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means a+b = a+c and a+c = b and ]o-c (i.e. the conjunction of three equalities). This is quite different from
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(a+b = a+c) = (b = c)
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which means that the boolean a+b = a+c is equal to the boolean b = c. Later, we introduce another symbol for equality between booleans (symbol '=') partly to avoid confusion with continued equalities, but also for more significant reasons. Squaring of positive numbers is also 'invertible' with respect to the three ordering relations. This means that, whatever X is among the three relations, we have, for all numbers a, b and c, that the boolean a2 X b2 is equal to the boolean a X b. That is, for all positive numbers a and b, (a2Xb2) For example, (a2>b2) = (a>b) . = (aXb) . (3.3)
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The two properties (3.2) and (3.3) are all we need to know to calculate the relationship X between V3 + A/IT and \/5 + ^. Here is how it goes. It is a sequence of equalities, just like a calculation in algebra, in which the equality between the first and last terms is the result of the calculation. To help you understand the calculation, a hint has been added at each step, indicating the property that is being applied. (The parentheses needed to disambiguate in-line formulae are no longer needed because the layout of the calculations makes clear what is intended.)
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X V5 + v 7
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squaring is invertible with respect to X: (3.3) }
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3.2 Construction versus Verification (V3 + X/TT)2 X (75 +A/7) 2 = { arithmetic }
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14 + 2733 X 12 + 2V3I { = = = = {
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addition is invertible with respect to X: (3.2) } squaring is invertible with respect to X: (3.3) } arithmetic } addition is invertible with respect to X: (3.2) } squaring is invertible with respect to X: (3.3) and arithmetic }
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2 + 2V3I X 2v^ (2 + 2T33) X (2VJ5) 2 { { 8V33 X 4 { 136 + Sx/33 X 140
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2112 X 16 . Summarizing, we have established that
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(V3 + VUX A/5 + V7) = ( 2 1 1 2 X 1 6 ) . So, the boolean \/3 + \/TT < \/5 + \/7 is false, since 2112<16 is false; for the same reason, \/3 + vTT = \/5 + V is false. But, V3 + VTT > V5" + v/7 is true, since 2112 > 16. In this way, we have calculated that V3 + 7TT is greater than V5 + J7. Note that we could now repeat the calculation with X replaced everywhere by '>', adding the additional final step (2112>16)=true, as outlined below: LT > v^ + A/T { squaring is invertible with respect to X: (3.3) 2 vTT) > (V5 + V7) 2
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2112 > 16
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arithmetic }
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true . This would be a verification that \/3 + \/IT is greater than \/5 + \/7 because we start with the answer and verify its validity. A verification is undesirable because
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3: Calculation^ Proof
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it hides the process of discovery. Worse still would be to turn the calculation upside down:
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arithmetic }
2112 > 16
squaring is invertible with respect to X: (3.3) }
73 + vTT > V5 + V7 .
Now the process of discovery has been completely obliterated; a straightforward calculation has been turned into a piece of magic! Exercise 3.4. The arithmetic in the calculation could have been made simpler if 2 + 2N/33 X 2^ had been simplified to 1 + V33 X V35. State the rule that allows this simplification to be made. (Take care with positive and negative numbers.) D Exercise 3.5. Determine the ordering relation between \/3 + vT3 and \/5 + vTT. Try to copy the style of calculation used above. D Exercise 3.6. Below is an algorithm to determine the ordering relation between Ja + ^ and Jc + V5 for given natural numbers a, b, c and d. What is wrong with the algorithm Construct a counterexample to demonstrate the error. (Hint: examine each step to see whether it makes a valid use of a property of squaring or addition. Also, note that counterexamples do not need to be complicated!) Identify a suitable invariant property to be maintained by the algorithm in order to avoid the error; using the invariant, modify the algorithm so that it is correct. We refer to V# + Jb and ^c + \fd as the left and right sides of the relation, respectively. Step 1. Square the left side and simplify it to the form u + ^Jv. (For example, (V3 + VT3)2 is simplified to 16 + 2\/39.) Similarly, square and simplify the right side to the form x + ^/y. Step 2. Subtract u from both sides and simplify. The left side is now in the form Jv and the right side is in the form z + Jy. Step 3. Square both sides again. Simplify the right side to the form p + ^fq. The left side is simplified to v. Step 4. Subtract p from both sides and square again.