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Assignment 4

# Assignment 4 - Assignment#4 Section 1.7(pg 64 1a)Let n be...

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Assignment #4. Section 1.7 (pg. 64) 1a)Let n be an integer. Then n is either even or odd. If n is even, then there is an integer k so that n=2k. In this case, 5n^2+3n+1=2(10k^2+13k)+1. Since 10k^2+3k is an integer, 5n^2+3n+1 is odd. If n is odd there is an integer k such that n=2k+1, in which case 5n^2+3n+1=2(10k^2+13k+4)+1. Since 10k^2+13k+4 is an integer, 5n^2+3n+1 is odd. 1b) Prove that if x is an odd integer, then 2x^2+3x+4 is odd. If x is an odd integer, then x=2k+1 for some integer k. 4x^2+x+6 = 2(2k+1)^2+3(2k+1)+4 = 8x^2+14x+8+1 = 2( 4K^2 + 7K + 4) + 1 ==>odd. Therefore, if x is an odd integer, then 2x^2 + 3x + 4 is odd. 1c) Prove that the sum of 5 consecutive integers is always divisible by 5. Let x be an integers. Any five consecutive integers can then be respresented as x,x+1,x+2,x+3,x+4. Therefore the sum is x+(x+1)+(x+2)+(x+3)+(x+4)=5x+10=5(x+2) which is divisible by 5. 1e) Prove that between any two distinct rational numbers a,b with a < b there exists another rational number in between a and b. Suppose we have two distinct rational numbers, a and b. a = p/q for some integers p and q. b = n/m for some integers n and m. We can construct a rational number (a+b)/2. We know that (a+b)/2 must lie between a and b. We need to show (a+b)/2 is rational: a+b (p/q)+(n/m) (mp+nq)/(qm) (mp+nq) -----= ----------- = ------------ = --------==>rational number. 2 2 2 2qm Done. 3a) Suppose x is rational. Suppose x+y is rational. Then there exist integers pa nd q, q not equal to 0 such that x=p/q and integers r and s s not equal to 0 such that x+y=r/s. Then y=(x+y) -x = r/s -p/q = rq-ps/sq. Since rq-ps and sq are an integers and sq is not equal to 0, y is rational.

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