tag:blogger.com,1999:blog-2084641237928654834.post7297159646540717002..comments2017-12-15T03:38:41.639-05:00Comments on Algebra, Essentially: Why aren't exponents and roots always inverses?Emily Allmanhttp://www.blogger.com/profile/08966304042607333303noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2084641237928654834.post-81123763978925624992015-10-06T21:00:29.184-04:002015-10-06T21:00:29.184-04:00Thank you, Christopher Danielson, for pointing me ...Thank you, Christopher Danielson, for pointing me to your post about logs and non-commutative binary operations: https://christopherdanielson.wordpress.com/2011/12/07/enough-with-the-logs-already/<br /><br />I think this is getting me somewhere. Somehow, the fact that exponents are not commutative will throw off the ease with which inverse functions are grasped. For example, it is true that f(x) = x+3 and f(x) = 3+x will both yeild the same inverse function f-1(x) = x-3. What casues problems is this: if g(x) = x - 3, then g-1(x) = x+3... BUT if g(x) = 3-x, then g-1(x) is NOT 3 + x. Since the original g function is not commutative, we cannot use the same 'thought process' to invert them when the arguments are reversed. I like.Emily Allmanhttps://www.blogger.com/profile/08966304042607333303noreply@blogger.comtag:blogger.com,1999:blog-2084641237928654834.post-75758926194937683142015-10-06T20:41:13.909-04:002015-10-06T20:41:13.909-04:00Thanks Mike. I like how you ask another question t...Thanks Mike. I like how you ask another question that's also interesting. I HAVE done the compositions: f(g(x)) and also g(f(x)) and see of course that they are not x (although, you are right, they CAN be, which IS interesting). My students have only a rudimentary understanding of inverses and compositions at this point, so it would be useless to use this method as a rationale for them. They were satisfied with simply finding a counterexample: g(f(3)) is not equal to 3, so this cannot be the inverse. But to them, the xeth root of n seemed like such the logical choice for the inverse. After all, x^n does indeed yield nth root of x as its inverse. How can I explain this quirk to them so that they feel secure in their understanding of inverses overall (which, BTW, i think is excellent)?Emily Allmanhttps://www.blogger.com/profile/08966304042607333303noreply@blogger.comtag:blogger.com,1999:blog-2084641237928654834.post-40867257693602501132015-10-06T19:04:00.418-04:002015-10-06T19:04:00.418-04:00For the sake of being able to write this out in a ...For the sake of being able to write this out in a comment, let's call your inverse function g(x). If they're truly inverses, then g(f(x)) = x. The reason that this doesn't work becomes more obvious when you build the composite statement. Fleshing out the composite function, we get that (n^x)√(n) = x (if it isn't obvious, the n to the x-eth root of n equals x). Looking at it this way, it's clearly a very rare scenario where this would work. I think the more interesting question given those to functions is to ask when, if ever, g(f(x)) = x.Mikehttps://www.blogger.com/profile/17209841472729555694noreply@blogger.com