m^2 + (m + 7)^2 = (m + 8)^2

and move every term to the left.

❌ Not quite. Part of the hypothesis is unused:

m

is a side length, so

m > 0

, and

m = -3

is impossible.

❌ That arrow points the wrong way. The check shows

m = 5 \Rightarrow

the equation holds, so the case occurs; it does not show that the hypothesis forces

m = 5

❌ Not quite. The hypothesis says only that

m

is real, and

-3

is an integer anyway; positivity of a length is what rules out

-3

❌ Not quite. Use every part of the hypothesis to choose between the two roots.

Show solution

Assume $m$ is real and $m$ , $m + 7$ , $m + 8$ are the sides of a right triangle. The hypotenuse is the longest side, $m + 8$ , so by the Pythagorean Theorem

m^2 + (m + 7)^2 = (m + 8)^2

m^2 + (m^2 + 14m + 49) = m^2 + 16m + 64

2m^2 + 14m + 49 = m^2 + 16m + 64

m^2 - 2m - 15 = 0 \quad\text{so}\quad (m - 5)(m + 3) = 0.

Hence $m = 5$ or $m = -3$ . Since $m$ is the length of a side, $m > 0$ , which rules out $-3$ . Therefore $m = 5$ . ∎

The check $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ confirms only that the case really occurs; the proof is the argument above.

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Which Way the Arrow Runs: Choosing Answers and Finishing Two Direct Proofs

IKey Formulas

IIWhich Way Does the Arrow Run?

IIIFrom Six Examples to Every Odd mmm

IVSides mmm, m+1m + 1m+1, m+2m + 2m+2

VQuiz Questions

IIIFrom Six Examples to Every Odd $m$

IVSides $m$ , $m + 1$ , $m + 2$