Prove that if and are even integers, where , , and are integers, then is even. What kind of proof did you use?
Assume for integers , , and , and are even integers. That means that there are some integers a and b where = 2a and .
Adding those two together, we get
We see written as times an integer (), we conclude is even
Use a direct proof to show that every odd integer is the difference of two squares.
Consider the difference of two consecutive squares and .
a
Prove that if is irrational, then is irrational.
We prove the contrapositive. We prove "If is rational, then is rational".
We know that since a fraction cannot be divisible by . If is rational then there are some integers and with . Since can't be (if it were, we'd have the contradiction by multiplying both sides by , we know that . Now
We see can be written as the quotient of two integers with the denominator nonzero. Thus by definition, is rational.
Show that at least ten of any days chosen must fall on the same day of the week.
Proof by contradiction. Suppose we pick days and no more than fall on the same day. This means that we have chosen at most days, which is a contradiction.
Use a proof by contradiction to show that there is no rational number for which . (Hint: Assume that is a root, where and are integers and is in lowest terms. Obtain an equation involving integers by multiplying by . Then look at whether and are each odd or even.)
Proof by contradiction. Suppose is a rational number that is a root of . That means can be expressed as the ratio of two integers and and the ratio is in lowest terms. Plugging this root into the equation we get . Multiplying by we get . Since and are in lowest terms, they both can't be even. No matter which the different possible variations of and s parity, the lefthand side of the equation must be odd, and so it can't equal 0. This is a contradiction and shows no such root exists.
Prove that if is a positive integer, then is odd if and only if is odd.
Scratch: Need to show and are true.
For the domain of positive integers..
Need to prove "If is odd then is odd" and "If is odd then is odd".
If is odd then there is some where . Plugging that value in to we get or or or . Since is an integer, we see that is in the form of an odd number.
To prove that "If is odd then is odd" we prove the contrapositive, which is "If is even then is even". Suppose is even so there is some where . So we plug that in and we get or . That can be written in the form of an even number .
Prove that at least one of the real numbers is greater than or equal to the average of these numbers. What kind of proof did you use?
Proof by contradiction. Suppose that are less than their average, denoted by . In symbols, we have for all . If we add these inequalities, we see that .
By definition of the average, we see . The two formulae have a contradiction: they imply . This is a contradiction, and we conclude that at least one of the numbers is greater than or equal to their average.