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Eyad
 3 years ago
Note: This is NOT a question. This is a tutorial.
*How To Write Proof*
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Eyad
 3 years ago
Note: This is NOT a question. This is a tutorial. *How To Write Proof* ___________________

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Eyad
 3 years ago
Best ResponseYou've already chosen the best response.0All that is needed is some common sense and a basic understanding of a few trusted and easy to understand techniques. First we should know what is The Structure of a Proof : _______________________________________________________ The basic structure of a proof is just a series of statements, each one being either An assumption or A conclusion, clearly following from an assumption or previously proved result. A well written proof will flow. That is, the reader should feel as though they are being taken on a ride that takes them directly and inevitably to the desired conclusion without any distractions about irrelevant details. Each step should be clear or at least clearly justified. A good proof is easy to follow. [[[[[ Example_{1} ]]]]: The Irrationality of the Square Root of 2....... ___________________________________________________________________________ ==Before we begin the proof, let's recall a few definitions. A real number is called rational if it can be expressed as the ratio of two integers: p/q. The ancient Greeks thought that all numbers were rational. A number that is not rational would be called irrational. You probably believe that p is irrational. (It might surprise you that this is not easy to prove.) When the Greeks proved that the square root of 2 is not a rational number, the very foundations of arithmetic were called into question. This is one of the reasons that Greek geometry subsequently flourishedall numbers could be treated geometrically without reference to rationality. Another fact that we will need is the Fundamental Theorem of Arithmetic. This exciting sounding theorem is nothing more than the fact that every positive integer has a unique representation as a product of prime numbers. The technique of proof we will use is proof by contradiction . You do not need any specialized knowledge to understand what this means. It is very simple. We will assume that the square root of 2 is a rational number and then arrive at a contradiction. Make sure you understand every line of the proof. ===>Theorem:The square root of 2 is an irrational number. ===>Proof: Let's represent the square root of 2 by s. Then, by definition, s satisfies the equation :\[\Large s^2=2\] If s were a rational number, then we could write \[\Large s=\frac{p}{q}\]. where p and q where p and q are a pair of integers. In fact, by dividing out the common multiple if necessary, we may even assume p and q have no common multiple (other than 1). If we now substitute this into the first equation we obtain, after a little algebra, the equation :\[\Large p^2=2q^2\]. But now, by the Fundamental Theorem of Arithmetic, 2 must appear in the prime factorization of the number p2 (since it appears in the same number 2 q2). Since 2 itself is a prime number, 2 must then appear in the prime factorization of the number p. But then, 22 would appear in the prime factorization of p2, and hence in 2 q2. By dividing out a 2, it then appears that 2 is in the prime factorization of q2. Like before (with p2) we can now conclude 2 is a prime factor of q. But now we have p and q sharing a prime factor, namely 2. This violates our assumption above (see if you can find it) that p and q have no common multiple other than 1. [[[[ Example_{2} ]]]] : _______________________ Recall that a natural number is called composite if it is the product of other natural numbers all greater than 1. For example, the number 39481461 is composite since it is the product of 15489 and 2549. Theorem==> The number 100...01 (with 3n1 zeros where n is an integer larger then 0) is composite. Proof==> We can rewrite our number as \[\Large100...01=10^{3n} + 1\] where n is an integer larger than 0. Now use the identity \[\Large a^3 + b^3=(a+b)(a^2 ab+ b^2)\] with \[\Large a=10^n\] and b = 1, to get \[\Large (10^n)^3 + 1=(10^n + 1)(10^{2n}10^n + 1)\] We will be done once we have shown that both factors \[\Large (10^n+1),(10^{2n} 10^n+1)\] are greater than 1. In the first case, this is clear since \[\Large 10^n>0\] when n>0. In the second case, \[\Large 10^{2n}10^n=10^n(10^n1)>0\], when n>0.This completes the proof. ======================================================================================================================== Here is Some T!ps: ================== 1Writing multiple drafts for your proofs is not uncommon. Considering some homework sets will comprise 10 pages or more, you will want to make sure you got it right. 2The best thing about most proofs: they've already been proven, which means they are usually true! If you come to a conclusion that's different than what you were to prove, then you more than likely messed up somewhere. Just go back and carefully review each step. 3Proofs are difficult to learn to write. One excellent way to learn proofs is to study related theorems, and how those were proved. What looks like failure, but is more than you started with, is actually progress. It can inform the solution. 4There are thousands of "heuristics" or good ideas to try. Polya's book has two parts, a how to, and an encyclopedia of heuristics. 5Try to apply your proof to a case where it should fail, and see whether it actually does. For example, here's a possible proof that: The square root of a number (that means any number) tends to infinity as that number tends to infinity. ****"For all positive n, the square root of n+1 is greater than the square root of n. So if that is true as n increases, then its square root also increases; and as n tends to infinity, its square root tends to infinity for all n." (That might sound okay at first.) But, although the statement you are attempting to prove is true, the deduction is false. This proof should apply equally well to the arctan of n as it does to square root of n. Arctan of n+1 is always greater then arctan of n for all positive n. But arctan does not tend to infinity, it tends to pi/2. Instead, we prove it as follows. To prove something tends to infinity, we need that for all numbers M there exists a number N such that for all n bigger than N, the square root of n is bigger than M. There does exist such a number  it's M^2. This example also shows that you should carefully check the definition of the thing you are trying to prove. ================================================================== *_* : If U need any proof in mathematics ,Please Inform me and i will make a Simple One If I Can.Ty
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