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 = rqps/sq.
Since rqps and sq are an integers and sq is not equal to 0, y is
rational.
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 Winter '10
 Staff
 Rational number, Rolle, DA

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