Prove that if $\frac ab$ cannot be simplified further, then $\frac{a-b}{a+b}$ also cannot be simplified further

Question

Let $a,b \in \mathbb{Z}^{+}$ be numbers with different parity. Prove that if $\frac ab$ cannot be simplified further, then $\frac{a-b}{a+b}$ also cannot be simplified further.

Let $a$ be even, so $a=2k$ and let $b$ be odd, so $b=2k+1$ for any $k \in\mathbb Z$.

We can show that $\frac{a}{b}=\frac{2k}{2k+1}$, which cannot be simplified further.

Then,

$$\frac{a-b}{a+b}=\frac{2k-(2k+1)}{2k+2k+1}=\frac{1}{4k+1}$$

and from this we can clearly see that expression $\frac{a-b}{a+b}$ cannot be simplified further. Is this a correct approach?

That’s not what coprime means. You mean that the numerator and denominator are coprime. — Randall, Oct 11 '21 at 10:56
They have to be numbers of different parity. I used $2k+1$ and $2k$ to show this. — wizard844, Oct 11 '21 at 11:08
You can't use the same $k$ then it should be $2k+1$ and $2m$ for example. — kingW3, Oct 11 '21 at 11:12
From how to ask a good question: "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." — jjagmath, Oct 11 '21 at 11:37
@TonyK, oh, that restriction is not in the body of the question. I missed it in the title. Thanks. Sorry for the noise. — lhf, Oct 11 '21 at 12:40
@Ihf +1. I also missed this point and got downvote. Then, I added the proof to my answer. Initially, I had stated that the problem was wrong. — lone student, Oct 11 '21 at 13:18

lone student · Accepted Answer · 2021-10-11T12:06:01.650

1

If you mean $\frac ab$ is the lowest term, then your attempt is not correct. The correct attempt can be considered as follows:

"Let, $a=2m,\thinspace b=2n-1$ or $a=2m-1,\thinspace b=2n$ where $m,n\in\mathbb Z^{+}.$

But, we can continue in a simpler way:

Suppose that,

$$\gcd(a-b,a+b)=k,\thinspace k>1$$

where $\gcd(a,b)=1$ and $a,b$ have different parity with $k=2z-1, \thinspace z\in\mathbb Z^{+}$. Then we have,

$$a-b=mk, ~a+b=nk$$

where $\gcd(m,n)=1$.

If $k≥3$ and $k=2z-1,\thinspace z\in\mathbb Z^{+}$, then we conclude that

$$2a=k(m+n)\implies k\mid a\\ 2b=k(n-m)\implies k\mid b$$

This means $\gcd(a,b)≥k>1$, which gives a contradiction.

edited Oct 11 '21 at 12:06

answered Oct 11 '21 at 11:10

lone student

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But then $a$ and $b$ don't have different parity. So the statement is correct. – TonyK Oct 11 '21 at 11:28
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@TonyK The word "different parity" was just in the title of the question and I missed it, unfortunately. I just concentrated on the content of the question. Thank you for point. – lone student Oct 11 '21 at 11:53
@TonyK I got a quick downvote, but still good because I didn't get too many $\ddot\smile$. – lone student Oct 11 '21 at 12:03
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I apologize for my bad formatting and thank you for your answer. Helped me a lot! – wizard844 Oct 11 '21 at 14:29
@wizard844 You're welcome ( ◜‿◝ ) – lone student Oct 11 '21 at 14:47
The person who downvoted my answers, the moderators will catch you one day. – lone student Oct 14 '21 at 05:16

jjagmath · Answer 2 · 2021-10-11T11:40:29.637

The main problem with your answer is this:

One can say

If $a$ is even then $a = 2k$ for some integer $k$.

and

If $b$ is odd then $b = 2k+1$ for some integer $k$.

BUT the $k$'s involved are not necessarily the same.

Take $a = 6$ and $b=11$, for $a$ we have $6=2\times 3$ so here $k$ is $3$ and for $b$ we have $11 = 2\times 5 + 1$ so now $k$ is $5$. It's a different $k$.

For obvious reasons is a very bad idea to use the same letter to represent both values.

So what one needs to do is pick a different letter: $a = 2k$, $b=2l+1$ for some integers $k$ and $l$.

Another problem is that been of different parity mean $a$ is even and $b$ is odd or $a$ is odd and $b$ is even and you didn't even mention the second case.

score 0 · Answer 3 · answered Oct 11 '21 at 14:38

Proof by contrapositive is quick.

If $p|a+b,a-b$ then $p|2a$ and $p|2b$.

If $p \nmid 2$ then $p|a,b$ and $\frac{a}{b}$ can be simplified further.

If $p|2, p=2$ so from before, $2|a+b \implies a+b = 2k \implies a,b$ are the same parity which contradicts our assumption.

Prove that if $\frac ab$ cannot be simplified further, then $\frac{a-b}{a+b}$ also cannot be simplified further

3 Answers3