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5 fix misstatement

Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the Disquisitiones Mathematicae. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."

The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and whether $p$ is or is not a quadratic residue modulo $q$. He actually casts his proof as if it were a proof by minimal counterexample, so he further assumes in some instances that the result does not hold (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $p$ q$is a quadratic residue modulo$q$). p$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a quadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first subcase itself breaks into four separate sub-subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ or $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."

On the one hand, the proof is pretty much the first proof that one might think to try when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because one is so intently staring at the beetles in the bark of the tree directly in front.

Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out).

4 grammar

Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the Disquisitiones Mathematicae. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."

The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and whether $p$ is or is not a quadratic residue modulo $q$. He actually casts his proof as if it were a proof by minimal counterexample, so he further assumes in some instances that the result does not hold (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $p$ is a quadratic residue modulo $q$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a quadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first subcase itself breaks into four separate sub-subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ nor or $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."

On the one hand, the proof is pretty much the first proof that one might think to try when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because one is so intently staring at the beetles in the bark of the tree directly in front.

Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out).

3 added 3 characters in body

Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the Disquisitiones Mathematicae. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."

The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and whether $p$ is or is not a quadratic residue modulo $q$. He actually casts his proof as if it were a proof by minimal counterexample, so he further assumes in some instances that the result does not hold (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $p$ is a quadratic residue modulo $q$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a quadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first case subcase itself breaks into four separate sub-subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ nor $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."

On the one hand, the proof is pretty much the first proof that one might think to try when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because one is so intently staring at the beetles in the bark of the tree directly in front.

Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out).

2 spelling, grammar; added 1 characters in body

Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the Disquisitiones Mathematicae. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."

The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and the ways in which whether $p$ is or is not a quadratic reciprocity might fail residue modulo $q$. He actually casts his proof as if it were a proof by minimal counterexample, so he further assumes in some instances that the result does not hold (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $p$ is a quadratic residue modulo $q$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a qauadratic quadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first case itself breaks into four separate *sub*-subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ nor $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."

On the one hand, the proof is pretty much the first proof that one might think upon to try when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because you are one is so intent on intently staring at the beetles in the bark of the tree directly in frontof you.

Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out).

1 [made Community Wiki]

Not from measure theory, alas, but the example that jumps to my mind is Gauss's first proof of Quadratic Reciprocity. It appears in the Disquisitiones Mathematicae. The proof occupies arts. 135 through 144 (five and a half pages in the English edition published by Springer); the proof is by strong induction on $q$ (when $p\lt q$). I don't recall who, but someone once called it a proof by "mathematical revulsion."

The proof is quite messy. Gauss argues by cases, considering the congruence classes of $p$ and $q$ modulo $4$, and the ways in which quadratic reciprocity might fail (e.g., for $p\equiv q\equiv 1 \pmod{4}$, either $p$ is a quadratic residue modulo $q$ and $q$ is not one modulo $p$; or $p$ is not a quadratic residue modulo $q$ and $p$ is a quadratic residue modulo $q$). They fall into eight cases, though some of those cases themselves break into subcases. For example, Gauss looks at the case when $p$ and $q$ are both congruent to $1$ modulo $4$, and $\pm p$ is not a residue modulo $q$; then he takes a prime $\ell\neq p$ less than $q$ for which $q$ is not a qauadratic residue, and considers the cases in which $\ell\equiv 1 \pmod{4}$ or $\ell\equiv 3 \pmod{4}$ separately; the first case itself breaks into four separate *sub*subcases: since $p\ell$ is a quadratic residue modulo $q$, it is the square of some even $e$; then he considers the case when $e$ is not divisible by either $p$ nor $\ell$, when it is divisible by $p$ but not $\ell$; when it is divisible by $\ell$ but not $p$; and when it is divisible by $\ell$ and $p$. And so on. By the time Gauss finally gets to the eighth and final case, he is clearly somewhat exhausted, writing merely "The demonstration is the same as in the preceding case."

On the one hand, the proof is pretty much the first proof that one might think upon when encountering the problem. But the different cases are just way too messy, and one quickly loses sight of the forest because you are so intent on staring at the beetles in the bark of the tree in front of you.

Plenty of other proofs would follow (including five more by Gauss), ranging from the clever to the almost magical (do this, do that, and oops, quadratic reciprocity falls out).
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