M135F08A1S

M135F08A1S - MATH 135 Assignment #1 Solutions Hand-In...

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MATH 135 Fall 2008 Assignment #1 Solutions Hand-In Problems 1. Counterexample If n = 20, then n 2 + ( n + 1) 2 = 20 2 + 21 2 = 400 + 441 = 841 which is a perfect square, so the given statement is FALSE. How did we find this? We could find this by trying positive integer values starting at n = 4 until we obtained a value for n 2 + ( n + 1) 2 that was a perfect square. Alternatively, we might remember that (20 , 21 , 29) forms a Pythagorean triple, because 20 2 + 21 2 = 29 2 . 2. (a) Treating the given equation as a quadratic equation in x with coefficients in terms of y , 4 x 2 + x (4 y - 4) + (2 y 2 - 2 y + 1) = 0 Using the quadratic formula, we get x = - (4 y - 4) ± p (4 y - 4) 2 - 4(4)(2 y 2 - 2 y + 1) 8 = - (4 y - 4) ± p - 16 y 2 8 = - ( y - 1) ± p - y 2 2 (b) Counterexample If ( x,y ) = ( 1 2 , 0 ) , 4 x 2 +4 xy +2 y 2 - 4 x - 2 y +1 = 4 ( 1 2 ) 2 +4 · 1 2 · 0 - 2 · 0 2 - 4 · 1 2 - 2 · 0+1 = 0, so the given statement is FALSE. How did we find this? From (a), for a real value for x to exist, we must have - y 2 0, so y = 0. If y = 0, then x = 1 2 , so ( x,y ) = ( 1 2 , 0 ) makes the expression equal to 0. 3. LS = sin 4 x + 2 cos 2 x - cos 4 x = sin 4 x + 2(1 - sin 2 x ) - (1 - sin 2 x ) 2 = sin 4 x + 2 - 2 sin 2 x - (1 - 2 sin 2 x + sin 4 x ) = 1 = RS as required. 4. Consider ( x 2 - 3 x ) - 10 = x 2 - 3 x - 10 = ( x - 5)( x + 2). Since 2 x 5, then x - 5 0 and x + 2 0, so ( x - 5)( x + 2) 0. Therefore, x 2 - 3 x - 10 0 or x 2 - 3 x 10. If x 2, then x - 1 1 and x - 2 0, so x 2 - 3 x = ( x 2 - 3 x + 2) - 2 = ( x - 1)( x - 2) - 2 1(0) - 2 = - 2 Thus, if 2 x 5 then - 2 x 2 - 3 x 10.
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5. Suppose that n has units digit 5. Then n = 10 k + 5 for some non-negative integer k . Thus, n 2 = (10 k + 5) 2 = 100 k 2 + 100 k + 25 = 100( k 2 + k ) + 25. In other words, n 2 is a multiple of 100 plus 25, so its last two digits are 25. 6. First, we factor to obtain n 3 - n = n ( n 2 - 1) = n ( n - 1)( n + 1). This tells us that n 3 - n is the product of three consecutive positive integers, namely n - 1, n and n + 1. Not all three of these integers can be odd, so at least one is even, so n 3 - n is even (ie. is a multple of 2). Also, at least one of n - 1, n , n + 1 must be a multiple of 3, since they are three consecutive integers. Thus, n 3 - n is a multiple of 3. Therefore, n 3 - n is a multiple of 2 and of 3, so is a multiple of 2(3) = 6, as required. (We will see later that it is important that 2 and 3 have no common factors larger than 1.) 7. (a) This “if and only if” statement is equivalent to “If x 2 + y 2 - 2 y = - 1, then x = 0 and y = 1.” and “If x = 0 and y = 1, then x 2 + y 2 - 2 y = - 1.” (b) “ Suppose that x = 0 and y = 1.
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M135F08A1S - MATH 135 Assignment #1 Solutions Hand-In...

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