Math Help - Modus Ponens

1. Modus Ponens

Hi

Here we go, we are told the first two propositions are true. We're asked is it valid and if so give a deduction by modus ponens and proof by contradiction.

(i)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave cannot solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is not quadratic.

(ii)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is quadratic.

This is probably really simple but I just keep going around in circles, can anyone explain?

Thanks

2. Originally Posted by bobred
We're asked is it valid and if so give a deduction by modus ponens and proof by contradiction.
(i)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave cannot solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is not quadratic.

(ii)
We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
We conclude that:
The equation f(x) = 0 is quadratic.
This is not correct. You cannot make any conclusion here.

3. Hi

Sorry for being vague, we are asked which of the two is a valid deduction.
For the valid one give a deduction involving modus ponens and proof by contradiction. For the invalid one explain why not.

Thanks

4. Can anyone help?

Thanks

5. Originally Posted by bobred
Can anyone help?
The first is valid.
The second is invalid. The fallacy is call affirming the consequent.

6. I agree with the above, but I'll add a bit more to illustrate Plato's point

We know that:
If the equation is quadratic, then Dave can solve it;
Dave can solve the equation f(x)=0.
What if f(x) is, for example, a linear equation? If Dave can solve the equation f(x) = 0, where f(x) is a linear equation, then it is (obviously) not quadratic.

Arguement

X implies Y
Y
--
X <-- false. There could be a third cause, say, Z for Y.