# Problem involving rational number

• Oct 14th 2010, 09:18 PM
hellfire127
Problem involving rational number[Solved]
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.(Headbang)
• Oct 14th 2010, 09:49 PM
undefined
Quote:

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.(Headbang)

Solve for x..

ax + b = cx + d

ax - cx = d - b

x (a - c) = d - b

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

which is rational..
• Oct 14th 2010, 09:53 PM
hellfire127
wow, you made it look easy. thank you.
• Oct 14th 2010, 10:25 PM
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.
• Oct 14th 2010, 10:36 PM
undefined
Quote:

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.
• Oct 17th 2010, 09:39 PM
hellfire127
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?
• Oct 18th 2010, 01:17 AM
emakarov
Quote:

All n=integer, there does not exist a k=integer such that n^2 - 2=4k.
This should be written either in symbolic form: $\displaystyle \forall n:\text{integer}\,\neg\exists k:\text{integer}.\,n^2-2=4k$ or in words: "For all integers n, there does not exist an integer k such that $\displaystyle n^2 - 2=4k$". 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 $\displaystyle n\in\mathbb{Z}$ where $\displaystyle \mathbb{Z}$ 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 $\displaystyle n^2 - 2=4k$. 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] :)
• Oct 18th 2010, 08:02 PM
hellfire127
Quote:

Originally Posted by emakarov
This should be written either in symbolic form: $\displaystyle \forall n:\text{integer}\,\neg\exists k:\text{integer}.\,n^2-2=4k$ or in words: "For all integers n, there does not exist an integer k such that $\displaystyle n^2 - 2=4k$". 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 $\displaystyle n\in\mathbb{Z}$ where $\displaystyle \mathbb{Z}$ 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 $\displaystyle n^2 - 2=4k$. 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.
• Oct 18th 2010, 11:01 PM
emakarov
If $\displaystyle n^2 - 2=4k$, then $\displaystyle n^2 = 2(2k + 1)$, so $\displaystyle n^2$ is even. This implies that $\displaystyle n$ is even. Indeed, if $\displaystyle n$ were odd, then $\displaystyle n = 2m + 1$ for some integer $\displaystyle m$. Then $\displaystyle n^2=2(2m^2+2m) + 1$, which is odd. Since $\displaystyle n$ is even, $\displaystyle n = 2i$ for some integer $\displaystyle i$. Therefore, $\displaystyle n^2=4i^2$, i.e., $\displaystyle n^2$ is divisible by 4. On the other hand, the first equation of this post implies that $\displaystyle n^2 = 4k+2$, i.e., $\displaystyle n^2$ produces the remainder 2 when divided by 4. A contradiction. Therefore, $\displaystyle n^2 - 2=4k$ is false.