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I have no background in mathematical analysis or the like, but I am interested to know how to prove the uniqueness of the solution of $ax+b=0$? Perhaps your answers will help me to prove other uniqueness problems.

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Thanks for answering. I accept Arturo Magidin's answer rather than TonyK's answer because of the extra explanation. – kiss my armpit Jul 9 '12 at 16:55
up vote 3 down vote accepted

A standard way of showing that a certain object is unique is two assume that you have two objects that satisfy the desired properties, and deduce that they must be equal (when we say "two objects", we mean two "names", but which may refer to the same object).

In the case of the solutions to the equation $ax+b=0$, you have to distinguish two cases: if $a=0$, then the equation either has no solutions (if $b\neq 0$), or it has infinitely many solutions (if $b=0$).

So uniqueness really only exists when $a\neq 0$. The uniqueness is based on the following fact about real numbers:

For any real numbers $r$ and $s$, if $rs=0$, then $r=0$ or $s=0$.

Once you have that:

Claim. If $a\neq 0$, then there is at most one solution to $ax+b=0$.

Proof. Suppose that both $x$ and $y$ are solutions. We aim to show that $x=y$. Since $x$ is a solution, $ax+b=0$. Since $y$ is a solution as well, $ay+b=0$. That means that $ax+b=ay+b$. Adding $-b$ to both sides we conclude that $ax=ay$. Adding $-ay$ to both sides, we obtain $ax-ay = 0$. factoring out $a$, we have $a(x-y)=0$. Since the product is $0$, then $a=0$ or $x-y=0$. Since $a\neq 0$ by assumption, we conclude that $x-y=0$, so $x=y$. Thus, if $x$ and $y$ are both solutions, then $x=y$, so there is at most one solution. $\Box$

Note that this argument works in the context of the real numbers, or other kinds of "numbers" where $rs=0$ implies $r=0$ or $s=0$. There are other situations where this is not the case. For example, if you work with "integers modulo 12" ("clock arithmetic", where $11+3 = 2$), then $2x+8 = 0$ has many different solutions $0\leq x\lt 12$: one solution is $x=2$ (since $2(2)+8 = 4+8=12=0$ in clock arithmetic), and another solution is $x=8$ since $2(8)+8 = 16+8=24 = 0$ in clock arithmetic).

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Suppose $x$ and $y$ are both solutions to your equation. Then we have:

$ax + b = 0$


$ay + b = 0$

Subtracting the two equations gives

$a(x - y) = 0$

So if $a$ is non-zero, then $x-y$ must be zero, i.e. $x = y$.

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Sorry, why is the subtracting step valid? – kiss my armpit Jul 9 '12 at 16:41
If $r=s$ and $t=u$, then $r-t = s-u$. – Arturo Magidin Jul 9 '12 at 16:45
@HiggsBoson It's just applying the concept that if i=j and m=n, then i-m = j-n; in this case, i is ax+b and m is ay+b, and j and n are both 0. Even in more abstract cases than just numbers, the subtraction law here has to hold; in some sense it expresses the very nature of what it means for two things tobe equal. – Steven Stadnicki Jul 9 '12 at 16:45
Higgs makes a good point here. I'm not even sure whether this is an axiom of some given field or other algebraic structure, though I think it should: if we've an equality $\,a=b\,$ in some alg. structure, the equality remains when we apply the very same operation to both sides of the equality. – DonAntonio Jul 9 '12 at 16:53
@DonAntonio: It's first order logic (substitution of equal terms into functions/expressions). If $t=u$, then the result of performing a function to $t$ is the same as performing it to $u$, by definition of function. So if $f(x)=r-x$, then $t=u$ implies $r-t = f(t) = f(u) = r-u$. If $r=s$, then $g(x)=x-u$, then $r-t = r-u = g(r) = g(s) = s-u$. Now use transitivity of =. – Arturo Magidin Jul 9 '12 at 16:56

If you're working on a system with a unique inverse for every (or some) elements, then $$ax+b=0\Longleftrightarrow x=-ba^{-1}=-\frac{b}{a}$$ is equivalent to "a has a unique inverse". For example, fields (like $\,\mathbb Q\,,\,\Bbb R\,,\,\Bbb C\,$ , etc.) , where the unique inverse element axiom is true for anyn non-zero element.

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Where is the minus sign? – kiss my armpit Jul 9 '12 at 16:45
Thanks, it's been added now. +1 – DonAntonio Jul 9 '12 at 16:51
beautiful, algebra always do it in the nice way – Mohamez Jul 9 '12 at 17:07
What do you mean by "for every (or some) elements"? Also, your "$P \Longleftrightarrow Q$ is equivalent to R" would benefit from a prepended "$\forall x$", I think. but in any case it is pitched far above the original question! – TonyK Jul 9 '12 at 17:09
It may be we're working on a field or a division ring and then every non-zero element has a unique inverse; it may be we're on a general ring (say commutative) and then only some elements (the units) have an inverse, which is then unique as the set of all units in a (commutative) ring is an abelian group...In my answer, the $\,\Longleftrightarrow\,$ is assuming what is writtent below that line in my answer: $\,a\,$ has a unique element. Since this was a mathematical question I tried to give a mathematical answer without assuming more than what was in the OP. – DonAntonio Jul 9 '12 at 17:25

As Tony wrote, if it had two different solutions $\rm\,r\ne s,\,$ then $\rm\: a\, x = 0\,$ would have root $\rm\, x = r-s\ne 0,\:$ contra $\rm\:x,y\ne0\:\Rightarrow\:xy\ne 0.\:$ This implication ("no zero-divisors") is equivalent to the fact that polynomials have no more roots than their degree (so a linear polynomial has at most one root).

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"No zero divisors" is only equivalent to polynomials having no more roots than their degree in the presence of commutativity of multiplication, though; $x^2+1$ has infinitely many solutions in the real quaternions, even though there are no zero divisors. – Arturo Magidin Jul 9 '12 at 16:58
Of course, but rings are normally assumed commutative by default. If one objected to all the things that failed in noncommutative cases the site would be a mess. – Bill Dubuque Jul 9 '12 at 17:00
Still, the bottom line of Tony's answer is false say if we're in a ring with zero divisors: $$2x+4=0\Longrightarrow x=1\,\,or\,\,x=4\,\,,\,\,x\in\Bbb Z/6\Bbb Z$$ – DonAntonio Jul 9 '12 at 17:01
In this context, it makes sense, yes; I would not go so far as to say "ring" normally implies "commutative". Really depends what context you are used to. – Arturo Magidin Jul 9 '12 at 17:01
@Don Yes, that's essentially the trivial direction of the equivalence that I mentioned. – Bill Dubuque Jul 9 '12 at 17:08

Would the following be valid?

if $a=0$ and $b\neq 0$ then there is no solution and if $a=0$ and $b=0$ then there are infinite solutions. (So we won't be using these conditions in our proof because the do not give us a unique solution)


if $a\neq 0$ then there is a unique solution to $ax+b=0$


let $x,y \in \Bbb R: ax+b=0$ and $ay+b=0$ meaning $x=y$ $$\begin{align} &ax+b=ay+b\\ \implies &(ax+b)+(-b)=(ay+b)+ (-b)\\ \implies &ax+(b+(-b))=ay+(b+(-b))\tag {A2}\\ \implies &ax+0=ay+0\tag{A4}\\ \implies &ax=ay\\ \implies &ax(a^{-1})=ay(a^{-1})\\ \implies &x(a(a^{-1}))=y(a(a^{-1}))\tag{M2}\\ \implies &x(1)=y(1)\tag{M4}\\ \implies &x=y \end{align}$$

Proving that the solution is true

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Thanks for the edit @SubhadeepDey, I'm new to this site. Is the proof ok though? – Trial and error Mar 20 at 13:04
Yes, it is totally OK. – user249332 Mar 20 at 13:07

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