Trying to understand the mistake in my logical reasoning

[issacnewton]

Hello
I am in middle on solving problem 17 from Chapter 3 of Spivak's Calculus. We have a function $f(x)$, which is a non-zero function and it obeys the following properties.
$$\forall \, x \, y \, [f(x+y) = f(x) + f(y)]$$
$$\forall \,x \, y \, \left[f(x \cdot y) = f(x)\cdot f(y)\right]$$
We have to prove that $f(x) = x$ for all x. Using the quantifier, the goal becomes
$$\forall\, x [ x \in \mathbb{R} \Longrightarrow (f(x) = x)]$$
Now here is my proof, which I am sure is not correct, but I am trying to understand my flaw.
Let $x$ be arbitrary. Assume $x \in \mathbb{R}$. Now using the first given property of f , we can see that $f(0+0) = f(0) = f(0) + f(0)$. It follows that $f(0) = 0$. Now our goal is to prove that $f(x) = x$ for some arbitrary x. I am going to try indirect proof here. Assume $f(x) \ne x$. Then due to the property of trichotomy, we have either $f(x) > x$ or $f(x) < x$. For the case 1, we will assume $f(x) > x$. Since $x$ is arbitrary, this is also valid for $x=0$. So we have
$f(0) > 0$. Since $f(0) = 0$, it follows that $0 > 0$. We reach a contradiction here, so our assumption that $f(x) > x$ is wrong. Similarly, we can prove that $f(x) \nless x$. So it must follow that $f(x) = x$. The problem with this proof is that I have proven $f(x) = x$ only with the knowledge that $f(0) = 0$. But all kinds of functions have the property that $f(0) = 0$, and it should not necessarily follow that $f(x) = x$.

I think I am using the universal quantifier in a wrong way here. Any guidance will help...

 

[Orodruin]

Since x is arbitrary

This is a logical flaw. It is enough that there exists one x for which f(x) is not equal to x. You cannot just pick 0.

 

[Orodruin]

In logical terms, the negation of the statement you made is:
$\exists x : f(x) \neq x$
not
$\forall x : f(x) \neq x$
The former is the one you need to show results in a contradiction in order to have a proof by contradiction.

 

[issacnewton]

Hello Orodruin
My initial goal is $\forall\, x [x\in \mathbb{R} \Longrightarrow (f(x) = x)]$. In such case, we choose arbitrary x and then assume antecedent. So I assume tat $x \in \mathbb{R}$ and hence my new goal becomes $f(x) \ne x$. Now can't I use the method of indirect proof for this new goal ?

 

[Orodruin]

Hello Orodruin
My initial goal is $\forall\, x [x\in \mathbb{R} \Longrightarrow (f(x) = x)]$. In such case, we choose arbitrary x and then assume antecedent. So I assume tat $x \in \mathbb{R}$ and hence my new goal becomes $f(x) \ne x$. Now can't I use the method of indirect proof for this new goal ?

In principle, but x = 0 is not arbitrary, it is a specific choice of x. By your reasoning, you could also prove that x = x^2 for all x. It is you task to show that the existence of an x for which the statement f(x) != x leads to a contradiction. You do not get to arbitrarily assume that that x is zero.

 

[issacnewton]

I think I am getting what you are trying to say... What university you are professor at ?

 

[Orodruin]

I think I am getting what you are trying to say... What university you are professor at ?

I fail to see how this is relevant for the issue at hand. Please stay on topic.

 

[issacnewton]

Please don't get angry. I was inquiring because I liked your reasoning in mathematics but you have degree in physics.

 

[PeroK]

Please don't get angry. I was inquiring because I liked your reasoning in mathematics but you have degree in physics.

He is a man of many talents.

Let me use your logic to prove something quite interesting.

Let $f$ be a function with $f(0) = 0$

I'm going to show that $f(x) = 0$ for all $x$.

Suppose not, then we have an $x$ for which $f(x) \ne 0$, but as $x$ as arbitrary this must hold for $x = 0$ hence $f(0) \ne 0$. Which is a contadiction.

So, $f(0) = 0 \ \Rightarrow \ \forall x \ f(x) = 0$

Now, where's the flaw in that?

 

[issacnewton]

You are negating an universal quantifier and you get an existential quantifier. That gives you one particular x. This x may not be equal to zero. So we can't equate it to zero. Is that right ?

 

[PeroK]

You are negating an universal quantifier and you get an existential quantifier. That gives you one particular x. This x may not be equal to zero. So we can't equate it to zero. Is that right ?

Yes. One way to avoid this mistake is to use $x_0, x_1$ etc. or $a, b$ for specific values of $x$ and reserve $x$ for when you mean all $x$.

So, to edit my post above:

Let $f$ be a function with $f(0) = 0$

Let's try to show that $f(x) = 0$ for all $x$.

Suppose not, then we have an $x_0$ for which $f(x_0) \ne 0$ ...

And there the attempted proof grinds to a halt. There is no reason to suppose that $x_0 = 0$. In fact, the one thing we do know is that $x_0 \ne 0$.

 

[issacnewton]

Thanks Perok... it makes sense now. I just have a general question about the quantifiers. These quantifiers were introduced to the mathematics in 19^th century. But people have been proving theorems before that too. So how would people like Gauss, Laplace, Jacobi prove those theorems. ?

 

[mfb]

Probably with words instead of those symbols.