Thread: Problem involving rational number

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.

Originally Posted by hellfire127

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.

Solve for x..

ax + b = cx + d

ax - cx = d - b

x (a - c) = d - b

x = (d - b) / (a - c)

which is rational..

wow, you made it look easy. thank you.

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.

Originally Posted by hellfire127

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.

Proof by contradiction starts by assuming that the premise is true and the conclusion is false. You have done the opposite.

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.

All n=integer, there does not exist a k=integer such that n^2 - 2=4k. To proof this by contradiction I would assume that n=integer, k does not equal an integer, and n^2 - 2 does not equal 4k. What would be the best way to start it?

All n=integer, there does not exist a k=integer such that n^2 - 2=4k.

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.

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]

Originally Posted by emakarov

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.

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]

Ok, thanks for the rules, I'll keep that in mind. I see my mistake in the negation of the second part, but I still don't see how you can just state that n^2 is even. Should't have taken discrete math lol, I blame my adviser. Cal is so much easier.

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.