Ben

If you you truly believe all these delusions, I strongly suggest you seek medical help

What? You never heard of the [imath]\frac{1}{2}[/imath]-th dimension? Shame on you! Next you'll say you never heard of the [math]\pi[/math]-th dimension or the [math]e[/math]-th dimension.... What a troll. PS This a Math and Physics forum. "Dimension" here has a rather exact meaning.

So now both Qfwfq and Turtle refer to me as "bin"? How very rude. Surely there is some forum rule about mutating a member's name in a derogatory sense? There certainly should be....

OK, let's do it! First I apologize to anyone who find the following patronizing, but it illustrates an important general point in mathematics. If, for integers [math]x,\,y,\,a[/math] I say that [math]x = y \mod a[/math] I mean simply that [math]x[/math] and [math]y[/math] differ by the integer [math]a[/math]. That is [math]x-y=a[/math], say. This is called a "congruence" and has the following properties. [math]x=x \mod a[/math] (reflexivity) [math] x= y \mod a \Rightarrow y=x \mod a[/math] (symmetry) [math]x = y \mod a,\,\,\, y = z \mod a \Rightarrow x = z \mod a[/math] (transitivity) These three properties define the modulo relation as an equivalence relation, and these crop up all over the place in mathematics (usually masquerading as isomorphisms or even equalities!). So now let's set [math]a = 9[/math], and we easily see that any member of the set [math]\{....,-18,-9,0,9,18,.....\}[/math] can be written as [math]x=y \mod 9[/math]. This set is called an "equivalence class" and it is customary to elect a class representative, and write [math]\{....-18,-9,0,9,18,....\} \equiv [0][/math] although this choice is entirely arbitrary. Likewise we will have that [math]\{....-17,-8,1,10,19,....\} \equiv [1][/math] and so on up to the class [math][8][/math]. Right, so the number, say [math]456[/math] is really just shorthand for [math]400+50+6[/math], and since we easily see [math]400 =4 \mod 9,\,\,\,50 = 5 \mod 9,\,\,\,6=6 \mod 9[/math] then it is no great shock to learn that [math]4+5+6 = 456 \mod 9[/math]. Likewise it is expected that [math]4+5+6 = 15 = 1+5 = 6[/math] is in the same equivalence class as [math]456[/math] so that [math]6 = 456 \mod 9[/math], it doesn't matter which class we are in. Now by the definition of this particular sort of equivalence relation, we must have that subtracting one member of an equivalence class from another member of the same class sends the result to the class [math][0][/math] of which [math]9 [/math] is a member. So from the above we will have, no matter how many digits we start with, proceeding recursively we will always have the result to be 0 or 9. There are any number of "mind reading" party tricks you can devise using this sort of arithmetic, say mod 2, since it will be a closed book to most of your awe-struck audience

Now, now, there is no need to be rude. First this: you realize, of course that it is NOT true in general that [math]Ta^x = (Ta)^x[/math]? I know you never asserted as much, just want to clear the decks, as it were. Let's quote your monster so-called "identity" so we can keep our bearings. Here it is: [math] T*a^x= \left(T*a\right)^{\frac{\frac{x*\ln(a)}{\ln(T)}+1}{\frac{\ln(a)}{\ln(T)}+1}} [/math] I claim this is false when [math]x = 0[/math]. For, [math] Ta^0= (Ta)^0 =T,\,\,\, \frac{0\ln(a)}{\ln(T)}+1= 1[/math] in the numerator, and [math]T[/math] can be whatever we want. Agreed? So let's assume, for a contradiction, that [math]T \ne T'[/math] and set [math]a =1[/math] So that [math]\frac{0}{ \ln T'} +1 =1[/math] in the denominator which implies that [math]T = T'[/math], contrary to our datum

It gets worse!! Try this party trick, amaze your friends.... 1. Take any number at all with more than 1 digit and add the digits. 2. If the result has more than one digit, add again 3. Now subtract this single-digit number from your starting number (not added) 4. Add the resulting digits again. 5. I bet it will be the number 9 Can you see why? Once you do, you will see that the restriction in (1) to more than a single digit is not required. Hint: What Sanctus calls "base 10 arithmetic" could equally well be called "modulo 9 arithmetic" - they are very very intimately related. In fact there is a sense in which they are the same thing

Something like this? Which I am sure all here will agree with. Oh wait, what's that noise? Sounds like an axe being ground - maybe my hearing isn't all that it used to be......

No, it is merely a convention; you do not need to adopt if you don't want to. So, suppose for example we ditch 1 as the multiplicative identity and choose, say, 2. So that 2x = x. This is OK, since there exists a one-to-one correspondence between the infinite sets, say, [math]\{1,2,3,...\}[/math] and [math]\{2,4,6,...\}[/math] and we have that [math] x = \frac{x}{2}[/math]. This is still OK. But the problems start here: Let's agree that [math] \ln(xy) = \ln(x) + \ln(y)[/math], so that [math]\ln(2y) = \ln (y)=\ln(2) + \ln(y)[/math]. So we are still left with the problem that we MUST have that [math]\ln(2) =0[/math], unless you are prepared to ditch the additive identity also. Are you?

Oddly enough I don't posses a calculator so I cannot accept your invitation. So is it your contention that an identity can be absolutely true mod some exceptions? This is the thrust of Don's claim. My logic says otherwise; maybe "conditionally true" might be better? Hardly interesting though. PS I confess I earlier overstated my case: "not absolutely true" is not equivalent to "absolutely false"

Then show it. In my ignorance I get that [math]7+7 = 14[/math] and yet [math]49-49=0,\,\,\, 7-7=0 ,\,\, \frac{0}{0} \ne 14[/math]. Or anything else, as far as I can see In fact it is absolutely false! Have you any idea how to do natural logarithmic arithmetic? Yes, but it's not interesting; [math] Ta^x = a^x[/math] when [math]T=1[/math]. Let's write home to mother about this Earth-shattering discovery. But you still have a problem with your RHS - it is complete and utter nonsense. Can you not see why?

I agree with with Q. If [math]T = 1[/math] then [math]Ta^x = a^x[/math], so that [math] x={\frac{\frac{x*ln(a)}{ln(T)}+1}{\frac{ln(a)}{ln(T)}+1}}= 1[/math] and therefore [math]Ta^x = a^1 = a[/math] Proof: Given that [math]\ln 1 = 0[/math] then this implies that the numerator [math]\frac{x \ln a}{0} +1[/math] is crap. Likewise the denominator [math] \frac{ \ln a}{0}+1 =[/math] crap. And the quotient of crap by crap is quite clearly one. This completes the proof. Hint: Division by zero is not legal under any jurisdiction outside of mental institutions

Blooped.

Sorry, I don't fully understand this statement. First by "we" I assume mean you and your fellow scientists. Then you and I are colleagues! Second by "not adding a special effect in time" I assume you mean that when you and I, as scientists, perform an experiment on Monday,we expect the same result of the same same experiment on Tuesday. Or did we, as scientific equals, misunderstand each other? I don't know about you in your daily practising scientific life, but if I cannot repeat on Tuesday the result I got on Monday, the last thing I would blame is time. How do I "tweak time"? And what in the name of all that is holy is "time potential"? No. Because it is pure gibberish.

Answer: Because you have no choice. That is the definition of the domain of a binary operation (like multiplication. Or addition, if it comes to that). In spite of what else you might have read here...... Yes, it does take a little time, but not that much (with practice). Plus it makes one's posts sooooo much more readable

Lemme get this straight. Given [math]*:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z},\,\, 12 \in \mathbb{Z}[/math] you want that both [math](2,6)[/math] and [math](3,4)[/math] as pre-images [math]*^{-1}(12)[/math] to be defined as even in [math]\mathbb{Z} \times \mathbb{Z}[/math]? Bizarre. The usual definition of an even number [math]x[/math] is that there is some umber [math]y \ne x[/math] such that [math]x = 2y[/math] (or some variant of this). I repeat, in spite of your definition of the evens in the Cartesian product, I don't think the notion of oddness and evenness has any real meaning for ordered pairs. Look, you can mock me all you want - anyone can, I don't mind - but your humour doesn't exempt you from the obligation to use language unambiguously. Had you not earlier referred to something called "the Cartesian square"? Is 25 a Cartesian square? Is there some number [math]n \in \mathbb{Z}[/math] such that [math]n^2[/math] is a Cartesian square? Clearly this is not what you meant, but I don't read minds, just words.