Proof by Cases and Existence Proofs

Moving the point

(x, y)

shades its quadrant and unfolds that case's computation of

|xy| = |x|\cdot|y|

— the four cases (and the axes) cover the whole plane. And from

c = \sqrt{2}^{\sqrt{2}}

two branches fork and rejoin at "irrational

a, b

exist with

a^b

rational" without ever revealing which branch is true — the flavor of non-constructive existence.

Examples & Exercises

Prove completely by cases: $|xy| = |x|\,|y|$ for all real $x, y$ . Cover all four sign cases (the zero case folds into case 1) and apply the definition of $|t|$ ( $t$ if $t \ge 0$ , $-t$ if $t < 0$ ) directly in each case.

|xy| = |x|\,|y| \qquad \forall x, y \in \mathbb{R}

x < 0 < y \;\Rightarrow\; |xy| = -(xy) = (-x) \cdot y = |x|\,|y|

(x, y) = (-2, 3)

: case 3 (

x < 0 < y

) —

|xy| = |-6| = 6 = 2 \cdot 3 = |x|\,|y|

Show proof

Case 1: $x \ge 0$ , $y \ge 0$ . Then $xy \ge 0$ , so $|xy| = xy$ . Also $|x| = x$ and $|y| = y$ , so $|x|\,|y| = xy$ — both sides agree. (The case $x = 0$ or $y = 0$ folds in here: $xy = 0$ gives $|xy| = 0 = |x|\,|y|$ .)
Case 2: $x > 0$ , $y < 0$ . Then $xy < 0$ , so $|xy| = -(xy)$ . Since $-xy = x \cdot (-y)$ and $|x| = x$ , $|y| = -y$ , we get $|x|\,|y| = x \cdot (-y)$ — both sides agree.
Case 3: $x < 0$ , $y > 0$ . Then $xy < 0$ , so $|xy| = -(xy) = (-x) \cdot y$ . Since $|x| = -x$ and $|y| = y$ , we get $|x|\,|y| = (-x) \cdot y$ — both sides agree.
Case 4: $x < 0$ , $y < 0$ . Then $xy > 0$ , so $|xy| = xy$ . Since $|x| = -x$ and $|y| = -y$ , we get $|x|\,|y| = (-x) \cdot (-y) = xy$ — both sides agree. The four cases cover every $(x, y) \in \mathbb{R}^2$ with nothing left out, so $|xy| = |x|\,|y|$ holds for all real $x, y$ . ∎

Prove non-constructively: there exist irrational numbers $a, b$ with $a^b$ rational. Put $c = \sqrt{2}^{\sqrt{2}}$ and split into "if $c$ is rational / if $c$ is irrational". Rigorously justify the exponent step $\big(\sqrt{2}^{\sqrt{2}}\big)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^{2} = 2$ from the power law $(u^v)^w = u^{vw}$ ( $u > 0$ ), and explain why this proof is non-constructive.

c = \sqrt{2}^{\sqrt{2}} \approx 1.6325269194381528

c \in \mathbb{Q} \Rightarrow a = b = \sqrt{2}; \qquad c \notin \mathbb{Q} \Rightarrow a = c,\ b = \sqrt{2},\ a^b = 2

The existence fork: from

c = \sqrt{2}^{\sqrt{2}} \approx 1.6325

two branches split and rejoin at "existence".

c^{\sqrt{2}} \approx 2.0000000000

Show proof

That $\sqrt{2}$ is irrational is known (see the 'sqrt2-irrational' scene problems). Put $c = \sqrt{2}^{\sqrt{2}}$ — an irrational power of an irrational. Without asking whether $c$ is rational or irrational, split into two cases.
Case 1: $c$ is rational. Put $a = b = \sqrt{2}$ : then $a, b$ are irrational and $a^b = \sqrt{2}^{\sqrt{2}} = c$ is rational — done.
Case 2: $c$ is irrational. Put $a = c$ , $b = \sqrt{2}$ : then $a, b$ are irrational, and applying the power law $(u^v)^w = u^{vw}$ (valid for real $u > 0$ and real $v, w$ ) with $u = \sqrt{2}$ , $v = w = \sqrt{2}$ gives $a^b = \big(\sqrt{2}^{\sqrt{2}}\big)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^{2} = 2$ — rational. Numerically too: $c^{\sqrt{2}} \approx 2.0000000000$ .
In either case irrational $a, b$ exist with $a^b$ rational, completing the proof. The proof is non-constructive: it never says which pair $(a, b)$ works — $(\sqrt{2}, \sqrt{2})$ if case 1 holds, $(c, \sqrt{2})$ if case 2 holds — because whether $c$ is rational or irrational is never decided anywhere in the proof. (The Gelfond–Schneider theorem does imply $c$ is in fact transcendental, hence irrational, so case 2 is the true one — but the proof above never uses that.) ∎

Prove by cases ( $n$ even / $n$ odd): every integer square is congruent to $0$ or $1$ modulo $4$ . Use this to show that $x^2 + y^2 = 3$ has no integer solutions.

n = 2k \Rightarrow n^2 = 4k^2 \equiv 0 \pmod 4; \qquad n = 2k+1 \Rightarrow n^2 = 4(k^2+k)+1 \equiv 1 \pmod 4

x^2 + y^2 \pmod 4 \in \{0, 1, 2\} \not\ni 3

(x, y) = (2, 2)

: case 1 (

x, y \ge 0

) —

|xy| = 4 = 2 \cdot 2 = |x|\,|y|

Show proof

Case 1: $n$ is even, $n = 2k$ ( $k \in \mathbb{Z}$ ). Then $n^2 = 4k^2$ , so $n^2 \equiv 0 \pmod 4$ .
Case 2: $n$ is odd, $n = 2k+1$ . Then $n^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$ , so $n^2 \equiv 1 \pmod 4$ . The two cases cover every integer with nothing left out, so every integer square is $0$ or $1$ modulo $4$ .
Now suppose $x^2 + y^2 = 3$ had an integer solution (proof by contradiction). Since $x^2 \pmod 4, y^2 \pmod 4 \in \{0, 1\}$ , their sum $x^2 + y^2 \pmod 4$ lies in $\{0, 1, 2\}$ : $0+0 = 0$ , $0+1 = 1$ , $1+1 = 2$ . But $3 \equiv 3 \pmod 4$ is not in this set — a contradiction. Hence $x^2 + y^2 = 3$ has no integer solutions. ∎