# Proving Rationality of Squares of 9th Day 2 ARO 2004/2005 Numbers

• MHB
• Smb
In summary: Qb+c(+Qab+bc+ac+b^2+b+c+d+e+f+g+h+i+j+k+l+m+n+o+p+q+r+s+t+u+v+w+x+y+z+w+x+y+z+0+1In summary, all ten of the nonzero reals given are rational. Squares of all of these numbers are also rational.
Smb
Ten mutually distinct non-zero reals are given such that for any two, either their sum or their product is rational. Prove that squares of all these numbers are rational.
I tried using 3 of those numbers - a, b and c. And I checked each of the possible situations but I'm not sure if my maths teacher is going to accept it.
The problem is from the All-Russian Olympiad 2004/2005 9th Day 2.

Last edited:
Is this a homework assignment? If so, I recommend you post as much working as possible.

Joppy said:
Is this a homework assignment? If so, I recommend you post as much working as possible.

I worked out the answer easily but the method is slow.
I'll write for the first one because I don't have much time now:
a+b(-Q
b+c(-Q
a+c(-Q
ab+bc+ac+b^2(-Q
ab+bc+ac+c^2(-Q
(b-c)(b+c)(-Q
b-c(- Q Analogically we can do for each pair of numbers.
b^2-2bc+c^2-B^2-2bc-c^2(-Q
=> bc (- Q Analogically we can do for each pair of numbers. =>
b^2(- Q Analogically we can do for each pair of numbers.

Smb said:
I worked out the answer easily but the method is slow.
I'll write for the first one because I don't have much time now:
a+b(-Q
b+c(-Q
a+c(-Q
ab+bc+ac+b^2(-Q
ab+bc+ac+c^2(-Q
(b-c)(b+c)(-Q
b-c(- Q Analogically we can do for each pair of numbers.
b^2-2bc+c^2-B^2-2bc-c^2(-Q
=> bc (- Q Analogically we can do for each pair of numbers. =>
b^2(- Q Analogically we can do for each pair of numbers.

Hi Smb! Welcome to MHB! (Smile)Whatever way I can think of, it seems to be a number of cases, although I think we can do with fewer than you are suggesting.
Still, if we could cover all 8 similar cases we would be done, although I don't think we can.For starters, what you have now cannot occur, since if we start with $\def\inQ{\in\mathbb Q} a+b, a+c \inQ$, then it follows that $(a+b)-(a+c) = (b-c) \inQ$.
Now suppose $b+c\inQ$, then $\frac 12((b+c)+(b-c))=b\inQ$ and similarly $c \inQ$, so that $bc\inQ$, which contradicts the either-or.
Therefore $b+c\not\inQ$.Let's start with assuming that we can find a triplet $(a,b,c)$ with $ab,ac\inQ$.
Then $\frac {ab}{ac}=\frac bc \inQ$, isn't it?
Now suppose $b+c\inQ$, then $\frac bc\cdot c + c = (\frac bc + 1)c \inQ$.
Can we figure out from there if any of $b$, $c$, and in particular $bc$ is rational? (Wondering)
And if so, can we find that $a^2\inQ$?

I like Serena said:
Hi Smb! Welcome to MHB! (Smile)Whatever way I can think of, it seems to be a number of cases, although I think we can do with fewer than you are suggesting.
Still, if we could cover all 8 similar cases we would be done, although I don't think we can.For starters, what you have now cannot occur, since if we start with $\def\inQ{\in\mathbb Q} a+b, a+c \inQ$, then it follows that $(a+b)-(a+c) = (b-c) \inQ$.
Now suppose $b+c\inQ$, then $\frac 12((b+c)+(b-c))=b\inQ$ and similarly $c \inQ$, so that $bc\inQ$, which contradicts the either-or.
Therefore $b+c\not\inQ$.Let's start with assuming that we can find a triplet $(a,b,c)$ with $ab,ac\inQ$.
Then $\frac {ab}{ac}=\frac bc \inQ$, isn't it?
Now suppose $b+c\inQ$, then $\frac bc\cdot c + c = (\frac bc + 1)c \inQ$.
Can we figure out from there if any of $b$, $c$, and in particular $bc$ is rational? (Wondering)
And if so, can we find that $a^2\inQ$?
I forgot to rewrite it! I'm sorry for the misunderstanding it can be and/or.Both b +c(-Q and bc(-Q could be a case..

Smb said:
I forgot to rewrite it! I'm sorry for the misunderstanding it can be and/or.Both b +c(-Q and bc(-Q could be a case..

Fair enough. The proof doesn't really change though. Just a couple more cases to consider.

## 1. How can we prove the rationality of squares of 9th day 2 ARO 2004/2005 numbers?

To prove the rationality of squares of 9th day 2 ARO 2004/2005 numbers, we can use mathematical induction. This method involves proving the base case (n = 1) and then assuming the statement is true for n = k and proving it for n = k+1. This will show that the statement is true for all natural numbers and therefore the rationality of squares of 9th day 2 ARO 2004/2005 numbers can be proven.

## 2. What is the significance of proving the rationality of squares of 9th day 2 ARO 2004/2005 numbers?

Proving the rationality of squares of 9th day 2 ARO 2004/2005 numbers is important as it helps us understand the patterns and properties of these numbers. It also allows us to make accurate calculations and predictions based on these numbers.

## 3. Can we use a different method to prove the rationality of squares of 9th day 2 ARO 2004/2005 numbers?

Yes, there are other methods that can be used to prove the rationality of squares of 9th day 2 ARO 2004/2005 numbers, such as the Euclidean algorithm or using mathematical properties such as the distributive property. However, mathematical induction is a commonly used and efficient method for this type of proof.

## 4. What is the difference between rational and irrational numbers?

Rational numbers are numbers that can be expressed as a ratio of two integers, such as 1/2 or 3/4. Irrational numbers, on the other hand, cannot be expressed as a ratio of two integers and have decimal representations that do not terminate or repeat, such as π or √2.

## 5. Are there any real-life applications of proving the rationality of squares of 9th day 2 ARO 2004/2005 numbers?

Yes, there are real-life applications of proving the rationality of squares of 9th day 2 ARO 2004/2005 numbers. For example, in physics, rational numbers are commonly used in calculations involving measurements and ratios. Additionally, in finance, rational numbers are used in calculating interest rates and making financial projections.

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