psi20 Posted February 25, 2004 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
psi20 Posted February 25, 2004 Author 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.
Dave Posted February 26, 2004 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]
psi20 Posted March 7, 2004 Author 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.
NSX Posted March 7, 2004 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.
Dave Posted March 7, 2004 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.
psi20 Posted March 7, 2004 Author 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
fafalone Posted March 7, 2004 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.
NSX Posted March 7, 2004 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].
fafalone Posted March 8, 2004 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.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now