i+n ^ j+n <= i^j . in .NET

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i+n ^ j+n <= i^j .
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In addition, we may require the following property.
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In programming language jargon, we say that the division operator '/' is overloaded. The computation that is executed depends on the type of the arguments. When both its arguments are integers, the result is an integer; if both arguments are floating point values, the result is a floating point value.
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Integer division rounds towards 0. In particular, m+n^m/n <= O^ra A O ^ n .
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(We will not need to consider the case when either of m or n is negative.) The first two of these properties are properties that one can reasonably expect to hold of any implementation of integer division, independently of how one chooses to round the real division. The third property is a design choice. Other choices are to round away from zero, or round up, or round down, and other programming languages may choose differently. For this reason, we are careful to avoid using the property in our calculations wherever possible. See Exercise 4.4 for further evidence of why the exploitation of this property should be avoided. Now, recall that what we have to prove is I ^ (l+r-l)+2 < r This is the same as showing that <= 0 ^ I < Y ^ N .
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The proof of the first of these properties proceeds as follows. First, we begin with the property I ^ (i+r-l)-r2 and calculate a simpler formula that guarantees its truth (simpler in the sense of not involving integer division). { I = (2x0-2 (division by 2 is introduced here in order to eliminate it at the next step) }
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<= { division by 2 is monotonic }
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2x1 ^ l+r-l = = { l^r-l { I and r are integers } addition is monotonic }
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We have thus shown that I ^ U+r-l)-2
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So it follows that I ^ (l+r-l)+2
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as required. The calculational proof style used above was introduced in Section 3.3. Since this is the first significant application, let us briefly recall the main elements of the format. The calculation has four steps beginning with [ ^ U+r-l)-2 and ending with Each step asserts either an equality between two properties or that the upper property is true if ('<=') the lower property is true. A reason or hint why a step is valid is stated between braces. For example, the hint in the first step is that I and (2xiH2 are equal so that I ^ (J+r-l)-=-2isthesameas (2x[)-=-2 ^ (l+r-l)+2. The second and third hints state the property being used. The final hint states that the step is valid because I and r are integers. (It would not be valid if I and r were real values.) The fact that the four step calculation allows us to conclude that I ^ (l+r-l)+2 <= I <r ,
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i.e. that the first line follows from the last line, is a consequence of simple properties of equality. The second step asserts that (2xl) + 2 < (l+r-l)+2 <= 2x1 ^ l+r-l ,
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but the first step asserts that the left side of this proposition is equal to I ^ (l+r-l)+2, and the third and fourth steps assert that the right side is equal to I < r. So the conclusion is obtained from the second step by 'substitution of equals for equals' the replacement of subexpressions by equal subexpressions. Note that we use plain old '=' for equality of booleans (and not, for example, ' *'). To avoid ambiguities, we often use the symbol '=' for equality of booleans in in-line formulae. For example, the final step in the above proof uses the property7
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I <r-l = l<r .
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(Read this as: 'the boolean I ^ r-1 is equal to the boolean l<r\) Here, we need to distinguish between equality of booleans and equality of integers. It would be confusing to write
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