Proof by infinite descent

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In mathematics, a proof by infinite descent is a particular kind of proof by contradiction which relies on the facts that the natural numbers are well ordered and that there are only a finite number of them that are smaller than any given one. One typical application is to show that a given equation has no solutions.

Typically, one shows that if a solution to a problem existed, which in some sense was related to one or more natural numbers, it would necessarily imply that a second solution existed, which was related to one or more 'smaller' natural numbers. This in turn would imply a third solution related to smaller natural numbers, implying a fourth solution, therefore a fifth solution, and so on. However there cannot be an infinity of ever-smaller natural numbers, and therefore by mathematical induction (repeating the same step) the original premise—that any solution exists—must be incorrect. It is disproven because its logical outcome would require a contradiction.

An alternative way to express this is to assume one or more solutions or examples exists. Then there must be a smallest solution or example—a minimal counterexample. We then prove that if a smallest solution exists, it must imply the existence of a smaller solution (in some sense)—which again proves that the existence of any solution would lead to a contradiction.

The method of infinite descent was developed by Fermat, who often used it for Diophantine equations.[1] Two typical examples are showing the non-solvability of the Diophantine equation r2 + s4t4 and proving Fermat's theorem on sums of two squares, which states that any prime p such that p ≡ 1 (mod 4) can be expressed as a sum of two squares (see proof). In some cases, to a modern eye, what he was using was (in effect) the doubling mapping on an elliptic curve. More precisely, his method of infinite descent was an exploitation in particular of the possibility of halving rational points on an elliptic curve E by inversion of the doubling formulae. The context is of a hypothetical rational point on E with large co-ordinates. Doubling a point on E roughly doubles the length of the numbers required to write it (as number of digits): so that a 'halved' point is quite clearly smaller. In this way Fermat was able to show the non-existence of solutions in many cases of Diophantine equations of classical interest (for example, the problem of four perfect squares in arithmetic progression).

Number theory[edit]

In the number theory of the twentieth century, the infinite descent method was taken up again, and pushed to a point where it connected with the main thrust of algebraic number theory and the study of L-functions. The structural result of Mordell, that the rational points on an elliptic curve E form a finitely-generated abelian group, used an infinite descent argument based on E/2E in Fermat's style.

To extend this to the case of an abelian variety A, André Weil had to make more explicit the way of quantifying the size of a solution, by means of a height function – a concept that became foundational. To show that A(Q)/2A(Q) is finite, which is certainly a necessary condition for the finite generation of the group A(Q) of rational points of A, one must do calculations in what later was recognised as Galois cohomology. In this way, abstractly-defined cohomology groups in the theory become identified with descents in the tradition of Fermat. The Mordell–Weil theorem was at the start of what later became a very extensive theory.

Application examples[edit]

Irrationality of √2[edit]

The proof that the square root of 2 (√2) is irrational (i.e. cannot be expressed as a fraction of two whole numbers) was discovered by the ancient Greeks, and is perhaps the earliest known example of a proof by infinite descent. Pythagoreans discovered that the diagonal of a square is incommensurable with its side, or in modern language, that the square root of two is irrational. Little is known with certainty about the time or circumstances of this discovery, but the name of Hippasus of Metapontum is often mentioned. For a while, the Pythagoreans treated as an official secret the discovery that the square root of two is irrational, and, according to legend, Hippasus was murdered for divulging it.[2][3][4] The square root of two is occasionally called "Pythagoras' number" or "Pythagoras' Constant", for example Conway & Guy (1996).[5]

The ancient Greeks, not having algebra, worked out a geometric proof by infinite descent (John Horton Conway presented another geometric proof (no. 8 ' ' ' ) by infinite descent that may be more accessible). The following is an algebraic proof along similar lines:-

Suppose that √2 were rational. Then it could be written as

\sqrt{2} = \frac{p}{q}

for two natural numbers, p and q. Then squaring would give

2 = \frac{p^2}{q^2},
2q^2 = p^2, \,

so 2 must be a factor of p2, and therefore 2 must also be a factor of p itself (if 2 did not divide p, then the prime factorization of p (the product of its primes) would contain no 2's. So when one squares p by squaring all its factors, there still would be no 2's in the resulting prime factorization of p2. But since p2 has been found to be divisible by 2, p must be divisible by 2 as well.)

As 2 is a factor of p, we can now express p as 2 x some number r; thus

p=2r

But then

2q^2 = (2r)^2 = 4r^2, \,
q^2 = 2r^2, \,

so 2 must be a factor of q2, and therefore 2 must also be a factor of q itself, and q can be written as 2 x s for some whole number s (same reasoning as above). Therefore p/q can be written as (2 x r)/(2 x s), and we find that p and q are not the smallest natural numbers making √2: we can write √2 as r/s where r<p and s<q. Therefore if √2 could be written as a rational number, it could always be written as a natural number with smaller parts, which itself could be written with yet-smaller parts, ad infinitum. But this is impossible in the set of natural numbers. Since √2 is a real number, which can be either rational or irrational, the only option left is for √2 to be irrational.

(Alternatively, this proves that if √2 were rational, no "smallest" representation as a fraction could exist, as any attempt to find a "smallest" representation p/q would imply a smaller one existed, which is a similar contradiction).

Irrationality of √k if it is not an integer[edit]

For positive integer k, suppose that √k is not an integer, but is rational and can be expressed as mn for natural numbers m and n, and let q be the largest integer no greater than √k. Then

\begin{align}
\sqrt k&=\frac mn\\[8pt] &=\frac{m(\sqrt k-q)}{n(\sqrt k-q)}\\[8pt]
&=\frac{m\sqrt k-mq}{n\sqrt k-nq}\\[8pt] &=\frac{nk-mq}{m-nq} \text{ } (\text{replacing the first }m\text{ in the numerator with }n \sqrt k\text{ and }\sqrt k\text{ in the denominator with }m/n)
\end{align}

( The numerator and denominator were each multiplied by a positive expression less than 1, and then simplified independently, to show both products were still integers. Therefore, no matter what natural numbers m and n are used to express √k, there can always be smaller natural numbers m' <m and n' <n that have the same ratio. But infinite descent on the natural numbers is impossible, so this disproves the original assumption that √k can be expressed as a ratio of natural numbers.[6]

Non-solvability of r2 + s4 = t4[edit]

The non-solvability of r^2 + s^4 =t^4 in integers is sufficient to show the non-solvability of q^4 + s^4 =t^4 in integers, which is a special case of Fermat's Last Theorem, and the historical proofs of the latter proceeded by more broadly proving the former using infinite descent. The following more recent proof demonstrates both of these impossibilities by proving still more broadly that a Pythagorean triangle cannot have any two of its sides each either a square or twice a square, since there is no smallest such triangle:[7]

Suppose there exists such a Pythagorean triangle. Then it can be scaled down to give a primitive (i.e., with no common factors) Pythagorean triangle with the same property. Primitive Pythagorean triangles' sides can be written as x=2ab, y=a^2-b^2, z=a^2+b^2, with a and b relatively prime and with a+b odd and hence y and z both odd. There are three cases, depending on which two sides are postulated to each be a square or twice a square:

  • y and z: Neither y nor z, being odd, can be twice a square; if they are both square, the right triangle with legs \sqrt{yz} and b^2 and hypotenuse a^2 also would have integer sides including a square leg (b^2) and a square hypotenuse (a^2), and would have a smaller hypotenuse (a^2 compared to z=a^2+b^2).
  • y and x: If y is a square and x is a square or twice a square, then each of a and b is a square or twice a square and the integer right triangle with legs b and \sqrt{y} and hypotenuse a would have two sides (b and a) each of which is a square or twice a square, with a smaller hypotenuse than the original triangle (a compared to z=a^2+b^2).
  • z and x: If z is a square and x is a square or twice a square, again each of a and b is a square or twice a square and the integer right triangle with legs a and b and hypotenuse \sqrt{z} also would have two sides (a and b) each of which is a square or twice a square, and a smaller hypotenuse (\sqrt{z} compared to z).

In any of these cases, one Pythagorean triangle with two sides each of which is a square or twice a square has led to a smaller one, which in turn would lead to a smaller one, etc.; since such a sequence cannot go on infinitely, the original premise that such a triangle exists must be wrong. This implies that r^2 + s^4 =t^4 cannot have a solution, since if it did then r, s2, and t2 would be the sides of such a Pythagorean triangle.

For other proofs of this by infinite descent, see[8] and.[9]

Non-solvability of a2 + b2 = 3(s2 + t2)[edit]

Infinite descent can be used to show that there are no integer solutions to

a^2+b^2=3 \cdot (s^2+t^2),

other than a=b=s=t=0.

Suppose there is a nontrivial integer solution of the equation. Then there is a nontrivial nonnegative integer solution obtained by replacing each of a,b,s,t by its absolute value. So it suffices to show that there are no nontrivial nonnegative integer solutions.

Suppose that a_1, b_1, s_1, t_1 is a nonnegative solution. We have

 3 \mid a_1^2+b_1^2 \,

This is only true if both a_1 and b_1 are divisible by 3. Let

3 a_2 = a_1 \text{ and } 3 b_2 = b_1. \,

Thus we have

 (3 a_2)^2 + (3 b_2)^2 = 3 \cdot (s_1^2+t_1^2) \,

and

 3(a_2^2+b_2^2) = s_1^2+t_1^2, \,

which yields a new nontrivial nonnegative integer solution s1, t1, a2, b2. Under a suitable notion of size of the solutions, e.g. the sum of the four integers, this new solution is smaller than the original one.[citation needed] This process can be repeated infinitely, producing an infinite decreasing sequence of positive solution sizes. This is a contradiction, because no such sequence exists. This shows that there are no nonzero solutions for this Diophantine equation.

See also[edit]

References[edit]

  1. ^ Weil, André (1984), Number Theory: An approach through history from Hammurapi to Legendre, Birkhäuser, pp. 75–79, ISBN 0-8176-3141-0 
  2. ^ Stephanie J. Morris, "The Pythagorean Theorem", Dept. of Math. Ed., University of Georgia.
  3. ^ Brian Clegg, "The Dangerous Ratio ...", Nrich.org, November 2004.
  4. ^ Kurt von Fritz, "The discovery of incommensurability by Hippasus of Metapontum", Annals of Mathematics, 1945.
  5. ^ Conway, John H.; Guy, Richard K. (1996), The Book of Numbers, Copernicus, p. 25 
  6. ^ Sagher, Yoram (February 1988), "What Pythagoras could have done", American Mathematical Monthly 95: 117 
  7. ^ Dolan, Stan, "Fermat's method of descente infinie", Mathematical Gazette 95, July 2011, 269–271.
  8. ^ Grant, Mike, and Perella, Malcolm, "Descending to the irrational", Mathematical Gazette 83, July 1999, pp. 263–267.
  9. ^ Barbara, Roy, "Fermat's last theorem in the case n = 4", Mathematical Gazette 91, July 2007, 260–262.

Other reading[edit]