# I Trying to understand the mistake in my logical reasoning

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1. Sep 19, 2016

### 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...

2. Sep 20, 2016

### Orodruin

Staff Emeritus
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.

3. Sep 20, 2016

### Orodruin

Staff Emeritus
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.

4. Sep 20, 2016

### 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 ?

5. Sep 20, 2016

### Orodruin

Staff Emeritus
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.

6. Sep 20, 2016

### issacnewton

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

7. Sep 20, 2016

### Orodruin

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

8. Sep 20, 2016

### issacnewton

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

9. Sep 20, 2016

### PeroK

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?

10. Sep 20, 2016

### 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 ?

11. Sep 20, 2016

### PeroK

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 $x_0 \ne 0$.

12. Sep 20, 2016

### issacnewton

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

13. Sep 21, 2016

### Staff: Mentor

Probably with words instead of those symbols.