Prove for all $x$ there exists a $y$ such that $x + y = 1$ within the domain of all real numbers

Question

So we can simplify $x + y = 1$ to $y = 1 -x$ and no matter what value of $x$, the value of $y$ will always be $-x + 1$.

Is that an ample proof? It just seems too easy.

Answer 1

The fact that $$ x+y=1\iff y=1-x $$ can be proven this way.

$(\Longrightarrow)$ Suppose $x+y=1$. Then $$ -x+(x+y)=-x+1 $$ By associativity of addition (in $\mathbb{R}$), $-x+(x+y)=(-x+x)+y$. By definition of $-x$ as the additive inverse of $x$, we have$-x+x=0$. Hence $(-x+x)+y=0+y$. Since $0$ is the additive neutral element, we have $0+y=y$. Hence $$ y=-x+1 $$ By commutativity of the addition, $-x+1=1+(-x)$. The notation $1-x$ means $1+(-x)$, so $-x+1=1-x$.

All in all, we have $y=1-x$.

$(­\Longleftarrow)$ A similar argument works.

Note 1: Most of the time we take many properties of $\mathbb{R}$ for granted, namely the properties of a complete ordered field. One example of a property we often take for granted (that we take as an axiom) is that the addition is associative.

However, it is possible to take more general axioms and deduce those properties from them. The Peano axioms are sufficient to prove every property for $\mathbb{N}$ and from $\mathbb{N}$ we may construct rigorously the number systems $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$ and prove the properties as consequences of the Peano axioms (which are, in a sense, more minimal).

More general axioms than those of Peano would be axioms for a set theory, for example the axioms of the Zermelo-Fraenkel set theory. It is interesting that from the axioms of ZF we can actually deduce the Peano axioms and prove the properties of a complete ordered field that we wish $\mathbb{R}$ would have.

Note 2: Obviously all those steps are rarely shown. But if you're interested in an axiomatic approach, then it can't be bad to be aware of what's going on.

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If all you want to do is prove that for all $x$ there exists a $y$ such that $x+y=1$, then you can proceed like this.

Fix $x$. Let $y:=-x+1$ (we take $y$ to be this and, as user @Pragnya Jha mentioned in the comments, $-x+1$ is a real number). Then \begin{align} x+y&=x+(-x+1)&&\text{by def of }y\\ &=(x+(-x))+1&&\text{by associativity of }+\\ &=0+1&&\text{by def of }-x\\ &=1&&\text{neutral property of }0 \end{align}

Hence we have exhibited a $y$ that works. This completes the proof.

Answer 2

_{Note To avoid confusion, I'm using round parentheses to denote sequences, and square brackets for grouping.}

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We prove the following:

1. $(\mathbb R, +)$ is a group
2. For every group $(G, \circ)$ and for every $a,b\in G$, there is $y\in G$ such that $a\circ y = b$.

Then the claim follows with $G=\mathbb R, \circ = +, a=x, b=1$.

Proof:

2. Let $a,b\in G$. Since $G$ is a group, there is an inverse Element $a^-$ such that $a\circ a^- = a^-\circ a = 0$. Let $y:=a^-\circ b$. Then $a\circ y = a\circ[a^-\circ b]=[a\circ a^-]\circ b = 0\circ b = b$.

1. This is the hard part. The set $\mathbb R$ is defined as the set of equivalence classes of Cauchy sequences in $\mathbb Q$, so:

   $$ \mathbb R := \left\{(a_n)_{n\in\mathbb N}:a_n\in\mathbb Q,\forall\epsilon>0\exists N\in\mathbb N:\left|a_n-a_m\right|<\epsilon\text{ for }n,m > N\right\}/\sim $$

   with $$(a_n)\sim(b_n):\Leftrightarrow\left|a_n-b_n\right|\underset{n\rightarrow\infty}\longrightarrow0$$

   and the addition defined as $(a_n)+(b_n):=(a_n+b_n)$. [It must be proven that this is well-defined, which you probably have done already.]

   Sloppily speaking, everything in $\mathbb R$ is just a sequence of rational numbers, and addition happens member by member.

   Now, to prove that $(\mathbb R, +)$ is a group, we must show

   1. $a+[b+c] = [a+b]+c$ for all $a,b,c\in\mathbb R$, and
   2. $\color{red}{\text{For all } x \in \mathbb R\text{ there is }h\in\mathbb R\text{ such that }x+h=h+x=0}$. We write $-x$ to denote this element.

   It is very important to understand that you cannot blindly assume that $-x$ exists, just because you can write it down. The notation "$-x$" means: "This is an element that when added to $x$, yields the sum $0$". Again, it must be proven that this exists.

   Now, to prove the above points, write $a=(a_n), b=(b_n), c=(c_n), x=(x_n)$. Then, $$a+[b+c] =\\ (a_n)+[(b_n)+(c_n)] =\\ (a_n)+(b_n+c_n) =\\ (a_n+[b_n+c_n]) =\\([a_n+b_n]+c_n)=\\ (a_n+b_n)+(c_n) =\\ [(a_n)+(b_n)]+(c_n) =\\ [a+b]+c$$

   Remember that $a_n, b_n, c_n\in\mathbb Q$, so the associativity from $\mathbb Q$ can be used here.

   Likewise let $h:=(-x_n)$. We know that $-x_n\in\mathbb Q$ exist. Now,

$$x+h = (x_n)+(-x_n) = (x_n+[-x_n]) = (0) = 0$$

and

$$h+x = (-x_n)+(x_n) = (-x_n+x_n) = (0) = 0$$ $\square$

The basic idea here is to use the properties of $\mathbb Q$, which translate naturally to $\mathbb R$ via the members of the Cauchy sequences.

Answer 3

If we assume $\mathbb R$ is a field the OP'S argument is sufficient:

For all $x \in \mathbb R$ there is a number $y=1 + (-x) $ and $x +y = x + [1+(-x)] =[x+(-x)]+1=0+1=1$

(i.e. for all elements $x $ there is an additive inverse (-x) so that $x+(-x)=0$ and that there exists and additive identity so that $a+0=0+a $ for all $a $, and that $1$, the multiplicative identity, exist, and "$+$" is a binary operation so $a+b $ is a real number for all $a,b$)

So perhaps the exercise boils down to proving $\mathbb R $ is a field. Or maybe not.

This does come down to, just what class is this for, and what introduction of the reals has the student had at this point in the class.

Somehow, I doubt the idea of this exercise is actually to define and elucidate the properties of the reals. But if it is and that single line proof is not enough, perhaps we should be asking the OP to clarify, how the Reals have been introduced so far before going into such excruciating detail which is most likely overkill.

(Although, I have to admit the other answers do do the details very well.)