Hints to exercises
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 \[ \ldots -3, -2, -1, 0, 1, 2, 3, \ldots \] is a doubly infinite listing.
rat, rat, irr, irr, rat
Experiment!
What would it mean for this number to be rational? If we were to run into an element of the “see and say” sequence that is its own description, then from that point onward the sequence would get stuck repeating the same thing over and over (and the number whose digits are found by concatenating the elements of the ``see and say’’ sequence will enter into a repeating pattern.)
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). I would recommend the Sage computer algebra system because like this book it is free — you can download Sage and run it on your own system or you can try it out online without installing. Check it out at http://www.sagemath.org.
You can get sage to output \(\pi\) to high accuracy by typing
pi.N(digits=21)at the Sage \(>\) prompt.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: \[ \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}. \]
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 like Sage, or you can learn how to enter complex numbers on your graphing calculator. (On my TI-84, 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 next 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.
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.
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.
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 not-too-large number is to use the “table” feature of your graphing calculator. If you enter y1=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!
An even quicker way is to type
factor(2047)in Sage.Part of what makes the “prime-producing-power” 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.
How about \(a=2\cdot5\) and \(b=3\cdot7\). Now you come up with a different pair!
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 \(g(16)=2.\)
Four.
The chemical symbol for an element that is an exception is Hg which stands for “Hydro-argyrum” 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?
Write your own sentences containing four quantifiers. One sentence in which the quantifiers appear (\(\forall \exists \forall \exists\)) and another in which they appear (\(\exists \forall \exists \forall\)).
You’re on your own here. Be inventive!
Answers: yes, 0,4 and 8.
Even numbers have a zero in their units place. What digit must also be zero in a doubly-even 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 base-\(10\). 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 \[\begin{align*} 3267 & = 466 \cdot 7 + 5 \\ 466 & = 66 \cdot 7 + 4. \end{align*}\]
You do the rest.
One possibility is to mimic the result for base-10 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 \(\mathsf{A}\) is represented by \(1010\) in binary. (\(10_{10} = \mathsf{A}_{16} = 1010_{2}\)
Answers for the first three:
- \(757_8 = 111 101 111_2\)
- \(1007_8 = 001 000 000 111_2 = 0010 0000 0111_2 = 207_{16}\)
- \(100 101 010 110_2 = 4526_8\)
Answer for the first one: \(11110_2\)
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!
You have to try a bunch of examples. You should try to make sure the examples you try cover all the possibilities. The pairs that provide counterexamples (i.e. show the statement is false in general) are relatively sparse, so be systematic.
\(\pi^2 = 9.8696\)
I just can’t bring myself to spoil this one for you, you really need to work this out on your own.
The even numbered ones are 2, 1, 500000.
The even numbered ones are 1 and 1287. The TI-84 calculates binomial coefficients. The symbol used is
nCr(which is placed between the numbers – i.e. it is an infix operator). You getnCras the 3rd item in thePRBmenu underMATH. In Sage, the command isbinomial(n,k).You’re choosing three things out of a set of size seven.
r=27 q=0 r=27-5=22 q=0+1=1 r=22-5=17 q=1+1=2 r=17-5=12 q=2+1=3 r=12-5=7 q=3+1=4 r=7-5=2 q=4+1=5 return r is 2 and q is 5.
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 formula \[ \operatorname{lcm} (a, b) = \frac{ab}{\operatorname{gcd} (a, b).} \]
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 \(p^f\) ?}
The quotients you obtain should alternate between 1 and 2.
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 x2 was even but x itself was odd?}
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.
You can mostly just copy the argument for \(\sqrt{2}\).
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.)
\[ f = \{(0,0), (1,1), (2,4), (3,2), (4,2), (5,4), (6,1)\} \] and \[ \operatorname{Rng}(f) = \{0,1,2,4\}.\]
Less-than-or-equal-to
Yeah, hmmm. Forty is kind of a lot… Let’s look at the points (E,F,G and B) on the horizontal line in Figure 1.7. 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).
Figure 1.7: Star.
Is this the region above or below the curve \(y=x^2\)?
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?
Depending on your personal experience level with fruit there may be different answers. Certainly (orange, orange) will be one of the pairs, but (orange, green) happens too!