# A Circular reasoning and proof by Contradiction

1. Sep 29, 2016

### e2m2a

I need to understand something about proof by contradiction. Suppose there is an expression "a" and it is known to be equal to expression "b". Furthermore, suppose it is conjectured that expression "c" is also equal to expression "a". This would imply expression "c" is equal to expression "b".

Now here is where I might be naive. I think that one direct way to prove that expression "c" is not equal to expression "b" is to simply subtract the two expressions. That is, I assert that expression "b" is equal to expression "c" if and only if you get zero when you subtract them. If not, then they are not equal.

In fact, what if you get the equality expression, after subtracting "b" and "c", an expression that reads b = c.

Well, I think this is circular reasoning. The answer gives what we are trying to prove that which we assume to be true. Is this a proof by contradiction that expression "c" cannot be equal to expression "a"?

2. Sep 29, 2016

### andrewkirk

I don't understand what this is saying. What did you intend it to mean?

3. Sep 29, 2016

### e2m2a

I need to be more specific. Suppose it is known to be true that "a" equals a function a(w,x,y). Next we solve for y, getting a function y(w,x,a) Now we want to test if expression "c" is equal to "a". So we assume it is true, then proceed and find that "c" equals a function c(uw,vx,y). Then we solve for y, getting a function y(uw,vx,c) Now both expression "a" and expression "c" have functions for y. In expression "a", it is y(w,x,a), in expression "c", it is y(uw,vx,c).

Now, we are trying to prove if expression "a" is equal to expression "c". This would imply that y(u,x,a) should be equal to y(uw,vx,c)... So, we subtract the two functions to see if it always equals zero. I assert this would prove that expression "a" is equal to expression "c". But, if instead you get the answer a = c, and not zero, then this is a circular argument. My question is, is this a proof by contradiction that "a" cannot be equal to "c", the fact that it involves circular reasoning?

4. Sep 29, 2016

### andrewkirk

I still can't understand it. The problem is the notation. You cannot use a symbol like $a$ or $y$ to denote both an amount and a function. An amount is a number. A function is a rule that, given a specified number of input numbers, gives an output number. Any given symbol must be one or the other. It cannot denote both.

So it creates chaos to write things like 'Then we solve for y, getting a function y(uw,vx,c)' because you are trying to use $y$ to refer to both an amount and a function. What you should write is something like 'we solve for $y$, thereby deriving a function $f:\mathbb R^3\to \mathbb R$ such that $y=f(uw,vx,c)$'.

Try rewriting your question in a way that clearly distinguishes between amounts and functions. Then it may be clear enough to get help.

Last edited by a moderator: Sep 30, 2016
5. Sep 30, 2016

### e2m2a

Ok. Let me try to give a concrete example. Suppose we know that it is true that a = ku - v. This would imply v = ku - a. Next, suppose we conjecture that c = a. This would also imply that c =ku -v. And this would imply that v = ku - c. Finally, this would imply that ku - a = ku - c. But this would only be true if and only if c = a. But this is assuming to be true what needs to be proven. Circular reasoning. By this line of reasoning, does this prove that c could never be equal to a?

6. Sep 30, 2016

### andrewkirk

No. To prove that, a contradiction needs to be deduced, and your post does not deduce a contradiction.