Mu Prime Math is a library of math explanation videos.
Many of my videos cover interesting challenge problems with solutions that require unexpected shifts in perspective. I have playlists for challenging pre-calculus problems, integrals, and calculus problems like infinite series and differential equations!
I also have several series dedicated to intuitively explaining ideas in specific areas of math. I've made playlists on differential equations, linear algebra, and vector calculus. These videos emphasize understanding where formulas and methods come from and why they work, so you can build a base of knowledge that extends into higher mathematics!
Good video, but this demonstration rather than a proof because you don't show how this works in general, instead you show how it applies to a specific example: 924. If you were to prove this it would most likely involve an if-and-only-if statement or perhaps a proof by contradiction. Nevertheless, good insights are provided here. Thank you!
@@MuPrimeMath Yes, but this is not a proof. The title of the video suggests that we will be seeing a mathematical proof. My expectation is perhaps not appropriate for this particular kind video. Thank you again for your hard work.
I have finally come across this video, it's been several years trying to answer what seemed the million dollar question and you answered it! Thanks for showing this information, i wonder why this isn't more extended
You are an amazing teacher. I taught linear algebra at a community college for almost 40 years. I knew about the geometic interpretation of the determinant, but never broke it down to see why it works. I went and watched your 3d determinant video too. I'm retired now, but one of my former students asked about this on facebook, and when I looked I found your excellent videos. Thank you!
The fundamental theorem of arithmetic states that every natural number greater than one can either be a prime number or a finite product of primes, which is unique. ( or the number is 1) It doesn't state that for integers. Natural numbers include (1,2,3,....), but integers include (...,-2,-1,0,1,2,...). Right? or am I thinking of something else?
Yes, this video is concerned with positive integers. Zero is a perfect square, so the statement still holds. The square root of a negative number is imaginary, therefore not rational. However, mathematicians would not call the square root of a negative number "irrational" because "irrational" refers specifically to real numbers.
this feels like factor groups in abstract algebra, each subset of permutations feel like a coset. but what is the normal subgroup? or is this just completely different?
This is by far the best explanation of the linearity of lorentz transformation, which I dont know why people just assume implicitly, but there is still a scope of improvement for this video
Yo my guy, next time, actually use those variables and put some values in them. Writing a ton of letters all over the board is DRY. I get that you're explaining the laws and being very technical, but you have to throw some practical examples in there to show how the laws are applicable. For Christ's sake. This is where people fall out of mathematics. APPLICATION, APPLICATION, APPLICATION.
I ❤ these videos. He is great at elucidating novel perspectives on techniques to attack problems. That is ultimately what mathematics is: trailblazing.
This video is still relevant even today (2024). Thank you for making the video. It has made me appreciate the concept of surface integral of a vector field
here is the simplest proof of all. If the square root of 2 (or any other number that is not a perfect square) is rational, then there are only finitely many digits to the right of the decimal point. So we go from 2 to 200, and move the decimal point one place to the right. And we do that until we reach the last digit of the number. And here is where we can absolutely conclude that this number cannot be multiplied by itself to equal any number that ends in 0, because only numbers that end in 0 have powers that end in 0. QED.
¡Eso fue muy sexy! Explicar algo tan complicado como calculo en otra lengua es extremadamente difícil. ¡ Te felicito y te lo agradezco. ¡Las matemáticas y las lenguas extranjeras son muy sexys, y por eso, tú lección lo es también! Simplemente para que lo sepas, decimos "otra forma" en vez de "una otra forma" por ejemplo. Es como decís en inglés "another" en inglés como una sola palabra en vez de "an other." ¡Gracias, chico!
You can extend this to all rationals in canonical form by noticing that if p^2 / q^2 implies that every square root is going to only have half the prime factors in p and q. That means if the rational is in canonical form (gcd(p, q) = 1), then if either p or q has a prime factorization where one factor is repeated an odd number of times (i.e. it's not a perfect square), then it must be irrational. Using properties of the gcd, we can see if that if gcd(p, q) = 1, then gcd(p^2, q^2) = 1. So we can see that the new numerator in canonical form is going to just inherit the factors from p, but since squaring ensures all prime factors have an even multiplicity, there's no possible way that we can square p or q to get a rational number where either p or q has a prime factorization containing factors with an odd multiplicity. You can keep going in fact to show this for nth roots where if p or q have a prime factorization where a factor has multiplicity not divisible by n, it must be irrational.
You look pretty young to be doing that well in Algebra. Just give it 20 years, you will be a wizard at it. And you will never run out of new things to learn about it. It just goes on and on.
Man, hope you come back. But take your time. The video is great! Love your videos. But I need to add something. This does not show that it only works for f(x)=K exp(c x^2), it just provide a solution. You can see this if you try the trick on 2^(-x^2) or any w^(-x^2). Again, hope you the best
We're considering expressions of the form a*sin(x) + b*cos(x). We can write sin(x) + cos(x) as 1*sin(x) + 1*cos(x), which is simply a*sin(x) + b*cos(x) with a=1 and b=1.
The "volume" of an n-dimensional shape scales with r^n, so working through the math as in the video we find that the length must be r = nV/A, where A is the "surface area". The video is the special case n=2. For example, in 3 dimensions we obtain r = 3V/A. One can compute directly that this equation is satisfied for a sphere, which is why the derivative of a sphere's volume is its surface area when we use the radius as the length measure.
Ok I spent ages trying to get an intuitive proof of this theorem and I have it. This proof is a classic - takes a non-intuitive result and makes it seem obvious. The result is just a projection of one vector oscillating on another. It also makes the negative term on the arctan obvious - to make the phase a positive angle you need to take the negative of the tan of alphs.
