Fill in the blanks:
- p is a _________ number, and
- p is greater than _______.
You want to argue about the sum of two generic rational numbers. Maybe call them a/b and c/d. The definition of "rational number" then tells you that a, b, c and d are integers and that neither b nor d are zero. You add these generic rational numbers in the usual way -- put them over a common denominator and then add the numerators. A common denominator is bd, so we can express the sum as (ad+bc)/(bd). You can finish off the argument from here: you need to show that this expression for the sum satisfies the definition of a rational number (quotient of integers w/ non-zero denominator). Also, write it all up a bit more formally...
Hint: If we write x+y for the sum of two integers that is even (so x+y = 2k for some integer k), then we could subtract ____________ from it in order to obtain x-y. Once you fill in that blank properly the flow of the argument should become apparent to you.
You may be tempted to write "Since x is odd, it can be expressed as x = 2k+1 where k is an integer." This is slightly wrong since the variable k is already being used in the statement of the theorem. But, except for replacing k with some other variable (maybe m?) that is a good way to get started. From there it's really just algebra until, eventually, you'll find out what k really is...
evenness (n) = k ⇔ 2^{k} | n ∧ 2^{k+1} | n
State and prove a theorem concerning the evenness of products.
Well, the statement is that the evenness of a product is the sum of the evennesses of the factors...
This one is pretty straightforward. Be sure to not reuse any variables. Particularly, the fact that a | b tells us (because of the definition of divisibility) that there is an integer k such that b = ak. It is not okay to also use k when converting the statement "b | c."
(ax+b)/(cx+d) = 1.
Show that x is rational. Where is the hypothesis a ≠ c used?
Cross multiply and solve for x. If you need to divide by an expression, it had better be non-zero!
From the definition of divisibility, you get two integers j and k, such that a = jb and b = ka. Substitute one of those into the other and ask yourself what the resulting equation says about j and k. Can they be any old integers? Or, are there restrictions on their values?
A savings account where we are not depositing or withdrawing funds has a balance that is growing geometrically.
You don't need all the hypotheses. If a and c have different signs, then ac is a negative quantity.
This follows very easily by the method of working backwards from the conclusion. Remember that when multiplying or dividing both sides of an inequality by some number, the direction of the inequality may reverse (unless we know the number involved is positive). Also, remember that we can't divide by zero, so if we are (just for example, don't know why I'm mentioning it really...) dividing both sides of an inequality by x^{2} then we must treat the case where x=0 separately.
If you work backwards from the conclusion on this one, you should eventually come to the inequality (x-1)^{2} ≥ 0 . Notice that this inequality is always true -- all squares are non-negative. When you go to write-up your proof (writing things in the forward direction), you'll want to acknowledge this truth. Start with something like "Regardless of the value of x, the quantity (x-1)^{2} is greater than or equal to zero as it is a perfect square."
The best hint for this problem is simply to write down the contrapositive statement. It is trivial to prove!
The contrapositive is (p|x) ⇒ (p|x^{2}).
Well, if there was a largest integer -- let's call it L (for largest) -- then isn't L+1 an integer, and isn't it bigger?
Assume there was a smallest positive real number -- might as well call it s (for smallest) -- what can we do to produce an even smaller number? (But be careful that it needs to remain positive -- for instance s-1 won't work.)
Suppose that x is rational and y is irrational and there sum (let's call it z) is also rational. Do some algebra to solve for y, and you will see that y (which is, by presumption, irrational) is also the difference of two rational numbers (and hence, rational -- a contradiction.)
Hint: this result depends on knowing how squares behave mod 4. Odd squares are 1 mod 4. The sum of two odd squares is 2 mod 4 -- but even squares are 0 mod 4.
It sort of depends on what is meant by "a reasonably large range of inputs." For example the polynomial p(x) = 2x+1 gives primes three times in a row (at x=1,2 and 3).
The intent of the problem is that you find three numbers, a, b and c, that are all powers of 2 and such that a divides the product bc, but neither of the factors separately. For instance, if you pick a=16, then you would need to choose b and c so that 16 doesn't divide evenly into them (they would need to be less than 16...) but so that their product is divisible by 16.
1! = 1
2! - 1! = 1
3! - 2! + 1! = 5
4! - 3! + 2! - 1! = 19
et cetera
Are they all prime? (After the first two 1's.)
Here's some Maple code that would test this conjecture:
n:=1;
for i from 2 to 8 do
n := i! - n;
ifactors(n);
end do;
Of course it turns out that going out to 8 isn't quite far enough...
I would definitely seek help at your friendly neighborhood CAS.
Hint: Both √(2) and -√(2) are irrational numbers
See previous.
Is the inverse, 1/x, of an irrational number x rational?
See previous.
Hint: List all of the divisors of 36 = (2⋅3)^{2}. See if any of them are bigger than 6.
It is probably obvious that the "cases" will be the possible remainders mod 6. Numbers of the form 6q+0 will be multiples of 6, so clearly not prime. The other forms that need to be eliminated are 6q+2, 6q+3, and 6q+4.
The sum is n + (n+1) + (n+2).
Show that there are no 2-digit vampire numbers.
Show that there are seven 4-digit vampire numbers.
Note that 15 = 3^{2} + 2^{2} + 1^{2} + 1^{2}. Also, if 15 were expressible as a sum of fewer than 4 squares, the squares involved would be 1, 4 and 9. It's really not that hard to try all the possibilities.
You should really enlist a CAS (unless you're feeling masochistic...)
By trichotomy, x is either zero, negative, or positive. If x is zero, its square is zero. If x is negative, its square is positive. If x is positive, its square is also positive.
Can you say "Pythagorean triple"? I thought you could.
Hint: 6^{3} can be expressed as such a sum.
How about even integers? Is there a smallest one?
Consider the set {1, 1/2, 1/4, 1/8, ...}. Does it have a smallest element?
Still weaseling...
Unique existence proofs consist of two parts. First, just show existence. Then, show that if there were two of the things under consideration they must be equal.
If at first you don't succeed...
try googling "scissor paper rock lizard spock."