what if a negative number is raised to a negative exponent?

We know a negative number times a negative number will always result in a positive number. But what if a negative number is raised to a negative exponent?

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#justalgebra

[ryanmarcus]

basically when we have a negative base, an even exponent will give us a positive number, an odd exponent will give us a negative number, and a fractional exponent will give us an imaginary number

[Koheda]

am i the only one to notice how smoothly he changes the marker colour?

[oZqdiac]

What we can evaluate from this:
When a negative number is the base, the result could be either positive, negative or imaginary.
When a positive number is the base, the result can only be positive.

[tontonbeber]

A negative exponent is not a problem for a negative base. But a non integer exponent and especially an irrational one should be a problem.

[nicolasturek]

negative exponent has nothing to do with neg/pos result, it just inverts the number (1/n)

[Pacvalham]

pos^pos: there are no negatives, so the answer remains positive

[oida]

Simple a^(-b)=1/(a^b). So we have the sign of a^b.

[_marshP]

Sorry I wasn't listening I was too distracted by how fucking good he is with those markers

[savannahstewart]

thank god for this guy, i could not get this concept AT ALL but now i totally understand. math is not my subject and i have a test today, thank you for this video!!!

[mccauleybacalla]

thank you so much for the information sir. i love mathematics since it is my favorite subject!

[Sabagegah]

I did this (y=(-1)^(-x)) on a graphing calculator and it produced what I can only call the forbidden grilled cheese sandwich of death.

[mattyice]

I love learning stuff I’ll never use in life

[Md.Jaid]

a^b= ??? Conclusion
1. a is positive and b is positive
Then a^b= positive
2. a is positive and b is negative
Then a^b = positive
3. a is negative and b is positive
Then a^b=complex/positive/negative
4. a is negative and b is negative
Then a^b = complex/positive/negative
Then a^b = Complex
5. In 3, 4
If b is positive/negative rational no. with even denomination
a^b= positive /complex
6. In 3, 4
If b is positive/negative rational no. with odd denomination
a^b = negative/complex

[Kratos_TM]

Proud of myself, out of school for a year and a half and I was still able to figure out the solution to the imaginary one before he said it lol

[metekeles]

For when x is a positive variable and a is an arbitrarily selected positive real number, both a^x and x^a are always increasing functions the reason being there existing no real roots for the equation ln(a)*a^x=0 which is the derivative of a^x nor the equation a*[x^(a-1)]=0 which is the derivative of x^a since both x and a are greater than 0, which concludes there are no critical points for the function to change direction, and again since both x and a are greater than 0 and both functions at 0 is equal to 0(which is not in our domain), we prove that a positive number raised to the power of a postive number is always positive. I hope this is a valid proof, I'm a huge fan of your channel and thank you for everything I learned because of you, you really are one the best teachers out there :)

[mbapum]

WHAT HAPPENS WHEN A NEGATIVE IRRATIONAL NUMBER THAT CANNOT BE EXPRESSED AS A FRACTION IS RAISED TO ITSELF? (sorry for caps lock I just want you to read it dw I'm not shouting). The formula here is: Let's write our negative number as -z/v. In simplest form: If v is odd, we get a real number. If v is even, we get a complex number. If z is even, we get a positive number. If z is odd, we get a negative number. (or something like that I probably messed up I need coffee). This doesn't work with irrational numbers though. This is the reason why the graph of y=x^x is dotted on the negative side. How can we calculate this?

[cHrtzbrg]

If you define a^b = exp(b*log(a)) the problem with negative `a` is that the logarithm is not unique, e.g., log(-1) can be i*pi or -i*pi (or actually (2k+1)*i*pi for any integer `k`. If `b` is an integer, all give the same result, since `exp` is periodic with a period of `2pi*i`. But for rational b=p/q you get up to `q` possible solutions, and for irrational `b` you have arbitrarily many solutions.

[suryasrivastava]

Bro how many channels does this guy have i have seen him in over 10-12 different channels till now

[mbapum]

The formula here is: Let's write our negative number as -z/v. In simplest form: If v is odd, we get a real number. If v is even, we get a complex number. If z is even, we get a positive number. If z is odd, we get a negative number. (or something like that I probably messed up I need coffee). My question is: what happens with an irrational number that cannot be expressed as a fraction?

[esong]

Love your video, but here's a question for you: Take S = (1x2x3x4x...x n) + (4k+3), where n≥3. when k is an integer between 1 to 100, for how many different values of k such that S is a perfect square of an integer.