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s of the direct method in .NET Encoder ECC200 in .NET s of the direct method




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Examples of the direct method using barcode maker for visual studio .net control to generate, create ecc200 image in visual studio .net applications. Java We have already see gs1 datamatrix barcode for .NET n an example of the direct method in the proof of the if part of Theorem 18.1.

In this case we have a simple calculation to prove the statement. Another example is the following..

Theorem 20.1 Let m be an integer .NET datamatrix 2d barcode . If m is odd, then m2 is odd.

Proof. If m is odd, then m = 2r + 1 for some integer r. Then, m2 = (2r + 1)2 = (2r + 1)(2r + 1) = 4r 2 + 2r + 2r + 1 = 4r 2 + 4r + 1 = 2(2r 2 + 2r) + 1.

That is, m2 is odd. Recall the de nition of cardinality from 1 and examine the next theorem which is a solution to Problem (ii) on page 42 of 5..

Theorem 20.2 Let A and B be nite sets. Then A B = . A. + . B. A B . Apply the methods of 18 to the above. What happens to trivial examples, say A = What about extreme cases, such as A = B What about non-examples, for instance,.

CHAP T E R 20 Techniques of proof I: Direct method what happens when o ne or both of the sets are in nite Why is the theorem restricted to nite sets What are the assumptions and how good are they (Very good, since we only assume that the sets are nite.) How good is the conclusion (Very good, since it gives us an equation that allows us to calculate. Ways to calculate are always useful!) Can we draw a picture Yes we can because we can draw a Venn diagram.

Now let s look at the proof. Proof. If we count all the elements in A and then all the elements in B, then we will count the elements that are in A or are in B, but will count those that are in both A and B twice.

In the union of A and B, the elements that are in both are only counted once. Thus . A. + . B. A B counts elements that are in A or in B only once. Hence this is equal to A B . Let s try another gs1 datamatrix barcode for .NET simple (admittedly not particularly useful) example involving more steps and this time let us see the thinking behind how we might prove it.

. Theorem 20.3 Suppose that p Q VS .NET gs1 datamatrix barcode and p2 Z. Then, p Z.

What are our assumptions One is p Q. Well, we don t yet know many theorems about rational numbers, all we really know is the de nition, i.e.

we know p = a/b for some a, b Z, and we can assume that this is in its lowest form. The next assumption is that p2 Z. This means that (a/b)2 Z, i.

e. a 2 /b2 Z. This fraction is also in its lowest form.

But this can only happen if b = 1 because a 2 Z (as a Z). So we can see that p = a/( 1) = a Z. Let s write a polished version of the proof: Polished solution: Proof.

By assumption p = a/b for some integers a and b, where the fraction is in its lowest form. Thus p 2 = (a/b)2 = a 2/b2 . Since p2 Z and the fraction is also in its lowest form we have that b2 = 1.

Thus b = 1 and we can deduce that p = a/( 1) = a Z. Notice that we used the de nition of a rational number, i.e.

p = a/b. This is a good idea to apply when trying to nd a proof of a statement..

Theorem 20.4 Let m and n be real Visual Studio .NET DataMatrix numbers. If n > m > 0, then m+1 m > .

n+1 n One way to prove this would be to take the right-hand side to the left-hand side and prove something is bigger than 0. Instead we shall assume what had to be proved and work from there. This is a perfectly reasonable method of problem-solving.

It is not reasonable to use in the polished version, as we shall see later..
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