I can't seem to figure out a problem in one of my assignments. The question is:
Suppose a, b, c, and d are integers and a does not equal c. Suppose also that x is a real number that satisfies the equation
(ax+b)/(cx+d) = 1.
Must x be rational? If so, express x as a ratio of two integers.
So does that mean b can equal d? Confused on how to start it.
Instead of making a new thread, i have one more for the night which involves proofs. I can set it out but can't seem to get to the conclusion.
I need to proof this by contradiction:
If r>0 is irrational, then square root of r is irrational.
Assume r>0, r is rational, and sqrt of r is irrational.
Then sqrt of r is rational.
Assume r>0 is irrational and sqrt(r) is rational.
Consider (sqrt(r))^2. By closure of rational numbers, this number is rational. But it is equal to r, which is irrational. Contradiction.
This should be written either in symbolic form: or in words: "For all integers n, there does not exist an integer k such that ". In the first case, the notation "n : integer", as far as I know, comes from programming languages where it means "n is a variable of type integer". It is also possible to write where is the standard notation for the set of integers. Equality should not be used to write this.All n=integer, there does not exist a k=integer such that n^2 - 2=4k.
To prove this statement by contradiction, you assume that n is an integer and then assume the negation of what you have to prove, i.e., that there does exist an integer k such that . One way to continue is to note that in this case n^2 is even and then to use an auxiliary fact that if a square of an integer x is even (this happens only when x itself is even), then x^2 is divisible by 4. From here, a contradiction can be obtained.
Forum rules say, "9. Start a new thread for a new question. Don't tag a new question onto an existing thread. Otherwise the thread can become confusing and difficult to follow." I guess, another reason for this is that unless a new thread is opened, people don't know that there is a new question; they think that whoever answered your first question is better equipped to answer any follow-ups. Also, we love when a whole thread can be marked as [SOLVED]
If , then , so is even. This implies that is even. Indeed, if were odd, then for some integer . Then , which is odd. Since is even, for some integer . Therefore, , i.e., is divisible by 4. On the other hand, the first equation of this post implies that , i.e., produces the remainder 2 when divided by 4. A contradiction. Therefore, is false.