Direct Proof | Real Analysis

The sum of two rationals

r_1 = \frac{p_1}{q_1}

r_2 = \frac{p_2}{q_2}

on a number line, with the computation

\frac{p_1 q_2 + p_2 q_1}{q_1 q_2}

running straight from hypothesis to conclusion — that is a direct proof.

Examples & Exercises

Prove by a direct proof that the sum of two rational numbers is rational, and verify each step on this scene's default numbers $r_1 = \frac34$ , $r_2 = -\frac56$ , confirming the sum is $-\frac{1}{12}$ step by step.

r_1 = \frac{p_1}{q_1},\ r_2 = \frac{p_2}{q_2} \;\Rightarrow\; r_1 + r_2 = \frac{p_1 q_2 + p_2 q_1}{q_1 q_2}

p_1 = 3

q_1 = 4

p_2 = -5

q_2 = 6

: numerator

18 - 20 = -2

, denominator

24

\gcd(2, 24) = 2

, reduced

-\frac{1}{12} \approx -0.083333

Show proof

Take the hypothesis: since $r_1, r_2$ are rational, there exist integers $p_1, q_1, p_2, q_2$ with $q_1 \ne 0$ , $q_2 \ne 0$ such that $r_1 = \frac{p_1}{q_1}$ and $r_2 = \frac{p_2}{q_2}$ . A direct proof feeds this assumption straight into a computation.
Use a common denominator: $r_1 + r_2 = \frac{p_1 q_2}{q_1 q_2} + \frac{p_2 q_1}{q_1 q_2} = \frac{p_1 q_2 + p_2 q_1}{q_1 q_2}$ . The numerator $p_1 q_2 + p_2 q_1$ is a sum of products of integers, hence an integer; the denominator $q_1 q_2$ is a product of nonzero integers, hence a nonzero integer. So $r_1 + r_2$ is a ratio of integers — rational. That completes the direct proof. On the scene's numbers: $\frac{3 \cdot 6 + (-5) \cdot 4}{4 \cdot 6} = \frac{18 - 20}{24} = \frac{-2}{24} = -\frac{1}{12} \approx -0.083333$ .
$\frac{3}{4} + \frac{-5}{6} = \frac{-2}{24} = -\frac{1}{12}$

Prove by a direct proof that the product of two rational numbers is rational. Then prove that the square of a rational number is rational, and verify on the scene's numbers $r_1 = \frac34$ , $r_2 = -\frac56$ that the product is $-\frac58$ .

r_1 r_2 = \frac{p_1 p_2}{q_1 q_2}, \qquad r_1^2 = \frac{p_1^2}{q_1^2}

p_1 = 3

q_1 = 4

p_2 = -5

q_2 = 6

: product

-\frac{15}{24} = -\frac58 = -0.625

Show proof

Assume $r_1 = \frac{p_1}{q_1}$ , $r_2 = \frac{p_2}{q_2}$ (integers, $q_1, q_2 \ne 0$ ). Multiplying gives $r_1 r_2 = \frac{p_1 p_2}{q_1 q_2}$ . The numerator $p_1 p_2$ is an integer and the denominator $q_1 q_2$ is a nonzero integer, so the product is rational. On the scene's numbers: $\frac{3 \cdot (-5)}{4 \cdot 6} = \frac{-15}{24} = -\frac58 = -0.625$ (dividing numerator and denominator by 3).
The square is a special case of the product: if $r = \frac{p}{q}$ with $q \ne 0$ , then $r^2 = \frac{p^2}{q^2}$ , and $p^2$ is an integer while $q^2$ is a positive integer (hence $\ne 0$ ), so $r^2$ is rational. Example: $r_1^2 = \left(\frac34\right)^2 = \frac{9}{16}$ . Both claims are proved by a single computation running straight from hypothesis to conclusion — the typical shape of a direct proof.

Prove by a direct proof that if an integer $n$ is odd then $n^2$ is odd: write $n = 2k + 1$ and expand to show $(2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ .

n = 2k + 1 \;\Rightarrow\; n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1

This problem is about an algebraic proof for odd

n

— the preset is a valid scene combination,

r_1 = r_2 = \frac12

Show proof

Assume $n$ is odd. By the definition of odd, there is an integer $k$ with $n = 2k + 1$ . Now compute: $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ .
Group the even part: $4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ . Since $2k^2 + 2k$ is an integer, $n^2$ is "twice an integer, plus one" — odd. One expansion carried us from hypothesis to conclusion, so this is a direct proof. (Note: this lemma is exactly what the next scene, contrapositive, consumes — the proof of " $n^2$ even $\Rightarrow$ $n$ even" reduces to its contrapositive " $n$ odd $\Rightarrow$ $n^2$ odd", whose proof is precisely this direct proof.)