N  Z  Q  R  C  

17  
$\pi$  
22/7  
6  
e^{0}  
1+i  
$\sqrt{3}$  
i^{2} 
Note that these sets contain one another, so if you determine that a number is a natural number it is automatically an integer and a rational number and a real number and a complex number...
What the heck is meant by a "singly infinite listing "? To help you figure this out, note that... 3, 2, 1, 0, 1, 2, 3, ... is a doubly infinite listing.
If a decimal expansion terminates after, say, k digits, can you figure out how to produce an integer from that number? Think about multiplying by something...
A calculator will generally be inadequate for this
problem. You should try using a CAS (Computer Algebra System). In
the computer labs at SCSU you have access to Maple, a CAS from a
Canadian company called Waterloo Maple. Try searching Maple's help
facility for the variable DIGITS
.
You really need to actually sit down and do some long division problems. When in the process do you suddenly realize that the digits are going to repeat? Must this decision point always occur? Why?
Take for granted that the usual rule for multiplying two fractions is okay to use:(a/b)*(c/d) = (ac)/(bd). How do you know that the result is actually a rational number?
These are straightforward. If you really must verify these somehow, you can go to a CAS or you can learn how to enter complex numbers on your graphing calculator. (On my TI84, you get i by hitting the 2nd key and then the decimal point.)
This is really easy, but be sure to do it generically. In other words, don't just use examples  do the calculation with variables for the real and imaginary parts of the complex number.
Divide out the obvious factors in order to reduce the complexity of the remaining problem. The first number is divisible by 5. The last three are all even. Recall that a number is divisible by 3 if and only if the sum of its digits is divisible by 3.
1  2  3  4  5  6  7  8  9  10 
11  12  13  14  15  16  17  18  19  20 
21  22  23  24  25  26  27  28  29  30 
31  32  33  34  35  36  37  38  39  40 
41  42  43  44  45  46  47  48  49  50 
51  52  53  54  55  56  57  58  59  60 
61  62  63  64  65  66  67  68  69  70 
71  72  73  74  75  76  77  78  79  80 
81  82  83  84  85  86  87  88  89  90 
91  92  93  94  95  96  97  98  99  100 
The primes used in this instance of the sieve are
just 2, 3, 5 and 7. Any number less than 100 that isn't a multiple of
2, 3, 5 or 7 will not be crossed off during the sieving process.
If you're still unclear about the process, try a web search for
"Sieve of Eratosthenes" +applet
, there are several
interactive applets that will help you to understand how to sieve.
Remember that if a number factors into two multiplicands, the smaller of them will be less than the square root of the original number.
p  2^{p}1  prime?  factors 

2  3  yes  1 and 3 
3  7  yes  1 and 7 
5  31  yes  
7  127  
11 
You'll need to determine if 2^{11}1 = 2047 is prime or not. If you never figured out how to read the table of primes on page 15, here's a hint: If 2047 was a prime there would be a 7 in the cell at row 20, column 4.A quick way to find the factors of a nottoolarge number is to use the "table" feature of your graphing calculator. If you enter
y_{1}=2047/X
and select the table view (2ND GRAPH
). Now, just scan down the entries until you find one with nothing after the decimal point. That's an X that evenly divides 2047!
It might be helpful to write down a bunch of examples. Think about how the prime factorization of a number gets transformed when we square it.
Part of what makes the "primeproducingpower" of that polynomial impressive is that it gives each prime twice  once on the descending arm of the parabola and once on the ascending arm. In other words, the polynomial gives prime values on a set of contiguous natural numbers {0,1,2, ..., N} and the vertex of the parabola that is its graph lies dead in the middle of that range. You can figure out what N is by thinking about the other end of the range: (1)^{2} + 31 · (1) + 257 = 289 (289 is not a prime, you should recognize it as a perfect square.)
Well, we know that 6 really isn't a prime... Maybe its factors enter into this somehow...
It has to do with one of the numbers being divisible by 3. (Why is this forced to be the case?) If that number isn't actually 3, then you know it's composite.
If you don't like making graphs, a table of the values of g(n) would suffice. Note that we don't count sums twice that only differ by order. For example, 16 = 13+3 and 11+5 (and 5+11 and 3+13) but g(16)=2.
Well, I'll give you that there are four.
The chemical symbol for the element that is the exception is Hg which stands for "Hydroargyrum" it is also known as "liquid silver" or "quick silver".
Think about this: is there any way to (using a formula) find a number that lies in between two other numbers?
You're on your own here. Be inventive!
Yes. 0, 4 and 8.
Even numbers have a zero in their units place. What digit must also be zero in a doublyeven number's binary representation?
Eight is 10_{8}, nine is 11_{8}. The point of asking questions about 777, is that (in octal) 7 is the digit that is analogous to 9 in base10. Thus 777_{8} is something like 999_{10} in that the number following both of them is written 1000 (although 1000_{8} and 1000_{10} are certainly not equal!)
It is helpful to write something of the form n = qd+r at each stage. The first two stages should look like
3267 = 466·7 + 5
466 = 66·7 + 4
you do the rest...
One possibility is to mimic the result for base10 that an even number always ends in 0,2,4,6 or 8.
As a check, the tenth number after AB is B5.
The tenth hexadecimal number after FA is 104.


This is just counting in binary. Remember the sanity check that the hexadecimal digit A is represented by 1010 in binary. (10_{10} = A_{16} =1010_{2})
Answers for the first three:
757_{8} = 111 101 111_{2}
1007_{8} = 001 000 000 111_{2} = 0010 0000 0111_{2} = 207_{16}
100 101 010 110_{2} = 4526_{8}
Might this effect have something to do with 10 being just one bigger than 9 (a multiple of 3)?
Seven 50 pound bags would hold 350 pounds of sand. They'd also be able to handle 340 pounds!
I can't spoil this one for you, you really need to work this one out on your own.
The even numbered ones are 2, 1, 500000.
The even numbered ones are 1 and 1287. The TI84 calculates binomial coefficients. The symbol used is nCr (which is placed between the numbers  i.e. it is an infix operator). You get nCr as the 3rd item in the PRB menu under MATH.
You're choosing three things out of a set of size seven...
r=27
q=0
r=275=22
q=0+1=1
r=225=17
q=1+1=2
r=175=12
q=2+1=3
r=125=7
q=3+1=4
r=75=2
q=4+1=5
return r is 2 and q is 5.
a  b  gcd(a,b)  lcm(a,b) 

110  273  
105  42  
168  189 
For such small numbers you can just find their prime factorizations and use that, although it might be useful to practice your understanding of the Euclidean algorithm by tracing through it to find the gcd's and then using the formulalcm(a,b) = ab/gcd(a,b).
Suppose that one number's prime factorization contains p^{e} and the other contains p^{f}, where e < f. What power of p will divide both, p^{e} or p^{f} ?
You're on you own here.
One approach is to truncate a decimal approximation and then rationalize. E.g. $\sqrt{2}$ is approximately 1.4142, so 14142/10000 isn't a bad approximator (although naturally 7071/5000 is better since it involves smaller numbers).
Does the rule about rational numbers having terminating or repeating decimal representations carry over to other bases?
What if the lemma wasn't true? Can you work out what it would mean if we had a number x such that x^{2} was even but x itself was odd?
Hint: Saying "x is even" is the same thing as saying "x is evenly divisible by 2". Replace the 2 by p and you're halfway there...
Yeah, do that!
A pair is "in" the relation when the first number gazinta the second number. 1 gazinta anything, 2 gazinta the even numbers, 3 gazinta 3, 6 and 9, etc. (Also a number always gazinta itself.)
{(0,0), (1,1), (2,4), (3,2), (4,2), (5,4), (6,1)}
{0,1,2,4}
(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)  (1,7)  (1,8)  (1,9)  (1,10) 
(2,2)  (2,3)  (2,4)  (2,5)  (2,6)  (2,7)  (2,8)  (2,9)  (2,10)  
(3,3)  (3,4)  (3,5)  (3,6)  (3,7)  (3,8)  (3,9)  (3,10)  
(4,4)  (4,5)  (4,6)  (4,7)  (4,8)  (4,9)  (4,10)  
(5,5)  (5,6)  (5,7)  (5,8)  (5,9)  (5,10)  
(6,6)  (6,7)  (6,8)  (6,9)  (6,10)  
(7,7)  (7,8)  (7,9)  (7,10)  
(8,8)  (8,9)  (8,10)  
(9,9)  (9,10)  
(10,10) 
I'd say it's lessthanorequalto.
Yeah, hmmm. Forty is kind of a lot...Let's look at the points (E,F,G and B) on the horizontal line in the diagram below. The triples involving these four points are: (E,F,G), (G,F,E), (E,F,B), (B,F,E), (E,G,B), (B,G,E), (F,G,B), (B,G,F).
If f sends x to y, then we want f^{1} to send y back to x. So the inverse just has the pairs in f reversed. When is the inverse going to fail to be a function?