psi20 Posted February 25, 2004 Share Posted February 25, 2004 If you had b^(m/n), is it a) nthroot of (b to the mth) b) (nthroot of b) to the mth c) either d) both Link to comment Share on other sites More sharing options...
NSX Posted February 25, 2004 Share Posted February 25, 2004 All of the above. Link to comment Share on other sites More sharing options...
psi20 Posted February 25, 2004 Author Share Posted February 25, 2004 Now I'm really confused! When I say both, I mean both a and b. When I saw either, I mean either a or b. Link to comment Share on other sites More sharing options...
NSX Posted February 25, 2004 Share Posted February 25, 2004 Okay, then d Link to comment Share on other sites More sharing options...
Dave Posted February 26, 2004 Share Posted February 26, 2004 Use the laws of indices: [math]\(a^{b}\)^{c}=a^{bc}[/math]. So [math]a^{m/n}=\(a^{m}\)^{1/n}=\(a^{1/n}\)^{m}[/math] Link to comment Share on other sites More sharing options...
psi20 Posted March 7, 2004 Author Share Posted March 7, 2004 Oh yeah, that makes sense. But now this doesn't make sense. If 4^(1/2) = +/- 2 Then 8^(1/3) = +/- 2 too? 8^(1/3) = (8^(2/3))^(1/2) But if -2 is a root of 8^(1/3), then why doesn't -2^3 = 8? Or maybe -2^3 = 8? Also look at this: 2^2 = 2^(4 x 1/2) = etc. etc. and that means 2^2 = +/- 4. Link to comment Share on other sites More sharing options...
NSX Posted March 7, 2004 Share Posted March 7, 2004 psi20 said in post # :But now this doesn't make sense. If 4^(1/2) = +/- 2 Then 8^(1/3) = +/- 2 too? 8^(1/3) = (8^(2/3))^(1/2) But if -2 is a root of 8^(1/3), then why doesn't -2^3 = 8? Or maybe -2^3 = 8? [math]\sqrt{4}[/math] :neq: [math]\pm2[/math] psi20 said in post # :Also look at this: 2^2 = 2^(4 x 1/2) = etc. etc. and that means 2^2 = +/- 4. No it doesn't. I think the problem you're encountering is that the [math]\pm[/math] only arises when you're solving for the argument being squared, cubed, etc. ie. if [math]x^2 = 4[/math], then [math]x=\pm2[/math] Likewise, [math]\sqrt{4}=2[/math] & [math]-\sqrt{4}=-2[/math] -- i see mimeTeX doesn't like spaces. Link to comment Share on other sites More sharing options...
Dave Posted March 7, 2004 Share Posted March 7, 2004 This was a great source of confusion for my maths class a few years back when we were learning the basics. Say if you have [math]x^2-4=0[/math]. Then [math]x^2=4[/math]. So you want to find a value for [math]x[/math] such that when you square it, it decides to become 4. However, when you draw the graph of the function, it becomes fairly clear that there's actually 2 solutions. The problem is that when you square a negative number, the answer is positive. So in this case, the answer would involve a plus/minus sign. -- I noticed this a while back, all the math tags actually do is encapsulate the stuff you put in them and execute a cgi script from the looks of it. You can't really have spaces in the request you send to the webserver easily so I think mimeTeX just ignores them. Link to comment Share on other sites More sharing options...
psi20 Posted March 7, 2004 Author Share Posted March 7, 2004 Ah man, the pictures don't show up. They are boxes with the red x in them, can you show me it without pictures? I once got replies of the square root of 4 equalling +/- 2. So the square root of four isn't -2? I get that if x^2 = 4, x = +/- 2. But I'm wondering if an expression sqroot 4 = +/- 2 Link to comment Share on other sites More sharing options...
fafalone Posted March 7, 2004 Share Posted March 7, 2004 When you multiply two negative numbers together, you get a positive number. Hence -2 is also a solution for x2 = 4. Link to comment Share on other sites More sharing options...
psi20 Posted March 7, 2004 Author Share Posted March 7, 2004 Yup, I know that, but is 4^0.5 = +/-2? Link to comment Share on other sites More sharing options...
NSX Posted March 7, 2004 Share Posted March 7, 2004 NSX said in post # : [math]\sqrt{4}=2[/math] & [math]-\sqrt{4}=-2[/math] It's just like graphing [math]y=\sqrt{x}[/math] with a graphing calculator, or other graphing utility. In order to properly create [math]y=x^2[/math]'s inverse, you have to graph both [math]y=\sqrt{x}[/math] and [math]y=-\sqrt{x}[/math]. Link to comment Share on other sites More sharing options...
fafalone Posted March 8, 2004 Share Posted March 8, 2004 psi20 said in post # :Yup, I know that, but is 4^0.5 = +/-2? No. That's performing a power operation on a number, not solving an equation. Link to comment Share on other sites More sharing options...
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