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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-order|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.
| | It is very common to have a dental emergency -- a fractured tooth, an abscess, or severe pain when chewing. Over-the-counter pain medication is just masking the problem. Seeing an emergency dentist is critical to getting the source of the problem diagnosed and corrected as soon as possible.<br><br>Here are some common dental emergencies:<br>Toothache: The most common dental emergency. This generally means a badly decayed tooth. As the pain affects the tooth's nerve, treatment involves gently removing any debris lodged in the cavity being careful not to poke deep as this will cause severe pain if the nerve is touched. Next rinse vigorously with warm water. Then soak a small piece of cotton in oil of cloves and insert it in the cavity. This will give temporary relief until a dentist can be reached.<br><br>At times the pain may have a more obscure location such as decay under an old filling. As this can be only corrected by a dentist there are two things you can do to help the pain. Administer a pain pill (aspirin or some other analgesic) internally or dissolve a tablet in a half glass (4 oz) of warm water holding it in the mouth for several minutes before spitting it out. DO NOT PLACE A WHOLE TABLET OR ANY PART OF IT IN THE TOOTH OR AGAINST THE SOFT GUM TISSUE AS IT WILL RESULT IN A NASTY BURN.<br><br>Swollen Jaw: This may be caused by several conditions the most probable being an abscessed tooth. In any case the treatment should be to reduce pain and swelling. An ice pack held on the outside of the jaw, (ten minutes on and ten minutes off) will take care of both. If this does not control the pain, an analgesic tablet can be given every four hours.<br><br>Other Oral Injuries: Broken teeth, cut lips, bitten tongue or lips if severe means a trip to a dentist as soon as possible. In the mean time rinse the mouth with warm water and place cold compression the face opposite the injury. If there is a lot of bleeding, apply direct pressure to the bleeding area. If bleeding does not stop get patient to the emergency room of a hospital as stitches may be necessary.<br><br>Prolonged Bleeding Following Extraction: Place a gauze pad or better still a moistened tea bag over the socket and have the patient bite down gently on it for 30 to 45 minutes. The tannic acid in the tea seeps into the tissues and often helps stop the bleeding. If bleeding continues after two hours, call the dentist or take patient to the emergency room of the nearest hospital.<br><br>Broken Jaw: If you suspect the patient's jaw is broken, bring the upper and lower teeth together. Put a necktie, handkerchief or towel under the chin, tying it over the head to immobilize the jaw until you can get the patient to a dentist or the emergency room of a hospital.<br><br>Painful Erupting Tooth: In young children teething pain can come from a loose baby tooth or from an erupting permanent tooth. Some relief can be given by crushing a little ice and wrapping it in gauze or a clean piece of cloth and putting it directly on the tooth or gum tissue where it hurts. The numbing effect of the cold, along with an appropriate dose of aspirin, usually provides temporary relief.<br><br>In young adults, an erupting 3rd molar (Wisdom tooth), especially if it is impacted, can cause the jaw to swell and be quite painful. Often the gum around the tooth will show signs of infection. Temporary relief can be had by giving aspirin or some other painkiller and by dissolving an aspirin in half a glass of warm water and holding this solution in the mouth over the sore gum. AGAIN DO NOT PLACE A TABLET DIRECTLY OVER THE GUM OR CHEEK OR USE THE ASPIRIN SOLUTION ANY STRONGER THAN RECOMMENDED TO PREVENT BURNING THE TISSUE. The swelling of the jaw can be reduced by using an ice pack on the outside of the face at intervals of ten minutes on and ten minutes off.<br><br>For those who have virtually any queries concerning exactly where in addition to the best way to utilize [http://www.youtube.com/watch?v=90z1mmiwNS8 Best Dentists in DC], you'll be able to call us on the web-site. |
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| 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]].
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| 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.
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| The method of infinite descent was developed by and much used for [[Diophantine equation]]s by [[Fermat]].<ref>{{citation | last = Weil | first = André | author-link = André Weil | title = Number Theory: An approach through history from Hammurapi to Legendre | publisher = [[Birkhäuser Verlag|Birkhäuser]] | year = 1984 | pages = 75–79 | isbn = 0-8176-3141-0}}</ref> Two typical examples are showing the non-solvability of the Diophantine equation ''r''<sup>2</sup> + ''s''<sup>4</sup> = ''t''<sup>4</sup> and proving [[Fermat's theorem on sums of two squares]], which states that any prime ''p'' such that ''p'' ≡ 1 ([[Modular arithmetic|mod]] 4) can be expressed as a sum of two [[square number|squares]] (see [[Proofs of Fermat's theorem on sums of two squares#Euler.27s_proof_by_infinite_descent|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 point]]s on an elliptic curve ''E'' by [[Inversive geometry|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]]).
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| == Number theory==
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| 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-function]]s. 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''/2''E'' in Fermat's style.
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| 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 [[Glossary of arithmetic and Diophantine geometry#Height function|height function]] – a concept that became foundational. To show that ''A''(''Q'')/2''A''(''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.
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| == Application examples ==
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| === Irrationality of √2 ===
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| The proof that the [[square root of 2]] (√2) is [[irrational number|irrational]] (i.e. cannot be expressed as a fraction of two whole numbers) was discovered by the [[ancient Greek]]s, and is perhaps the earliest known example of a proof by infinite descent. [[Pythagoreanism|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 number|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.<ref>Stephanie J. Morris, [http://jwilson.coe.uga.edu/emt669/student.folders/morris.stephanie/emt.669/essay.1/pythagorean.html "The Pythagorean Theorem"], Dept. of Math. Ed., [[University of Georgia]].</ref><ref>Brian Clegg, [http://nrich.maths.org/2671 "The Dangerous Ratio ..."], Nrich.org, November 2004.</ref><ref>Kurt von Fritz, [http://www.jstor.org/pss/1969021 "The discovery of incommensurability by Hippasus of Metapontum"], Annals of Mathematics, 1945.</ref> The square root of two is occasionally called "Pythagoras' number" or "Pythagoras' Constant", for example {{harvtxt|Conway|Guy|1996}}.<ref>{{citation
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| | last1 = Conway | first1 = John H. | author1-link = John H. Conway
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| | last2 = Guy | first2 = Richard K. | author2-link = Richard K. Guy
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| | page = 25
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| | publisher = Copernicus
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| | title = The Book of Numbers
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| | year = 1996}}</ref>
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| The [[ancient Greek]]s, not having [[algebra]], worked out a [[Square_root_of_2#Geometric proof|geometric proof]] by infinite descent ([[John Horton Conway]] presented another [http://www.cut-the-knot.org/proofs/sq_root.shtml geometric proof (no. 8 ' ' ' )] by infinite descent that may be more accessible). The following is an [[algebra]]ic proof along similar lines:-
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| Suppose that √2 were [[rational number|rational]]. Then it could be written as
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| :<math>\sqrt{2} = \frac{p}{q}</math>
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| for two natural numbers, {{math|''p''}} and {{math|''q''}}. Then squaring would give
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| :<math>2 = \frac{p^2}{q^2}, </math>
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| :<math>2q^2 = p^2, \, </math>
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| so 2 must be a factor of ''p''<sup>2</sup>, 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 ''p''<sup>2</sup>. But since ''p''<sup>2</sup> ''has'' been found to be divisible by 2, ''p'' must be divisible by 2 as well.)
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| As 2 is a factor of ''p'', we can now express p as 2 x some number ''r''; thus
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| :<math>p=2r</math>
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| But then
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| :<math>2q^2 = (2r)^2 = 4r^2, \, </math> | |
| :<math>q^2 = 2r^2, \, </math>
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| so 2 must be a factor of ''q''<sup>2</sup>, 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 [[Well-ordering principle|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.
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| (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).
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| === Irrationality of √''k'' if it is not an integer ===
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| For positive integer ''k'', suppose that √k is not an integer, but is rational and can be expressed as <sup>''m''</sup>⁄<sub>''n''</sub> for natural numbers ''m'' and ''n'', and let ''q'' be the largest integer no greater than √k. Then
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| :<math>\begin{align}
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| \sqrt k&=\frac mn\\[8pt] &=\frac{m(\sqrt k-q)}{n(\sqrt k-q)}\\[8pt]
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| &=\frac{m\sqrt k-mq}{n\sqrt k-nq}\\[8pt] &=\frac{nk-mq}{m-nq}
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| \end{align}</math>
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| 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.<ref>{{Citation | last = Sagher | first = Yoram |date=February 1988 | journal = [[American Mathematical Monthly]] | volume = 95 | page = 117 | title = What Pythagoras could have done}}</ref>
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| ===Non-solvability of ''r''<sup>2</sup> + ''s''<sup>4</sup> = ''t''<sup>4</sup>===
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| The non-solvability of <math>r^2 + s^4 =t^4</math> in integers is sufficient to show the non-solvability of <math>q^4 + s^4 =t^4</math> 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:<ref>Dolan, Stan, "Fermat's method of ''descente infinie''", ''[[Mathematical Gazette]]'' 95, July 2011, 269–271.</ref> | |
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| 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 <math>x=2ab,</math> <math>y=a^2-b^2,</math> <math>z=a^2+b^2</math>, with ''a'' and ''b'' [[coprime|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:
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| *'''''y'' and ''z''''': Neither ''y'' nor ''z'', being odd, can be twice a square; if they are both square, the right triangle with legs <math>\sqrt{yz}</math> and <math>b^2</math> and hypotenuse <math>a^2</math> also would have integer sides including a square leg (<math>b^2</math>) and a square hypotenuse (<math>a^2</math>), and would have a smaller hypotenuse (<math>a^2</math> compared to <math>z=a^2+b^2</math>).
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| *'''''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 <math>b</math> and <math>\sqrt{y}</math> and hypotenuse <math>a</math> 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 (<math>a</math> compared to <math>z=a^2+b^2</math>).
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| *'''''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 <math>a</math> and <math>b</math> and hypotenuse <math>\sqrt{z}</math> also would have two sides (<math>a</math> and <math>b</math>) each of which is a square or twice a square, and a smaller hypotenuse (<math>\sqrt{z}</math> compared to {{nowrap|<math>z</math>)}}.
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| 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 <math>r^2 + s^4 =t^4</math> cannot have a solution, since if it did then ''r'', ''s<sup>2</sup>'', and ''t<sup>2</sup>'' would be the sides of such a Pythagorean triangle.
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| For other proofs of this by infinite descent, see<ref>Grant, Mike, and Perella, Malcolm, "Descending to the irrational", ''Mathematical Gazette'' 83, July 1999, pp. 263–267.</ref> and.<ref>Barbara, Roy, "Fermat's last theorem in the case ''n'' = 4", ''Mathematical Gazette'' 91, July 2007, 260–262.</ref>
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| === Non-solvability of ''a''<sup>2</sup> + ''b''<sup>2</sup> = 3(''s''<sup>2</sup> + ''t''<sup>2</sup>) ===
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| Infinite descent can be used to show that there are no integer solutions to
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| :<math>a^2+b^2=3 \cdot (s^2+t^2)</math>,
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| other than <math>a=b=s=t=0</math>.
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| Suppose there is a nontrivial integer solution of the equation. Then there is a nontrivial nonnegative integer solution obtained by replacing each of <math>a,b,s,t</math> by its absolute value. So it suffices to show that there are no nontrivial nonnegative integer solutions.
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| Suppose that <math>a_1, b_1, s_1, t_1</math> is a nonnegative solution. We have
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| :<math> 3 \mid a_1^2+b_1^2 \, </math> | |
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| This is only true if both <math>a_1</math> and <math>b_1</math> are divisible by 3. Let
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| :<math>3 a_2 = a_1 \text{ and } 3 b_2 = b_1. \, </math>
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| Thus we have
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| :<math> (3 a_2)^2 + (3 b_2)^2 = 3 \cdot (s_1^2+t_1^2) \, </math>
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| and
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| :<math> 3(a_2^2+b_2^2) = s_1^2+t_1^2, \, </math>
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| which yields a new nontrivial nonnegative integer solution ''s''<sub>1</sub>, ''t''<sub>1</sub>, ''a''<sub>2</sub>, ''b''<sub>2</sub>. 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|date=June 2013}} 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]].
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| == See also ==
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| *[[Vieta jumping]]
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| ==References==
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| {{reflist}}
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| ==Other reading==
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| *{{PlanetMath|urlname=InfiniteDescent|title=Infinite descent}}
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| *{{PlanetMath|urlname=ExampleOfFermatsLastTheorem|title=Example of Fermat's last theorem}}
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| {{DEFAULTSORT:Infinite Descent}}
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| [[Category:Mathematical proofs]]
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| [[Category:Mathematical terminology]]
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| [[Category:Diophantine equations]]
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It is very common to have a dental emergency -- a fractured tooth, an abscess, or severe pain when chewing. Over-the-counter pain medication is just masking the problem. Seeing an emergency dentist is critical to getting the source of the problem diagnosed and corrected as soon as possible.
Here are some common dental emergencies:
Toothache: The most common dental emergency. This generally means a badly decayed tooth. As the pain affects the tooth's nerve, treatment involves gently removing any debris lodged in the cavity being careful not to poke deep as this will cause severe pain if the nerve is touched. Next rinse vigorously with warm water. Then soak a small piece of cotton in oil of cloves and insert it in the cavity. This will give temporary relief until a dentist can be reached.
At times the pain may have a more obscure location such as decay under an old filling. As this can be only corrected by a dentist there are two things you can do to help the pain. Administer a pain pill (aspirin or some other analgesic) internally or dissolve a tablet in a half glass (4 oz) of warm water holding it in the mouth for several minutes before spitting it out. DO NOT PLACE A WHOLE TABLET OR ANY PART OF IT IN THE TOOTH OR AGAINST THE SOFT GUM TISSUE AS IT WILL RESULT IN A NASTY BURN.
Swollen Jaw: This may be caused by several conditions the most probable being an abscessed tooth. In any case the treatment should be to reduce pain and swelling. An ice pack held on the outside of the jaw, (ten minutes on and ten minutes off) will take care of both. If this does not control the pain, an analgesic tablet can be given every four hours.
Other Oral Injuries: Broken teeth, cut lips, bitten tongue or lips if severe means a trip to a dentist as soon as possible. In the mean time rinse the mouth with warm water and place cold compression the face opposite the injury. If there is a lot of bleeding, apply direct pressure to the bleeding area. If bleeding does not stop get patient to the emergency room of a hospital as stitches may be necessary.
Prolonged Bleeding Following Extraction: Place a gauze pad or better still a moistened tea bag over the socket and have the patient bite down gently on it for 30 to 45 minutes. The tannic acid in the tea seeps into the tissues and often helps stop the bleeding. If bleeding continues after two hours, call the dentist or take patient to the emergency room of the nearest hospital.
Broken Jaw: If you suspect the patient's jaw is broken, bring the upper and lower teeth together. Put a necktie, handkerchief or towel under the chin, tying it over the head to immobilize the jaw until you can get the patient to a dentist or the emergency room of a hospital.
Painful Erupting Tooth: In young children teething pain can come from a loose baby tooth or from an erupting permanent tooth. Some relief can be given by crushing a little ice and wrapping it in gauze or a clean piece of cloth and putting it directly on the tooth or gum tissue where it hurts. The numbing effect of the cold, along with an appropriate dose of aspirin, usually provides temporary relief.
In young adults, an erupting 3rd molar (Wisdom tooth), especially if it is impacted, can cause the jaw to swell and be quite painful. Often the gum around the tooth will show signs of infection. Temporary relief can be had by giving aspirin or some other painkiller and by dissolving an aspirin in half a glass of warm water and holding this solution in the mouth over the sore gum. AGAIN DO NOT PLACE A TABLET DIRECTLY OVER THE GUM OR CHEEK OR USE THE ASPIRIN SOLUTION ANY STRONGER THAN RECOMMENDED TO PREVENT BURNING THE TISSUE. The swelling of the jaw can be reduced by using an ice pack on the outside of the face at intervals of ten minutes on and ten minutes off.
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