71–80 of 136 posts
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What is true in mathematics is whatever leads to no logical contradictions.
There are cases when both a claim and its negation do not lead to contradiction, but them both being true obviously does.
I'll stick with the grade-school math approach, at least until I need to approach it differently. 4^2 = 2 fours multiplied = 4 * 4 = 16 divide by 4 - so you take away one of the 4s by division(canceling like terms like we do in grade school fraction math): 4^1 = 4*4/4 = 4 divide by 4 again 4^0 = 4*4/(4*4) = 1 divide by 4 again! 4^-1 = 4*4/(4*4*4) = 1/4 Now try it with 0: 0^2 = two zeros multiplied = 0*0 = 0 Divide by…
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What is "indeterminate form"? http://en.wikipedia.org/wiki/Indeterminate_form You see, 0^0 = 1, and it's obvious to a mathematician . . . we define 0^0 = 1, to be consistent with exponentiation rules Well, you're going to be inconsistent with them no matter how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.…
By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one. 0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this…
Mathematician here; we do not. See http://math.stackexchange.com/questions/11150/zero-to-zero-p....
More precisely, as Arturo Magidin points out at http://math.stackexchange.com/questions/11150/zero-to-zero-p..., if we view exponentiation in the 'discrete setting', then $0^0$ must be $1$; whereas, if we view it in the continuous setting, there is simply no good answer—unlike $e^{i\pi}$, which also lives in the continuous setting, but has a perfectly good, unambiguous answer. (lotharbot gives a nice explanation below of the ways that this is consistent with existing mathematics; but it can also be derived from the definition of the exponential function, with no further arbitrary conventions needed.)
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>most of us don't have a good conceptual model for what exponents really do Instead of matching math to real world objects (1= one banana, 2 = two bananas, 1+2 = 3 bananas etc. ) and building up to exponentiation, multiplication etc. thereby introducing all sorts of paradoxes, group theory dodges all that and treats the whole thing as a very consistent rule-based system. Things fall into place quickly once the rules…
Group theory is formed by abstracting out the observed properties of number systems. If you want to show that some collection is a group, you will have to do the computations to show that they follow the rules, in which case, it helps to have an understanding of the mechanics of the computation.
My claim is the exact opposite. I claim you don't need to understand the mechanics ( just blindly abide by the rules of the group or abelian group or finite simple group or whatever), which is why the approach is better. If you show a monkey red means stop and green means go and reinforce these rules by rewarding with a banana, eventually the monkey will stop when he sees the red. Not because he understands the mechanics of traffic management. Simply because he is abiding by the rules. Similarly, large portions of math can be approached by either the definitional route ( ie. rules ie. define propositions & theorems that logically follow if those props held ) or via trying to understand actual mechanics by mapping everything to real world phenomena ( x = distance, dx/dt = velocity, d/dt(dx/dt) = acceleration etc. ) which are problematic because the mapping breaks down due to the nature of physical reality ( like friction etc. )
How would one explain say Hilbert's 7th problem via the actual mechanics ?
If a is algebraic and b is irrational show a^b is transcendental.
What does that even mean when you map them to the real world ? Instead, the solution is to build upon theorems that logically follow from the axioms you start out with. Problem: http://en.wikipedia.org/wiki/Hilbert%27s_seventh_problem Solution: http://terrytao.wordpress.com/2011/08/21/hilberts-seventh-pr...
0 ^ 0 = 0, on all architectures.
Earlier quoted context omitted.
What is "indeterminate form"? http://en.wikipedia.org/wiki/Indeterminate_form You see, 0^0 = 1, and it's obvious to a mathematician . . . we define 0^0 = 1, to be consistent with exponentiation rules Well, you're going to be inconsistent with them no matter how you define it, since, as you point out, x^y should be zero if you approach (0,0) along the x=0 axis, and it should be one if you approach along the y=0 axis.…
By "exponentiation rules" I mean algebraic equalities, like a^x * a^y = a^(x+y). Most of them work no matter if you define 0^0 = 1 or 0, but some of them are cleaner with 0^0 = 1. It's also consistent with cardinal and ordinal exponentiation (look it up). "Approaching along x axis" is not algebraic notion, it's analytic one. 0^0 makes no less sense than, say, -e^(i pi). They're both 1 because we define them like this…
As an erstwhile mathematician, I don't know of anyone who would suggest that -e^(i pi) is defined as 1. It falls out, rather elegantly, from the definitions of the constants involved.
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There are cases when both a claim and its negation do not lead to contradiction, but them both being true obviously does.
Thus, mathematics unexpectedly turns into a Choose Your Own Adventure novel!
Technically, 0^0 is an indeterminate form and has no specific solution. Accurate but unhelpful. Practically, 0^0 highlights the issue that most of us don't have a good conceptual model for what exponents really do. How would you explain to a 10-year old why 3^0 = 1 beyond "it's necessary to make the algebra of powers work out". I use an "expand-o-tron" analogy http://betterexplained.com/articles/understanding-exponen…
X to the 0.5 is square root, X to the 0.3 is cube root, X to the 0.25 is fourth root, etc
Therefore X^0 is what you get if you're taking the 'infiniteth' root
= 1
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Your parent never says anything about arbitrary axioms. To your main point: are irrational numbers, say, "out there"? If so, where? Until a few centuries ago it was mathematical standard practice to fudge 1/2 as 25/49 when taking it's square root. But then mathematicians invented (some would argue) the notion of an irrational, because it was, well... useful. There are real metaphysical questions here; there have been…
To your main point: are irrational numbers, say, "out there"? If so, where? On the sheet of paper in front of me, as the hypotenuse of the isosceles right-angled triangle with unit length I drew a moment ago. Irrational numbers are probably a bad example of what you're talking about.
Doesn't the quantised nature of matter mean that 2^½ exists as a real measure of a material object only as much as a perfect circle actually is existent in our universe?