User talk:Eequor/Archives/vs. MIT

On August 27th, Eequor deleted what in her summary she called a "fallacious argument", the thought-experiment of the calculator that only multiplies. I am appalled, and I put it back. A number of mathematicians have edited this page without complaining about that argument. I've also used it in teaching basic combinatorics in probability courses I've taught at several universities. One of those was MIT, where some of the student are exceedingly mentally acute, and no one complained. Could those concerned venture their opinions here? Michael Hardy 23:49, 26 Sep 2004 (UTC)

The most glaring error is the idea that a calculator which can only multiply must necessarily continue to usefully function if its value is cleared. It may easily be claimed that, in fact, the calculator is functioning exactly as designed if 3 is entered after clear is pressed and 0 is the result. It is probably buggy, and will never again produce any number other than 0, but it does what you said it should do. No further conclusion can be drawn from this, anyway, because exactly two numbers are being multiplied every time, not zero numbers.

Another problem: the argument supposes that the displayed value must be identical to the value stored in memory. This may sometimes be true in real life, but it is not necessary. For example, calculators must often round fractional values so they will fit on the display, but many calculators continue to store the least significant digits. The imaginary calculator might display 0 while storing 1 internally.

Additionally, the argument is inconsistent with itself. It supposes a calculator which "can only multiply", and then adds that the calculator also has a clear function. There is also no definition of what clear might do. Because of this, the result of pressing the clear key after multiplying 21 by 4 is undefined, and so no conclusions about the future behavior can be made.

One could define the clear key to mean "remove the current number and display 1", but this leads to a circular argument, additionally neglecting to demonstrate why the result cannot be 0. Defining the key as replacing the previous value with 0 gives an apparent reductio ad absurdum, but not of a sort that makes any conclusion about the definition of the empty product (it only shows the clear key must not produce 0 if it is desired that the calculator will continue to function). The key must be defined to produce no numerical value. Call this value nil, and let the calculator display nothing at all if its value is nil.

As commonly understood, a calculator displaying nothing accepts the next-input value as its new value (entering 3 when the value is nil produces 3). Since the calculator is claimed to only multiply, and because the only value which when multiplied by a number n produces exactly n is the multiplicative identity, 1, nil must be numerically equivalent to 1.

Now, consider the behavior of the calculator immediately after pressing clear. It is blank, and will remain blank indefinitely until a number is entered, yet when a number is entered the blankness will be numerically equivalent to 1. No numbers are multiplied until enter is pressed. Therefore the empty product is equal to 1. --[[User:Eequor|ηυωρ]] 01:11, 27 Sep 2004 (UTC)

I find the arguments above largely correct if construed literally, but completely lacking in merit, for reasons I would have thought were obvious. They're written by someone who is too literal-minded. Michael Hardy 19:56, 27 Sep 2004 (UTC)

No, you're simply incorrect. You needn't be rude when you're shown to be wrong. If this is the quality of education provided by MIT, I consider myself fortunate to have not been a student there. --[[User:Eequor|ηυωρ]] 00:47, 30 Sep 2004 (UTC)

I was not shown to be wrong. A somewhat hand-waving argument was given; you showed that there were holes in it if construed literally, but it's not-quite-literal meaning should have been clear. Another user has taken some trouble to rephrase it in view of your comments. What exactly is it you're calling rude? My statement that the fact that the meaning should have been obvious? Some people consider excessive literal-mindedness rude. Michael Hardy 01:46, 30 Sep 2004 (UTC)

Ad hominem is inherently rude.

"Ronald Reagan is a Republican, therefore the argument he just gave is wrong." That's ad hominem. But I don't think it's inherently rude, although fallacious, nor that Ronald Reagan would be offended by being called a Republican. Michael Hardy 20:26, 30 Sep 2004 (UTC)

Whether a person is "too literal-minded" is a matter of opinion.

OK, your removing a very good argument is literal-mindedness. OK? Michael Hardy 20:26, 30 Sep 2004 (UTC)

As written, your previous statement implies literal-mindedness should have been obvious, not the meaning of the thought experiment.

While its general meaning should be obvious to laypeople, and most people will probably accept it without question, the experiment has been presented poorly and makes its conclusion for the wrong reasons. This is an encyclopedia, not a grade school textbook. Articles should be made readable without sacrificing accuracy. Science does not accept arguments simply because they are "good enough", and neither should Wikipedia. --[[User:Eequor|ηυωρ]] 02:13, 30 Sep 2004 (UTC)

Eequor, you are rude and gratuitously belligerent. I have been polite in addressing you and have tried consistently to reconcile, but you persist in belligerence. Michael Hardy 20:26, 30 Sep 2004 (UTC)

I see that Eequor has deleted the rest of this discussion.

For the record, I was not shown to be wrong, nor was I rude, nor was there an ad hominem. "Literal-minded" applied to the nature of the arguments, not to the person. Michael Hardy 21:25, 8 Oct 2004 (UTC)