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?

Answer 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:

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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.

Answer 2

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.

Answer 3

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.