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Exercise 27.14 in .NET Insert 2d Data Matrix barcode in .NET Exercise 27.14




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Exercise 27.14 using .net toaccess ecc200 in asp.net web,windows application isbn 13 Show that x 2 + 9x + 20 is divisible by 2 for all x Z. There exist an in nite number of primes The next theorem is another classic of mathematics. The rst known proof was by Euclid. Theorem 27.15 The number of prime numbe .net vs 2010 datamatrix 2d barcode rs is in nite. Proof.

Assume the contrary statement: there is a nite number of primes, p1 to pn . The number p1 p2 . .

. pn + 1 is larger than all of p1 to pn so by assumption cannot be a prime. On the other hand, if we divide this number by pj for any j , then we get remainder 1.

Now suppose that this number is divisible by a non-prime less than it, say b. By the Fundamental Theorem of Arithmetic pi divides b for some pi and so pi divides. CHAP T E R 27 Divisors p1 p2 . . .

pn + 1, which .NET Data Matrix we have seen cannot be true. This means that p1 p2 .

. . pn + 1 is only divisible by itself and 1.

That is, it is prime. Hence, we have shown that if there is a nite number of primes, then p1 p2 . .

. pn + 1 is not a prime and is a prime. This is an obvious contradiction.

. Exercise 27.16 Analyse the above proof. How many times did we use contradiction Greatest common divisor De nition 27.17 The greatest common divis or of two non-zero integers a and b, denoted gcd(a, b), is the largest positive integer that divides both numbers. It is sometimes known as the greatest common factor (gcf) or highest common factor (hcf)..

Examples 27.18 (i) The positive divisors of 20 are 1, 2, 4, 5, 10, and 20. The positive divisors of 12 are 1, 2, 3, 4, 6 and 12. Therefore the greatest common divisor is 4.

(ii) The greatest common divisor of 65 and 36 is 1.. Exercise 27.19 Create your own examples and non-examples. Include some with negative numbers, e.g.

gcd(50, 35). Let s look at some basic properties of the gcd..

Theorem 27.20 For any integers a and b we have (i) gcd(a, b) = gcd(b, a), (ii) gcd(a, b) 1, (iii) gcd(a, b) = gcd(. a. , . b. ), a b (iv) gcd , = 1, gc d(a, b) gcd(a, b) (v) gcd(a, b) = gcd(a + nb, b) for all integers n. Proof. (i) This is trivial.

We just use the de nition. (ii) The gcd is de ned to be a positive integer so it must be greater than or equal to 1. (iii) This is a simple exercise.

(iv) Let s begin by giving gcd(a, b) a name. Let gcd(a, b) = d. So we have a = rd and b = sd for some integers r and s as the greatest common divisor is obviously a divisor.

Hence, gcd a b , gcd(a, b) gcd(a, b) = gcd rd sd , d d = gcd(r, s).. A common mistake 193 Let s give this gcd a nam e too. Let gcd(r, s) = d . Then r = pd and b = qd for some p and q.

So we know that a = pd d and b = qd d. This though means that d d is a divisor of a and b. If d is greater than 1, then this would contradict that d is the greatest common divisor of a and b as d d would be greater than d.

(v) This involves saying that the greatest common divisors of a, b and the sum a + nb are connected, so in particular says something about their divisors. Hence I ask myself what theorems do I know that connect divisors and sums By looking back I notice that Theorem 27.5 has assumptions and conclusions involving a divisor and sums so I will use this and see what happens.

We have gcd(a, b). a and gcd(a, b). b by de nition of gcd. So gcd(a, b) divides a + nb by Theorem 27.5.

Hence gcd(a, b) is a common divisor for b and a + nb. It may not be the greatest though, therefore we have gcd(a, b) gcd(a + nb, b). This is not what we wanted to prove! We wanted an equality.

It s not a problem though since x = y is equivalent to x y and x y. Thus we need to prove gcd(a, b) gcd(a + nb, b). It seems reasonable that we can use an argument similar to the above.

The proof is almost identical: We have gcd(a + nb, b). a + nb and gcd(a + nb, b). b by de nition of gcd. So gcd(a + nb, b) divides (a + nb) + ( n)b by Theorem 27.5.

Hence gcd(a + nb, b) is a common divisor for b and a. It may not be the greatest, therefore gcd(a + nb, b) gcd(a, b). Hence, as gcd(a, b) gcd(a + nb, b) and gcd(a + nb, b) gcd(a, b), we have gcd(a + nb, b) = gcd(a, b).

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