EPTI search for it on google, found a lot of hows, but not whys, can some one explain why ? Teaching some math and was doing good till a student ask me why square root of 2 can be written as 2^(1/2) and I got stuck.
WHY!?!? fractional exponents for radicals
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rei
United States3594 Posts
I search for it on google, found a lot of hows, but not whys, can some one explain why ? Teaching some math and was doing good till a student ask me why square root of 2 can be written as 2^(1/2) and I got stuck. | ||
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Seronei
Sweden991 Posts
2^(1/2) is the square root of 2 and the definition of square root is x^(1/2) | ||
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Kazius
Israel1456 Posts
2^1 * 2^(-1) = 2^(1 + (-1)) = 2^0 2^(1/2) * 2^(1/2) = 2^(1/2 + 1/2) = 2^1 | ||
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Slayer91
Ireland23335 Posts
Since the Square root of 2 is a number that is multiplied by itself to give 2, to get 2, that's the same as adding the exponents of 1/2 and 1/2 to get 2^1 Remember the law of indices? 2^1/2 + 2^1/2 = 2^(1/2 + 1/2) =2^1 It's really the definition of square root, that's why its hard to explain. | ||
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Inzek
Chile802 Posts
a*a=a^(1+1) 2^(1/2)*2^(1/2)=2 go backwards and i think you can prove something.. also finally is just notation... like dx/dy is not dx/dy | ||
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Housemd
United States1407 Posts
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xlep
Germany274 Posts
By the definition of arithmetics when using powers (hope thats the correct word) in mathematics: x^a * x^b = x^(a+b) Considering that "square root of x" * "square root of x" = x; "square roof of x" must be x^(1/2) | ||
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MangoTango
United States3670 Posts
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starfries
Canada3508 Posts
ie for x and y =/= 0 x ^ 0 = 1 0 ^ y = 0 0 ^ 0 = 1 | ||
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SirKibbleX
United States479 Posts
2^3 = 8 2^2 = 4 2^1 = 2 2^0 = 1 (Just keep dividing by 2 each time you decrement the exponent) 2^(-1) = 1/2 2^1 * 2^1 = 2^2 2^0 * 2^1 = 2^1, etc. (Exponents multiplied "sum") 2^(1/2) * 2^(1/2) = 2^1 Also remembering that exponents raised to another power 'multiply': 2^(1/2) * 2^(1/2) = [2^(1/2)]^2 = 2^1 It probably makes most sense backwards, by saying something like "What number squared would make two? That number must be two to the one-half power, because exponents 'sum' when multiplied." x^(2) = 2 => x=sqrt(2) | ||
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nalgene
Canada2153 Posts
You can use that to check ( you can also type Engrish words into it if you need to ) is he looking for something like this? 2^3/2 = Sqrt ( 2^3 ) 2^5/2 = Sqrt ( 2^5 ) 2^4/3 = CubeRoot ( 2^4 ) 2^5/4 = 4th Root( 2^5 ) | ||
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andrewlt
United States7702 Posts
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Bibdy
United States3481 Posts
On February 05 2011 03:11 SirKibbleX wrote: This is a very simple explanation: 2^3 = 8 2^2 = 4 2^1 = 2 2^0 = 1 (Just keep dividing by 2 each time you decrement the exponent) 2^(-1) = 1/2 2^1 * 2^1 = 2^2 2^0 * 2^1 = 2^1, etc. (Exponents multiplied "sum") 2^(1/2) * 2^(1/2) = 2^1 Also remembering that exponents raised to another power 'multiply': 2^(1/2) * 2^(1/2) = [2^(1/2)]^2 = 2^1 It probably makes most sense backwards, by saying something like "What number squared would make two? That number must be two to the one-half power, because exponents 'sum' when multiplied." x^(2) = 2 => x=sqrt(2) Here's your answer. Its simply logical progression. If you take the root of x (x^1/2) and square it (multiply the exponent by 2) you get x^1, as expected. | ||
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rei
United States3594 Posts
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micronesia
United States24612 Posts
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Empyrean
16953 Posts
On February 05 2011 03:00 starfries wrote: the 0 exponents blew my mind more ie for x and y =/= 0 x ^ 0 = 1 0 ^ y = 0 0 ^ 0 = 1 The last assertion is incorrect. 0^0 is an indeterminate form. | ||
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Djzapz
Canada10681 Posts
On February 05 2011 05:36 Empyrean wrote: The last assertion is incorrect. 0^0 is an indeterminate form. It still gives 1 on calculators and maple so people don't know that  | ||
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rei
United States3594 Posts
On February 05 2011 05:22 micronesia wrote: Who were you teaching math? What math? Where Why How? 10th graders, advance algebra2,Thomas Jefferson high school Federal way Washington, My reason for teaching is to free students' minds and show them how deep the rabbit hole really goes. The way I teach them depends on the subject at the time and each students' prior experience that i can make a connection to, in other words, i flow like water, because if you put water in a cup it becomes the cup, if you put water in a bowl it becomes the bow, be like water my friend. | ||
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Lemonwalrus
United States5465 Posts
Do it. | ||
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starfries
Canada3508 Posts
On February 05 2011 05:36 Empyrean wrote: The last assertion is incorrect. 0^0 is an indeterminate form. yeah I know, but most people define it as 1 because of how nice it makes math edit: part of the mind-blowing comes from how hilariously complicated some of the arguments are for defining it that way it's one of the thermal exhaust ports in the death star of mathematics that we've nailed a bunch of planks over as a fairly effective but not really satisfying solution | ||
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sidr
United States55 Posts
One can define the function x^k for any k which is a positive integer quite naturally (ie x^k = x*x*...*x k-times). From this, the relation x^(k+m) = x^k * x^m is immediately derivable (x*x*...x k+m times is just x*x*...*x k-times followed by x*x*...*x m-times) and it is this idea that forms the basis for our extension of exponents. Indeed, for positive numbers x, to define x^(1/k) we find a positive real number a such that a^k = n (trying to define this for negative numbers leads to trouble as x^2, for example, never takes on a negative value). The fact that this number exists and is unique follows as the functions x^k are continuous and strictly increasing on [0,Infinity). One can then extend this to positive rationals (n/k) by taking integer powers as above and verify that x^((n/k) + (p/q)) = x^(n/k) * x^(p/q). By defining x^0 = 1 and x^(-a) = (1/ (x^a)), we can extend this to all rationals and guarantee that the additive property of the exponential is maintained (as x^(-a + a) = x^(-a) * x^a = x^a / x^a = 1. The next question is how to define something like 2^x for every real number x. One can check that the above definition of c^x (where c>0) is continuous in the rationals. By some continuity theorems (rationals are dense in [a,b] for every a<b and c^x is uniformly continuous on these intervals, as well as an extension theorem), there exists a unique function c^x defined for ALL real values of x that is continuous and gives us our original function for rationals. After some playing around, one may find this function: specifically, it is exp(x log(c)) (where exp(a) = e^a, e being Euler's number, and log is log "base-e" (or "ln" aka "natural log")). Note that exp and log can be defined independent of our above derivations: log(a) being the integral from 1 to a of (dx/x) for positive a and exp being its inverse function, defined for all reals. Moreover, exp obeys our addition rule. Indeed, the derivative of exp(x+y)/exp(x) with respect to x is just [exp(x)exp(x+y)-exp(x+y)-exp(x)] / [exp(x)^2] = 0, so exp(x+y)/exp(x) is a constant by the mean value theorem. Plugging in x=0 shows this constant is just exp(0+y)/exp(0) = exp(y). From the above, it makes sense to take the definition of a^b as exp(b log a) (for a>0). Knowing analytic expansion and absolute convergence of complex power series allows us to extend this notion to complex exponents. For any real y, we have exp(i y) = 1 + (i y) + (i y)^2 / 2! + (i y)^3 / 6! + ... which converges absolutely for every y. Hence, we may define exp(x+iy) by its series expansion for all complex numbers. A little more series manipulation shows exp(iy) = cos(y) + i sin(y). This, together with the fact that exp(x+iy) = exp(x)exp(iy) allows us to compute exp (and thus a^z for any complex z, positive a) quite easily. As mentioned above, for x+iy = i pi, this gives exp(0)[cos[pi] + i sin(pi)] = -1, ie e^(i pi) + 1 = 0. As for 0^0, it's undefined. From limits like lim(a->0+) a^0 = 1 (a goes to 0 from above) one might think this should be 1, but from the above interpolation with 0 in place of x, one should have 0^x = 0 for x>0 and hence 0^0=0. There is no right answer, the real question is why do you want to know 0^0 in the first place. Usually it's just a limit to be found, and as such should be evaluated without explicitly plugging in 0. On a similar note, 0! = 1 is defined mainly (from what i've seen) from convenience (indeed from the idea that the value of the empty product should be one, as the empty product multiplied by another product should be that product). | ||
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