Hello there,

I'm new to the kind of math you have to prove, so be patient ^^

Are there any improvements I could make here?

In the following, I will talk about $Z$ as a ring. More specifically I consider $(Z,+,\times)$ a commutative and unitary ring, where $Z = \{...,-2,-1,0,1,2,...\}$. I presuppose that every element of this ring has exactly one additive inverse and that 1 is the multiplicative neutral element. Implicitly, I presuppose that there is exactly one multiplicative inverse element for every element, so that $--a = a$

Let's consider the relation $R := \{(a,b)\in Z \times Z | a = -b\}$, which might be expressible as a function $f: Z \rightarrow Z , f(a) = -a$.

But is it a function? There are two conditions to be fulfilled:

Firstly, $ \forall a \in Z \; \exists \; b \in Z : (a,b) \in R$

And secondly, for

$a,b,c \in Z : (a,b) \in R \land (c,b) \in R \implies a = c $

1: Let's take an arbitrary constant $c \in Z$. For this $c$ there is exactly one additive inverse $-c$, since Z is a ring. Therefore $(c,-c) \in R \; \forall c \in Z$

2: Let's take three arbitrary constants $a,b,c \in Z $ and assume that $a = -b$, then $(a,b) \in R$

Let's also assume that $ c = -b $, then $(c,b) \in R$

Considering that $a = -b = (-1) \times b $ and that $c = -b = (-1) \times b$, it follows that $ a = c $ since Z is a ring which has exactly one multiplicative inverse element.

Therefore $R$ can be considered equivalent to $f$ and as a function.

$$$$

So is $f$ a bijective function? For that to be the case, it needs to injective as well as surjective.

Def. Injectivity: $$\forall a, a' \in Z : a \neq a' \implies f(a) \neq f(a')$$

Def. Surjectivity: $$\forall b \in Z \; \exists \; a \in Z : f(a)=b $$

Injectivity:

Let's take the contrapositive of the definition of injectivity:

$\forall a,a' \in Z : f(a) = f(a') \implies a = a'$

Additionally, let's take two arbitrary constants $c,c' \in Z$, then

$f(c) = -c$ and $f(c') = -c'$.

Let's suppose that $f(c) = f(c')$, then it follows $-c = -c'$ and therefore also $c = c'$, ergo $f$ is injective.

Surjectivity:

Let's take an arbitrary constant $c \in Z$, then $-c \in Z$ has to exist as well, since Z is a ring.

Since $f(c) = -c$ it follows that $f(-c) = c$ and therefore every $c \in Z$ has a $c' \in Z$ so that $f(c) = c'$. So $f$ is surjective.

Concluding, we can say that $R$ is a bijective function (on $Z$).

I'm new to the kind of math you have to prove, so be patient ^^

Are there any improvements I could make here?

In the following, I will talk about $Z$ as a ring. More specifically I consider $(Z,+,\times)$ a commutative and unitary ring, where $Z = \{...,-2,-1,0,1,2,...\}$. I presuppose that every element of this ring has exactly one additive inverse and that 1 is the multiplicative neutral element. Implicitly, I presuppose that there is exactly one multiplicative inverse element for every element, so that $--a = a$

Let's consider the relation $R := \{(a,b)\in Z \times Z | a = -b\}$, which might be expressible as a function $f: Z \rightarrow Z , f(a) = -a$.

But is it a function? There are two conditions to be fulfilled:

Firstly, $ \forall a \in Z \; \exists \; b \in Z : (a,b) \in R$

And secondly, for

$a,b,c \in Z : (a,b) \in R \land (c,b) \in R \implies a = c $

1: Let's take an arbitrary constant $c \in Z$. For this $c$ there is exactly one additive inverse $-c$, since Z is a ring. Therefore $(c,-c) \in R \; \forall c \in Z$

2: Let's take three arbitrary constants $a,b,c \in Z $ and assume that $a = -b$, then $(a,b) \in R$

Let's also assume that $ c = -b $, then $(c,b) \in R$

Considering that $a = -b = (-1) \times b $ and that $c = -b = (-1) \times b$, it follows that $ a = c $ since Z is a ring which has exactly one multiplicative inverse element.

Therefore $R$ can be considered equivalent to $f$ and as a function.

$$$$

So is $f$ a bijective function? For that to be the case, it needs to injective as well as surjective.

Def. Injectivity: $$\forall a, a' \in Z : a \neq a' \implies f(a) \neq f(a')$$

Def. Surjectivity: $$\forall b \in Z \; \exists \; a \in Z : f(a)=b $$

Injectivity:

Let's take the contrapositive of the definition of injectivity:

$\forall a,a' \in Z : f(a) = f(a') \implies a = a'$

Additionally, let's take two arbitrary constants $c,c' \in Z$, then

$f(c) = -c$ and $f(c') = -c'$.

Let's suppose that $f(c) = f(c')$, then it follows $-c = -c'$ and therefore also $c = c'$, ergo $f$ is injective.

Surjectivity:

Let's take an arbitrary constant $c \in Z$, then $-c \in Z$ has to exist as well, since Z is a ring.

Since $f(c) = -c$ it follows that $f(-c) = c$ and therefore every $c \in Z$ has a $c' \in Z$ so that $f(c) = c'$. So $f$ is surjective.

Concluding, we can say that $R$ is a bijective function (on $Z$).

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