+0  
 
+1
540
8
avatar+12 

Furthermore, how/why do powers of non-integers work? For example, why would (-1)^1.5 be -i?

 Nov 10, 2017
 #1
avatar+637 
0

(-1) ^ 1.5 is 1.

 

2^0.5 is basically just multiplying 2 0.5 times, which is square root. 

 Nov 10, 2017
 #2
avatar+12 
0

I'm not sure I understand. 2*0.5 is NOT sqrt(2).

Epic8009  Nov 10, 2017
 #4
avatar+99282 
0

2^0.5   IS   sqrt(2)

Melody  Nov 11, 2017
 #3
avatar+99282 
+2

(-1)^1.5 

 

\((-1)^{1.5}\\ = (-1)^{\frac{3}{2}}\\ \text{I can look at this in 2 different ways so lets see what happens.}\\ = [(-1)^3]^{\frac{1}{2}}\qquad or \qquad [(-1)^\frac{1}{2}]^{3}\\ = [-1]^{\frac{1}{2}}\qquad or \qquad [\sqrt{-1}]^{3}\\ =\sqrt{-1}\qquad or \qquad [i]^{3}\\ =\pm i \qquad \qquad or \qquad -i\\ \)

 

The only common answer is -i so I guess that is the only correct answer.

 

Maybe another mathematician would like to comment here......

 Nov 11, 2017
 #5
avatar+12 
0

How did you get from \(\left(-1\right)^{\frac{1}{2}}\)to \(\sqrt{-1}\)?

Epic8009  Nov 11, 2017
edited by Epic8009  Nov 11, 2017
 #6
avatar+27540 
+1

This is what WolframAlpha says Melody:

 

Alan  Nov 11, 2017
 #7
avatar+2340 
+1

\((-1)^{\frac{1}{2}}\Rightarrow\sqrt{-1}\)  is a law of fractional exponents. In general terms, \(x^{\frac{a}{b}}=\sqrt[b]{x^a}=\left(\sqrt[b]{x}\right)^a\)

 

Why is this the case? Well, I can attempt to explain it to you.

 

The law of exponents explains how to handle multiplication of exponents with identical bases. 

 

\(x^3*x^2=x^{3+2}=x^5\)

 

Let's try another example but with fractional exponents.

 

\(3^{\frac{1}{2}}*3^{\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}}=3^1=3\)

 

In this example here, \(3^{\frac{1}{2}}\) is a number that when multiplied by itself yields 3. That sounds like the definition of the square root, doesn't it? Therefore, \(3^{\frac{1}{2}}=\sqrt{3}\). One can also expound upon this.

 

\(3^{\frac{1}{3}}*3^{\frac{1}{3}}*3^{\frac{1}{3}}=3^{{\frac{1}{3}}+{\frac{1}{3}}+{\frac{1}{3}}}=3^1=3\)
\(3^{\frac{1}{4}}*3^{\frac{1}{4}}*3^{\frac{1}{4}}*3^{\frac{1}{4}}=3^{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=3^1=3\)
\(3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}=3^{\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}}=3^1=3\)
\(3^{\frac{1}{x}}*3^{\frac{1}{x}}*3^{\frac{1}{x}}*...*3^{\frac{1}{x}}*3^{\frac{1}{x}}=3^{\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+...+\frac{1}{x}+\frac{1}{x}}=3^1=3\)

 

The pattern continues. \(3^{\frac{1}{3}}\) is a number that when multiplied by itself 3 times yields 3. This is the definition of the cubic root. Then, as you can see, I made a generalization. \(3^{\frac{1}{x}}\), when multiplied x times, yields 3. 

TheXSquaredFactor  Nov 11, 2017
 #8
avatar+12 
0

Thank you!

Epic8009  Nov 11, 2017

43 Online Users

avatar
avatar
avatar
avatar