Is this correct? (sum of two rational numbers is rational)

[ephedyn]

Homework Statement

"Prove that the sum of two rational numbers is a rational number."

I just started on proof writing, so I'll just like to verify if I'm not missing anything here, and get some comments about the style.

The attempt at a solution

Theorem. If $$a,b \in \mathbb{Q}$$ then $$a+b \in \mathbb{Q}$$

Proof. Given $$a,b \in \mathbb{Q}$$
we have some $$p_1,p_2,q_1,q_2 \in \mathbb{Z}$$ satisfying

$$a=\dfrac{p_1}{q_1} \, b=\dfrac{p_2}{q_2}$$

which implies that

$$a+b =\dfrac{p_1}{q_1}+\dfrac{p_2}{q_2} =\dfrac{p_{1}q_{2}+p_{2}q_{1}}{q_{1}q_{2}}$$

By closure of $$\mathbb{Z}$$ under addition and multiplication, it follows from the assumption that $$p_1,p_2,q_1,q_2 \in \mathbb{Z}$$ that

$$p_{1}q_{2}+p_{2}q_{1}$$ and $$q_{1}q_{2}$$ are also integers.

Hence, $$a+b$$ can be expressed as the ratio of 2 integers. By definition, $$a+b$$ is rational. $$\square$$

 

[vela]

Looks good.

 

[ephedyn]

^Ah, great! Thanks.

One question though... is it immediately obvious that $$\mathbb{Z}$$ is closed under addition/multiplication? Is there a proof for this? I mean, this one's still rather easy, but I find it hard to verify that some given finite group has closure... say... {1, 9, 16, 22, 29, 53, 74, 79, 81} under multiplication modulo 91?

 

[D H]

$$a=\dfrac{p_1}{q_1} \, b=\dfrac{p_2}{q_2}$$

which implies that

$$a+b =\dfrac{p_1}{q_1}+\dfrac{p_2}{q_2} =\dfrac{p_{1}q_{2}+p_{2}q_{1}}{q_{1}q_{2}}$$

How does the first imply the second? It doesn't. Yes, I'm being picky here, but you are doing proofs, ephedyn. Pickiness is essential in proofs.

To prove that the sum of two rationals is rational, you need

1. A definition of what exactly constitutes a rational number.
   Hint: 0 and 1 are integers. 0/1 is a rational, but 1/0 is not. Why not?
2. A definition of the sum of two rationals.
   This is why I asked the question at the start of this post. How is the sum of two rationals defined?
3. Proof that this definition always yields a rational.
   You need to use the correct definition of what constitutes a rational (item 1 above) here.

 

[vela]

D H has brought up good points you need to address. I should have thought about it much more carefully before saying your proof was okay.

 

[ephedyn]

How does the first imply the second? It doesn't. Yes, I'm being picky here, but you are doing proofs, ephedyn. Pickiness is essential in proofs.

To prove that the sum of two rationals is rational, you need

1. A definition of what exactly constitutes a rational number.
   Hint: 0 and 1 are integers. 0/1 is a rational, but 1/0 is not. Why not?
2. A definition of the sum of two rationals.
   This is why I asked the question at the start of this post. How is the sum of two rationals defined?
3. Proof that this definition always yields a rational.
   You need to use the correct definition of what constitutes a rational (item 1 above) here.

Okay, thanks. I gave it some serious thought and rewrote it. But it doesn't look very different from my previous one, and looks much more long-winded so maybe I'm still wrong >_<. How would you change it:

Theorem The sum of any two rational numbers is rational.

Let $$\mathbb{Q}$$ denote the set of all rational numbers

Definition Any number $$n \in \mathbb{Q} \iff n = \dfrac{p}{q}$$, where $$p,q \in \mathbb{Z} : q>0$$

Consider the sum of any two numbers $$a, b$$,

$$a+b$$

Assuming $$a, b \in \mathbb{Q}$$
then by definition,

$$a+b=\dfrac{p_1}{q_1}+\dfrac{p_2}{q_2}=\dfrac{p_{1}q_{2}+p_{2}q_{1}}{q_{1}q_{2}}$$

for some $$p_{1},p_{2},q_{1},q_{2} \in \mathbb{Z}: q_{1},q{2}>0,$$

By closure of $$\mathbb{Z}$$ under addition and multiplication, it follows that $$p_{1}q_{2}+p_{2}q_{1}$$ and $$q_{1}q_{2}$$ are also integers. Also, since we have $$q_{1},q{2}>0$$, $$q_{1}q{2}>0$$. Hence, $$\left( a+b \right) \in \mathbb{Q}$$

 

[ephedyn]

^Also, I have a question... am I missing $$gcd(p,q)=1$$ from the definition? Does it affect the proof?

 

[D H]

Much improved. There is one minor issue: is 2/-1 a rational? It is not per your definition. The constraint on the denominator is that it is a non-zero integer rather than a positive integer.

My main objections to your original proof were

• Your use of the word "implies." p₁/q₁ + p₂/q₂ = (p₁q₂+p₂q₁)/(q₁q₂) by definition. The word "implies" has a special meaning in proofs.
• That the numerator and denominator are integers does not suffice. The denominator cannot be zero.

You have addressed those objections.

Here is a proof that is a bit shorter than your proof. Note well: This proof, like yours, is an informal proof.

Given p₁/q₁, p₂/q₂ ∈ ℚ, then p₁, q₁, p₂, q₂ ∈ ℤ and q₁, q₂ ≠ 0. By definition, p₁/q₁+p₂/q₂= (p₁q₂+p₂q₁)/(q₁q₂). The numerator p₁q₂+p₂q₁ and denominator q₁q₂ are integers by closure of multiplication and addition of the integers. Moreover, q₁q₂ ≠ 0 since neither q₁ nor q₂ is zero. Thus p₁/q₁+p₂/q₂ is a rational.

 

[ephedyn]

Your use of the word "implies." p1/q1 + p2/q2 = (p1q2+p2q1)/(q1q2) by definition. The word "implies" has a special meaning in proofs.

Much improved. There is one minor issue: is 2/-1 a rational? It is not per your definition. The constraint on the denominator is that it is a non-zero integer rather than a positive integer.

Noted!

OK, thanks a lot for your advice. That's much more concise indeed.