MedVision ad
F
Reaction score
107

Profile posts Latest activity Postings About

  • Bro change as soon as you can, trust me you will regret not choosing. I suggest that you speak with your English Head teacher tomorrow asap and move in the ADVENG class as soon you can.

    NB: We can help each other with identical subs we have! :)
    I've been doing lots of work on Circle Geo so I'm getting my head around it...
    Yeah I did, but I was in Advanced for a while now, so I forgot to change it. Thanks :)
    ATAR aim would be around 90-5+...
    Thinking of dropping 3U if it doesn't go well :l Its very hard.
    you are my favourite person right now :D
    thank you so much for the past nsg maths test papers
    im going into year 10 so it will be of great use to me
    ily person ^-^
    Yepp your on the ball, medicine, becoming a doctor one day is my ultimate dream bro!
    Honestly... i didn't find the work that great, it was tedious and repetitive. But due to doing standard i kept the same teacher for 3 years consecutively. This teacher knew about my general lack of motivation and the fact that i used to be socially awkward as fuck and worked around it, making me kind of enjoy coming to English, i also had good classmates who i joked around with. By the time i started doing daily essays and putting effort in i found that i just enjoyed coming to class... it was a good opportunity to have a discussion and joke around. 'Twas just a good experience i guess

    I honestly think that in my current state i could of handled 3U English, but i would likely still be an unmotivated UWSCollege candidate if it wasn't for my time in Standard
    e = 1/0! + 1/1! + 1/2! + ...

    let e = p/q, S(n) = 1/0! + 1/1! + ... + 1/n!

    e - S(n) = 1/(n+1)! + 1/(n+2)! + .... < 1/(n+1)! (1 + 1/(n+1) + 1/(n+1)^2 + ...) = 1/((n)(n!))

    e - S(n) > 0

    0< n n! e - n n! S(n) < 1

    Therefore n n! e - n n! S(n) is not an integer

    Since e = p/q, and take n > q, then n! e is an integer

    So n n! e - n n! S(n) is an integer

    So n n! e - n n! S(n) is an integer and is not an integer

    Contradiction!

    Therefore, e is irrational

    (I don't take credit for this proof, its the classic one)
  • Loading…
  • Loading…
  • Loading…
Top