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The following issue came up in my Intro to Proofs course and I wasn't sure how to explain my distaste of the student proof.

Prove that the solution for $x$ in $ax+b=c$ is unique ($a \neq 0$).

Student Proof: Solving gives \begin{align*} ax+b &= c\\ ax &=c-b\\ x &= \frac{c-b}{a}. \end{align*} Hence the solution must be $x=\frac{c-b}{a}.$

The Proof I Wanted: Suppose both $x$ and $y$ solve the equation, then \begin{align*} ax+b &= ay+b\\ ax&=ay\\ x&=y. \end{align*} Thus the solution is unique.

Is the student solution acceptable?

I feel like finding an algebraic solution is not a proof of uniqueness. My gut says that there should be a good example where algebraically solving implies a unique solution when really there isn't one. (But I can only think of simple cases where you forget a $\pm$ on a square root.)

Another way of explaining my distaste for this is that when solving choices are made to arrive at $x$. Maybe other choices would've gotten you to a different solution. Whereas in the desired proof since we assumed two solutions from the start, it's okay to make algebraic choices to show the two solutions are the same. But maybe I'm being too pedantic.

Am I justified in disliking the student solution?

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    $\begingroup$ I'd say the student's proof is valid in this case but not a good one for generalising. $\endgroup$ – Jessica B Apr 20 '15 at 13:54
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    $\begingroup$ The student's proof is correct. $\endgroup$ – NiloCK Apr 20 '15 at 13:54
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    $\begingroup$ Arguably, the student's solution is better since it doesn't just show uniqueness but also finds the value. $\endgroup$ – Adam Apr 20 '15 at 15:15
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    $\begingroup$ Another sense in which the student's proof is arguably better is that the OP's desired proof doesn't even show existence of solutions. $\endgroup$ – Dag Oskar Madsen Apr 20 '15 at 21:53
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    $\begingroup$ In the future it would be wise not to base any judgement of proofs on gut feelings (intuition, experience, preference, popularity, ...). If one cannot find any mistake in a proof, one is not justified in finding it distasteful! That said, as Dirk said both presentations do not make the logical structure clear, but the student's solution gives a stronger result than your expected solution as Dag said and hence I would prefer it. Furthermore, the question itself is totally flawed. What in the world are $a,b,x$? quid gives an excellent counter-example. $\endgroup$ – user21820 Apr 21 '15 at 8:07
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Both are valid ways to prove the uniqueness of solutions, i.e. that the set $\,S\,$ of solutions contains at most one element. Indeed, one easily proves that the following are equivalent for any set $\,S.$

$(1)\ \ \ |S| \le 1,\ $ i.e. $S\,$ contains at most one element.

$(2)\ \ \ s\in S\,\Rightarrow\, s = t,\ $ for some fixed $\,t$

$(3)\ \ \ s,t\in S\,\Rightarrow\, s = t$

The student's solution uses method $(2)$ whereas your solution uses method $(3)$.

In my experience, such doubts about the method used in $(2)$ are not unusual. One can find many such questions in any general-level math forum such as sci.math, e.g. see this thread about such a proof in Max Rosenlicht's Introduction to Analysis.

Remark $\ $ It would be interesting to better understand the source of this apparent bias against the method $(2).\,$ It may be related to the fact that the general method(s) of proving uniqueness seem to be rarely explicitly taught. Rather, the method(s) are typically learned en passant while working on specific problems. As a result, some students never master such method(s) and often miss opportunities for employing these powerful tools. To help rectify this, I often emphasize such applications, e.g. over 30 answers on math.SE, e.g. this interesting simple example occurred only a few days ago: $\,4ab\!-\!1\mid 4a^2\!-\!1\,\Rightarrow\, a=b\,$ for integers $\,a,b > 0.\,$ As I explain there, it is just a special case of the uniqueness of (multiplicative) inverses (in $\,\Bbb Z/m = $ integers modulo $\,m).$

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    $\begingroup$ I think you forgot to define $t$ in method (2). $\endgroup$ – David Z Apr 21 '15 at 6:57
  • $\begingroup$ @DavidZ $ $ In $(2),$ $\ t$ denotes any (fixed) element of the ambient universe (probably a field such as $\,\Bbb Q,\Bbb R\,$ or $\,\Bbb C\,$ in the OP). $\endgroup$ – Bill Dubuque Apr 21 '15 at 14:37
  • $\begingroup$ OK, I see what you mean. It would make the answer clearer, I think, if you mentioned that - perhaps saying there exists some $t$ such that $s\in S\implies s=t$ (if I've thought through it correctly). $\endgroup$ – David Z Apr 21 '15 at 14:40
  • $\begingroup$ @BillDubuque: You're right. I'm biased against (2). I feel like somewhere in grad school I was sternly steered away from (2). So my rational mind recognizes it's valid, but my gut hates it (similar to proofs by exhaustion that could've been done in two lines). $\endgroup$ – Aeryk Apr 21 '15 at 16:00
  • $\begingroup$ @BillDubuque: Interesting side note: the text I'm using for the course is Solow's "How to Read and Do Proofs" and he only gives methods, strategies, and examples of (3). He seems to also disfavor (2). $\endgroup$ – Aeryk Apr 21 '15 at 16:02
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Be careful here. You're teetering close to the edge of checking whether the student has guessed your solution to the problem rather than whether the student has solved the problem.

This is easy to do, especially in cases like this where a problem is constructed with the specific purpose of exercising a particular technique (eg, your solution).

This purposeful construction inflicts the problem setter with a bit of blindness for other potential solution paths - when we solve a problem, we stop looking for different angles on it. And being presented with an alternate solution path can also be a little blow to the ego - because of the implicit suggestion that the problem doesn't serve its purpose perfectly, eg, the desired technique was not required.

My suggestion would be to congratulate the student on their straightforward approach, and explain that you had tripped yourself up with the expectation of a different line of thought.

Moving forward, there's nothing wrong with prescribing a technique for problems when you want a particular technique to be used. (Excepting that one can get into a bit of a taxonomical nightmare when trying to put names on individual techniques for things. May be better to treat the technique names as disposable: eg, Tuesday's Method can be a labelled technique from the class notes).

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I do not see any issue $\iff$ it is clear what the sequence of equations means. (Sorry for the pun.)

The following is true (for $a \neq 0$, in the real numbers, say)

$$ax+b = c \iff ax =c-b \iff x = \frac{c-b}{a}$$ Hence the solution $x$ must be $x=\frac{c-b}{a}.$

Each transformation is an equivalence. The problem is that sometimes some do not pay enough attention and write sequences of equations like this that in fact are not equivalences. (Or perhaps they pay attention but just mean to express something different.)

For example the following might be written but does not give uniqueness: \begin{align*} x^2 +5 & = 9 \\ x^2 & = 4 \\ x & = 2 \end{align*} The issue is that the last equation is not an equivalence; it is false that $x^2 = 4 \iff x = 2$ (again in the reals, that is; in the positive reals it would be true).

The main issue I see, but this is the same for the students and your solution, is that it is not completely made clear (at least in what is written) what the sequence of equation means.

For example, would we work over the $\mathbb{Z}/4\mathbb{Z}$ your solution would not show uniqueness either as even for $a\neq 0$ the equation $ax=ay$ is not equivalent to $x= y$.


Added: The discussion above uses equivalence ($\iff$) that is "if and only if" throughout. Such transformations will preserve the set of solutions exactly. It can sometimes be more desirable to use implications only in one way. Then one also has a relation among the solutions, namely:

$$f(x)= a \implies g(x)=b$$ will yield that every solution to $f(x)=a$ is also a solution to $g(x)=b$. That is if the concern is only to show uniqueness it is even sufficient to have sequence of equations linked by $\implies$ and a unique solution for the final one.

Conversely, if one wants existence of a solution one needs a sequence of equations linked by $\Leftarrow$ and a solution for the last.

Thus (in the reals),
\begin{align*} x^2 +5 & = 9 \\ x^2 & = 4 \\ x & = 2 \end{align*} can be taken to show the existence of a solution of the first equation as $$x^2 +5 = 9 \Leftarrow x^2 = 4 \Leftarrow x = 2 $$ is true, yet not uniqueness as $$x^2 +5 = 9 \implies x^2 = 4 \not \Rightarrow x = 2 $$

Summary: To me the key point is to make clear how the different equations are logically linked, and to know what this means for the respective solutions.

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    $\begingroup$ I like your comment about $Z/4Z$, which makes very clear that the instructor's answer is no better than the student's. $\endgroup$ – Tom Church Apr 20 '15 at 20:04
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    $\begingroup$ Indeed, as a quick example, $2x + 2 = 0$ does not have a unique solution over $\mathbb{Z}/4\mathbb{Z}$: Both $x = 1$ and $x = 3$ are solutions. $\endgroup$ – Benjamin Dickman Apr 20 '15 at 20:59
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I suspect your discomfort with the student's proof is carryover from high school algebra where the logical underpinnings of what you are actually doing isn't often made clear.

The student's work is essentially a rigorous proof of the theorem

Theorem: If $a \neq 0$ and $a x + b=c$, then $x = (c-b)/a$.

and this sort of thing is the typical result of the work involved in solving a problem.

Uniqueness is an easy corollary:

Corollary: If $a \neq 0$ and $ax+b = c$ and $ay+b=c$, then $x=y$

Proof: $x = (c-b)/a$ and $y = (c-b)/a$, therefore $x=y$.


You should also note that any skepticism about the student's approach showing the solution for $x$ is unique should also apply to your approach of showing that $x=y$ is the unique solution for $x$. e.g. after all you made choices too; maybe if you made other choices you wouldn't have arrived at $x=y$?


All of the work demonstrated in your post can be thought of as simply applying transformations to a problem to derive simpler problems that are implied by it.

The student opted to simplify first, and produced a simpler problem for which it was trivial to prove uniqueness.

You opted to invoke a proof technique for showing uniqueness first, and then proceeded to simplify the ensuing system of equations.

Maybe all of your concerns would be obviated if you did as I did above and explicitly tacked on the remaining, trivial argument needed to arrive at that the conclusion that the solution (if it exists) is unique.

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There are two typical ways for uniqueness proofs: showing that all answers have the same form or that any two potential answers must be the same. What your student did was if $x$ is a solution to $ax +b = c$ ($a \neq 0$), then $x= \frac{c-b}{a}$ implying that if $y$ is another solution, then $y = \frac{c-b}{a} = x$.

Both ways should be explained as valid forms of proving uniqueness. While supposing if $x$ and $y$ are solutions, then $ax +b = c = ay+b$, the leap to setting both equations equal might be a leap for some. For instance, I personally would show that all solutions have the same form to get uniqueness. You could add that, since this is a proof class, that linear functions are one-to-one or injective to get the desired result as well. This allows for quite a bit of generalization for linear systems of equations.

In the end, if the student uses a valid proof for something that is not what you were considering, congratulate them and, perhaps, point out the method you were intending.

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Actually, I get a bit confused about this, each step he took was reversable. But I don't think that by itself is enough. The important thing seems to be that each step he took preserved the solution set. In which case he proved that the solution set is a singleton.

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    $\begingroup$ What do you mean by "reversible"? That there exists an inverse operation to recover the prior step? In the case of such a linear system, yes as these correspond with row operations. $\endgroup$ – Chris C Apr 20 '15 at 19:06
  • $\begingroup$ I'm also not quite sure what you mean, but certainly if you have a sequence of propositions each separated (correctly) by $\iff$ then you can follow the sequence from beginning to end or end to beginning. See quid's answer for one approach using iff in the given problem. $\endgroup$ – Benjamin Dickman Apr 20 '15 at 21:02
  • $\begingroup$ I tried to expand my answer a bit in the hope to address your concern. $\endgroup$ – quid Apr 20 '15 at 21:36
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    $\begingroup$ There is a sequence of operations performed on each equation to produce the next equation. The final equation being $x = \frac{c - b}{a}$. There is also a sequence of operations by which you can start at the bottom and produce $ax + b = c$. It can be shown that each of these operations does not altar the solution set. $\endgroup$ – Steven Gregory Apr 27 '15 at 16:49

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