Euler totient function made easy

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Solution to 4 typical exam/test questions. See other videos
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You are literally the best human on this planet

timothywahib
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Impeccable explanation I finally get it, thank you very much sir

HansFlamme
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only 299 views for such a master piece ?? :(

BHARATKOTESWARAO
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thank you for saving me from my useless number theory book.

badradish
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Hey bro this was exactly a masterpiece.hats off to you.

aishwaryaratnam
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Thank you a bunch. Been trying to solve Liber Primus and I think this will definitely help.

kidhinobi
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Finally got the sense of Euler's function
Good explanation Sir...

conradt.s
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I'm just a random youtube viewer that is to curious about Euler's Phi. I only got the solution for prime number. Thanks for the explanation. Now I got it.

sleeplesshead
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You deserve more views. Thanks for posting

Imperator-SPQR
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This guy is the man!!! please take over Deakin Uni maths!

urajunky
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I remember trying to use the 4th rule and hating it. Loved the simplicity of the first three rules. Just found your videos and wish I had watched them in when I was in college!

_koppy
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ahhh i got finals tomorrow and this helps me a lot!! thank you so much 💝💝

kneebrain
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Very well made! Many videos I watch simply mention the end algorithm, but I enjoy using the rules.

ListRav
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Excellent explanation! Thank you so much!

j.vonhogen
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Finally! The 27.8k views you deserve!!!

patrick-ccyu
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super easy to understand, ,,l loved it explained better than my lecturere would have :-)

ropafadzomaseko
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Thanks. This is a great video, but is there a function that takes into account all the things you take for granted in these rules? For example, you for 600, what's the function you used to figure out that you could use 2^3 * 3 * 5^2... Also, 41 is easy to identify as a prime, but what about large numbers that you may not be sure if they are prime... just curious about your thoughts on that. Thanks.

arizonajterm
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for phi(p^n) = p^n - p^(n-1), doesn't that equal (p^n)*[ (p-1)/(p)] ? Example phi(2^3) = 2^3 - 2^2 = 8-4 or (8 * (2-1)/2). And phi(3^2) = 9 - 3 = 9*[ (3-1)/3 ], phi(5^3) = 125 - 25 = 125 * [(5-1)/5]… am I wrong or doesn't this pattern hold in general?

stevenwilson
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Thanks for making this video. It helped me out alot.

DaveandEm
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0:04 no, that's the null sign, the symbol to denote the empty set. Phi is something else.

samueldeandrade