Mathematical-Reasoning · Unit 8 · Video 4 · Interactive Practice

Why No Fraction Squares to Two: Rational Numbers, Irrational Numbers and the Proof That √2 Is Irrational

IKey Formulas

Statement	Name	Role
$x \in \mathbb{Q} \iff x = \frac{m}{n}$ for some $m, n \in \mathbb{Z}$ with $n \neq 0$	Rational number	Irrational means not rational
$x, y \in \mathbb{Q} \Rightarrow x + y$ , $x - y$ , $xy \in \mathbb{Q}$ ; and $y \neq 0 \Rightarrow \frac{x}{y} \in \mathbb{Q}$	Closure of $\mathbb{Q}$	Computing with fractions gives fractions
$x \in \mathbb{Q}$ , $x \neq 0$ , $y$ irrational $\Rightarrow$ $xy$ irrational	Proposition	The proof divides by $x$
$r \in \mathbb{R}$ , $r^2 = 2$ $\Rightarrow$ $r$ irrational	Theorem	$r = \frac{m}{n}$ in lowest terms gives $m^2 = 2n^2$ , so $m$ and $n$ are both even

Key Insight: Irrational is defined by a negation, so a proof by contradiction assumes the number is rational, and that assumption supplies a fraction $\frac{m}{n}$ with $n \neq 0$ to compute with.

IIWhere the Proposition's Proof Needs Its Hypotheses

The proof divides by $x$ and uses closure of $\mathbb{Q}$ ; remove a hypothesis and a specific line fails.

x y

Hypotheses

Proof by contradiction

Outcome

IIIFractions as Lattice Points

A fraction $\frac{m}{n}$ with $\left(\frac{m}{n}\right)^2 = 2$ would be a lattice point $(m, n)$ on the line $m = \sqrt{2}\,n$ .

numerator m 7 denominator n 5

💡 The marked points $(1, 1)$ , $(3, 2)$ , $(7, 5)$ and $(17, 12)$ have $m^2 - 2n^2 = \pm 1$ . Since $m^2 - 2n^2$ is an integer, it can never be closer to $0$ without being $0$ , and the theorem shows it is never $0$ .

IVThe Proof That √2 Is Irrational

Writing $r = \frac{m}{n}$ in lowest terms and squaring forces both $m$ and $n$ to be even.

Step 1

Proof

1 · Assume the negation

Assume

r \in \mathbb{R}

r^2 = 2

and

r \in \mathbb{Q}

. In lowest terms,

r = \frac{m}{n}, \quad m, n \in \mathbb{Z}, \quad n > 0,

and

m

and

n

have no common factor greater than

1

2 · Square and clear the denominator

2 = r^2 = \frac{m^2}{n^2} \;\Rightarrow\; m^2 = 2n^2

3 · Name half of m

Since

m

is even,

m = 2p

for some integer

p

4 · Substitute and divide by 2

(2p)^2 = 2n^2 \;\Rightarrow\; 4p^2 = 2n^2 \;\Rightarrow\; n^2 = 2p^2

5 · Contradiction

2

is a common factor of

m

and

n

, which contradicts lowest terms. So

r

is irrational.

Parity notes

VQuiz Questions

Problem 1 · Apply the Proposition

Given: $\sqrt{2}$ is irrational, and the proposition: if $x \in \mathbb{Q}$ , $x \neq 0$ and $y$ is irrational, then $xy$ is irrational. Which number does the proposition prove to be irrational?

✅ Correct! Take

x = -\frac{3}{7}

and

y = \sqrt{2}

x

is rational because

-3, 7 \in \mathbb{Z}

with

7 \neq 0

x \neq 0

, and

y

is irrational, so

xy = -\frac{3}{7}\sqrt{2}

is irrational.

❌ The hypothesis $x \neq 0$ fails. Here

x = 0

, and

0 \cdot \sqrt{2} = 0 = \frac{0}{1}

is rational. This is exactly the counterexample that makes the condition

x \neq 0

necessary.

❌ The hypothesis $x \in \mathbb{Q}$ fails. Here

x = \sqrt{2}

is irrational, so the proposition does not apply, and in fact

\sqrt{2} \cdot \sqrt{2} = 2

is rational.

❌ This is a quotient of two irrationals, not a nonzero rational times an irrational. It equals

1 = \frac{1}{1}

, which is rational.

Show solution

The proposition needs a factor $x$ that is rational and nonzero, and a factor $y$ that is irrational.

(a) $x = 0$ violates $x \neq 0$ ; the product is $0$ , which is rational. (b) $x = \sqrt{2}$ is not rational; the product is $2$ , which is rational. (d) is a quotient, and it equals $1$ , which is rational.

(c) With $x = -\frac{3}{7}$ (integers $-3$ and $7$ , denominator $7 \neq 0$ , so $x \in \mathbb{Q}$ , and $x \neq 0$ ) and $y = \sqrt{2}$ (irrational), the proposition gives that $-\frac{3}{7}\sqrt{2}$ is irrational.

Problem 2 · Set Up the Contradiction

Given: the proposition "for all real numbers $x$ and $y$ , if $x \in \mathbb{Q}$ and $x \neq 0$ and $y$ is irrational, then $xy$ is irrational." Which statement does a proof by contradiction assume?

✅ Correct! The negation of

(\forall x, y)(H \to C)

(\exists x, y)(H \wedge \neg C)

: the hypotheses all hold and the conclusion fails.

❌ The quantifier did not flip. The negation of a "for all" statement is a "there exist" statement: one pair

x, y

where the conditional fails is enough.

❌ That keeps the conclusion. A pair with

xy

irrational is an instance of the proposition, not a failure of it. The negation needs

\neg C

xy

is rational.

❌ That negates the hypothesis.

\neg(H \to C)

H \wedge \neg C

, so the hypotheses must all hold. If a hypothesis fails, the conditional is true for that pair, and nothing contradicts it.

Show solution

Write $H(x, y)$ for " $x \in \mathbb{Q}$ and $x \neq 0$ and $y$ is irrational" and $C(x, y)$ for " $xy$ is irrational". The proposition is $(\forall x, y \in \mathbb{R})(H(x, y) \to C(x, y))$ .

\neg(\forall x, y)\big(H \to C\big) \equiv (\exists x, y)\,\neg\big(H \to C\big) \equiv (\exists x, y)\big(H \wedge \neg C\big).

Since $\neg C$ says that $xy$ is not irrational, that is, $xy$ is rational, the proof assumes: there exist real numbers $x$ and $y$ with $x \in \mathbb{Q}$ , $x \neq 0$ , $y$ irrational, and $xy \in \mathbb{Q}$ . That is option (d).

Problem 3 · The Second Even Integer

Given: in the proof that $\sqrt{2}$ is irrational, $m^2 = 2n^2$ , and $m = 2p$ for some integer $p$ . Which equation does the proof derive next to show that $n$ is even?

✅ Correct!

(2p)^2 = 2n^2

gives

4p^2 = 2n^2

, and dividing by

2

gives

n^2 = 2p^2

. So

n^2

is even, and

n

is even.

❌ Check the square.

(2p)^2 = 4p^2

, not

2p^2

. From

4p^2 = 2n^2

, dividing by

2

gives

n^2 = 2p^2

❌ That substitutes into the wrong letter. The integer

p

was defined by

m = 2p

, so

2p

replaces

m

, not

n

. Nothing is known yet about

n

being

2p

❌ True, but it says nothing about $n$ .

m^2 = 4p^2

only squares

m = 2p

. Substituting it into

m^2 = 2n^2

is the step that brings

n

in.

Show solution

Substitute $m = 2p$ into $m^2 = 2n^2$ :

(2p)^2 = 2n^2 \;\Rightarrow\; 4p^2 = 2n^2 \;\Rightarrow\; n^2 = 2p^2.

Since $p^2$ is an integer, $n^2 = 2 \cdot p^2$ is even. By the result from the previous unit (if $n^2$ is even, then $n$ is even), $n$ is even. Together with $m$ even, $2$ is a common factor of $m$ and $n$ , which contradicts lowest terms.

Problem 4 · Rational Plus Irrational

Given: $x \in \mathbb{Q}$ and $y$ is irrational. To prove that $x + y$ is irrational, a proof by contradiction assumes $x + y = q$ with $q \in \mathbb{Q}$ . Which line validly reaches a contradiction?

✅ Correct!

\mathbb{Q}

is closed under subtraction, so

y = q - x \in \mathbb{Q}

, contradicting that

y

is irrational.

❌ The algebra is wrong. From

x + y = q

you get

y = q - x

;

\frac{q}{x}

would come from

xy = q

. It would also need

x \neq 0

, which is not given.

❌ That is circular. "Rational plus irrational is irrational" is the statement being proved, so it cannot be used inside its own proof.

❌ That inference is false. A rational sum can have irrational terms:

\sqrt{2} + (-\sqrt{2}) = 0

, where

-\sqrt{2} = (-1)\sqrt{2}

is irrational by the proposition.

Show solution

Proof. We will use a proof by contradiction. Assume there exist $x \in \mathbb{Q}$ and an irrational $y$ such that $x + y$ is rational, say $x + y = q$ with $q \in \mathbb{Q}$ .

Subtracting $x$ from both sides gives $y = q - x$ . Both $q$ and $x$ are rational, and $\mathbb{Q}$ is closed under subtraction, so $y \in \mathbb{Q}$ .

A real number cannot be both rational and irrational, so this contradicts the assumption that $y$ is irrational. Therefore $x + y$ is irrational.

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