Show that an integer is unique

[Mr Davis 97]

Homework Statement

Suppose that a and b are odd integers with a ≠ b. Show that there is a unique integer c such that
|a - c| = |b - c|

Homework Equations

The Attempt at a Solution

What I did was this: Using the definition of absolute value, we have that ##(a - c) = \pm (b - c)##. If we choose the plus sign , then we have that ##a = b##, which contradicts our original assumption. So ##a - c = -b + c \implies c = \frac{a + b}{2}##. Does this solve the original problem? I have shown that there exists an integer ##c##, but is this sufficient to show that this c is unique?

 

[fresh_42]

Yes. If you like, you can assume, that ##c'## is another solution. But the same argument gives you ##c'=\frac{a+b}{2}=c##.

However, this isn't necessary, because to be exact, you only showed, that ##c=\frac{a+b}{2}## is necessary to be a solution and this is unique. You haven't shown, that it actually is a solution.

 

[Mr Davis 97]

Yes. If you like, you can assume, that ##c'## is another solution. But the same argument gives you ##c'=\frac{a+b}{2}=c##.

However, this isn't necessary, because to be exact, you only showed, that ##c=\frac{a+b}{2}## is necessary to be a solution and this is unique. You haven't shown, that it actually is a solution.

Are you saying that I have to plug it back into show that it is actually a solution? Why doesn't the algebra itself show that it is a solution?

 

[Ray Vickson]

Yes. If you like, you can assume, that ##c'## is another solution. But the same argument gives you ##c'=\frac{a+b}{2}=c##.

However, this isn't necessary, because to be exact, you only showed, that ##c=\frac{a+b}{2}## is necessary to be a solution and this is unique. You haven't shown, that it actually is a solution.

WLOG, let ##a < b##. Then there are three possible cases: (1) ##c < a##; (2) ##c > b##; or (3) ##a \leq c \leq b##. It is easy to show that cases (1) and (2) are impossible. In case (3) we MUST have ##c-a = b-c##, so ##c = (a+b)/2##. This is true whether or not ##a, b## are integers (odd or not), and whether or not we need ##c## to be an integer. Of course, when ##a,b## are odd integers, ##c## is an integer, and that is easily shown.

 

[Aufbauwerk 2045]

fresh_42 I am also a bit puzzled by your response. Can you elaborate?

 

[PeroK]

Are you saying that I have to plug it back into show that it is actually a solution? Why doesn't the algebra itself show that it is a solution?

You can check that your steps are reversible, or you can plug the solution you got into the initial equation.

One thing you did not do was to show that ##c## is an integer.

 

[fresh_42]

Are you saying that I have to plug it back into show that it is actually a solution?

Yes.

Why doesn't the algebra itself show that it is a solution?

Well, in this case, it is pretty much so, for it is easy to see. But not, if we want to be very rigorous.

fresh_42 I am also a bit puzzled by your response. Can you elaborate?

What @Mr Davis 97 has shown was:

If ##\mathcal{\, A}_1 := \left(\, |a-c| = |b-c| \,\text{ for integer } \; a,b,c \text{ is true } \right)## then ##\mathcal{A}_2 := \left(c = \frac{a+b}{2} \text{ is true }\right)##.

This means, for statement ##\mathcal{A}_1## to be true, it is necessary, that statement ##\mathcal{A}_2## is also true.
It does not say, that statement ##\mathcal{A}_2## implies ##\mathcal{A}_1##, because ##\mathcal{A}_2## could still be true, even if ##\mathcal{A}_1## wasn't.

Therefore, it remains to show, that statement ##\mathcal{A}_2## is also sufficient for statement ##\mathcal{A}_1## to hold.
(cp. @PeroK's remark above)

I admit, it can be done by simply looking at it and doing the calculations in mind, but there might be more difficult examples ahead, in which case it is not as obvious.

 

[Ray Vickson]

Yes.

Well, in this case, it is pretty much so, for it is easy to see. But not, if we want to be very rigorous.

What @Mr Davis 97 has shown was:

If ##\mathcal{\, A}_1 := \left(\, |a-c| = |b-c| \,\text{ for integer } \; a,b,c \text{ is true } \right)## then ##\mathcal{A}_2 := \left(c = \frac{a+b}{2} \text{ is true }\right)##.

This means, for statement ##\mathcal{A}_1## to be true, it is necessary, that statement ##\mathcal{A}_2## is also true.
It does not say, that statement ##\mathcal{A}_2## implies ##\mathcal{A}_1##, because ##\mathcal{A}_2## could still be true, even if ##\mathcal{A}_1## wasn't.

Therefore, it remains to show, that statement ##\mathcal{A}_2## is also sufficient for statement ##\mathcal{A}_1## to hold.
(cp. @PeroK's remark above)

I admit, it can be done by simply looking at it and doing the calculations in mind, but there might be more difficult examples ahead, in which case it is not as obvious.

Whether or not ##a## and ##b## are integers, the equation ##|c-a| = |c-b|## implies that ##c = (a+b)/2##, uniquely. That is true for all real numbers; see post #4.

 

[fresh_42]

Whether or not ##a## and ##b## are integers, the equation ##|c-a| = |c-b|## implies that ##c = (a+b)/2##, uniquely. That is true for all real numbers; see post #4.

Of course. But my mistake hasn't been this, it was that I didn't formulate ##\mathcal{A}_2## in a correct way, for it should have been ##\mathcal{A}_2 = (\exists c := \frac{a+b}{2} \in \mathbb{Z}\text{ is true })##. But even this wasn't really important for I spoke of the general logic of implications and the difference between a necessary and a sufficient condition. And, yes, it's pretty easy in this case.

 

[Mr Davis 97]

Yes.

Well, in this case, it is pretty much so, for it is easy to see. But not, if we want to be very rigorous.

What @Mr Davis 97 has shown was:

If ##\mathcal{\, A}_1 := \left(\, |a-c| = |b-c| \,\text{ for integer } \; a,b,c \text{ is true } \right)## then ##\mathcal{A}_2 := \left(c = \frac{a+b}{2} \text{ is true }\right)##.

This means, for statement ##\mathcal{A}_1## to be true, it is necessary, that statement ##\mathcal{A}_2## is also true.
It does not say, that statement ##\mathcal{A}_2## implies ##\mathcal{A}_1##, because ##\mathcal{A}_2## could still be true, even if ##\mathcal{A}_1## wasn't.

Therefore, it remains to show, that statement ##\mathcal{A}_2## is also sufficient for statement ##\mathcal{A}_1## to hold.
(cp. @PeroK's remark above)

I admit, it can be done by simply looking at it and doing the calculations in mind, but there might be more difficult examples ahead, in which case it is not as obvious.

So just to clarify, this is a problem of uniqueness, right? Thus, to prove that there is a unique solution, do we just have to prove ##\forall a \forall b \exists c(P(a,b,c) \land \forall r(P(a,b,r) \implies r = c))##, where ##P(a,b,c)## is the predicate that ##|a - c| = |b - c|##. So to show that ##P(a,b,c)## is true for some c, are you saying that I must show that ##|a - c| = |b - c| \Longleftrightarrow c = \frac{a+b}{2}## is true? Would it just be sufficient to show that ##c = \frac{a+b}{2} \implies |a - c| = |b - c| ##?

 

[fresh_42]

So just to clarify, this is a problem of uniqueness, right? Thus, to prove that there is a unique solution, do we just have to prove ##\forall a \forall b \exists c(P(a,b,c) \land \forall r(P(a,b,r) \implies r = c))##, where ##P(a,b,c)## is the predicate that ##|a - c| = |b - c|##.

No, and it is a bit of an overkill here. Formally ##\forall a \forall b \exists c(P(a,b,c)## is your goal. You showed ##P(a,b,c) \Longrightarrow c=f(a,b)##. And as @Ray Vickson pointed out, the uniqueness of expression and the way of deduction already prevents the existence of another solution, or formally: You have shown ##P(a,b,c) \Longrightarrow c=f(a,b)##, so any ##c'## with ##P(a,b,c')## also implies ##c'=f(a,b)=c##.
The missing step is ##P(a,b,f(a,b))##. Does ##f(a,b)## actually fulfill the predicate?

So to show that ##P(a,b,c)## is true for some c, are you saying that I must show that ##|a - c| = |b - c| \Longleftrightarrow c = \frac{a+b}{2}## is true? Would it just be sufficient to show that ##c = \frac{a+b}{2} \implies |a - c| = |b - c| ##?

Yes, this would be sufficient. You don't actually need equivalence, only that ##c = \frac{a+b}{2}## is actually a solution (and that it is an integer!).

Let me illustrate the principle with another example:

##\{ x \cdot (x-2) = 0 \Longrightarrow \text{ case 1: } x=2n \ldots \text{ case 2: } x=2n+1 \ldots \Longrightarrow x \text{ is even } \}## is a true statement, but it doesn't guarantee, that any even number solves the equation.
Of course one would write ## x \cdot (x-2) = 0 \Longrightarrow x \in \{0,2\}## and nobody would ask to check, whether ##0 \cdot (0-2) = 0## and ##2\cdot (2-2)=0## because everybody sees it. That's similar to your example: one can see it.

But imagine you solved a system of differential equations and found a function ##f## as a result. In such a case, one should make sure, that the derived function actually solves the problem. To say it over the top: ##x = 0 \Rightarrow 1 \in \mathbb{R}## is always true, but doesn't solve the equation for ##x##. However, as I said, this is only a principle issue and can often be done by simply mentioning it, because it's obvious or easy done.