Turn on thread page Beta
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    You cannot define a prime number like you would an even or odd number. Make sure you know how to prove \sqrt2 is irrational, or \sqrt3 is irrational etc...
    how would you do that??aha
    Offline

    3
    ReputationRep:
    (Original post by Alevelstudent678)
    how would you do that??aha
    I'll do the \sqrt2 one for you

    assume that \sqrt2 is rational, so it can be expressed in the form \frac{a}{b} where a and b share no common factors.

    \sqrt2 = \frac{a}{b}
    2 = \frac{a^2}{b^2}
    2b^2 = a^2 \Rightarrow a must be an even number.

    Because a is even, we can write in the the form 2n

    2b^2 = (2n)^2
    2b^2 = 4n^2
    b^2 = 2n^2 \Rightarrow b must be an even number.

    But this is a contradiction, because we said a and b share no common factors (and since they are both even they clearly have a factor of 2). Hence \sqrt2 is irrational.
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    I'll do the \sqrt2 one for you

    assume that \sqrt2 is rational, so it can be expressed in the form \frac{a}{b} where a and b share no common factors.

    \sqrt2 = \frac{a}{b}
    2 = \frac{a^2}{b^2}
    2b^2 = a^2 \Rightarrow a must be an even number.

    Because a is even, we can write in the the form 2n

    2b^2 = (2n)^2
    2b^2 = 4n^2
    b^2 = 2n^2 \Rightarrow b must be an even number.

    But this is a contradiction, because we said a and b share no common factors (and since they are both even they clearly have a factor of 2). Hence \sqrt2 is irrational.
    oh right V.clever, i'l learn those cheers mate
    Offline

    2
    ReputationRep:
    (Original post by Alevelstudent678)
    you know when doing proofs you can say even number is 2x, odd number is 2x+1, what others are there e.g. A prime number?
    Because i really struggle on these
    Any prime number above 3 is in the form 6n+/-1 but not everything in this form is a prime number, it is necessary but not sufficient; no other way to express primes


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by Alevelstudent678)
    oh right V.clever, i'l learn those cheers mate
    No problem

    (Original post by henrygriff28)
    Any prime number above 3 is in the form 6n+/-1 but not everything in this form is a prime number, it is necessary but not sufficient; no other way to express primes


    Posted from TSR Mobile
    Is this true? Wow, I did not know this. Have you got a proof?
    Offline

    2
    ReputationRep:
    (Original post by lizard54142)
    No problem



    Is this true? Wow, I did not know this. Have you got a proof?
    Yeah, it's by exhaustion though 😕

    Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n-1, 6n-2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n-2 = 2(3n-1) so is divisible by three so all primes must be 6+/-1. Not all are though as counter examples of 35 and 25 show. QED


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by henrygriff28)
    Yeah, it's by exhaustion though 😕

    Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n-1, 6n-2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n-2 = 2(3n-1) so is divisible by three so all primes must be 6+/-1. Not all are though as counter examples of 35 and 25 show. QED


    Posted from TSR Mobile
    I did not know this; you learn something every day! Awesome.
    Offline

    2
    ReputationRep:
    (Original post by henrygriff28)
    Yeah, it's by exhaustion though 😕

    Only possibilities are 6n, 6n+1, 6n+2, 6n+3, 6n-1, 6n-2. All real integers fall into one of these categories. 6n is clearly divisible by 6, 6n+2 = 2(3n+1) so is divisible by two, 6n+3 = 3(2n+1) so is divisible by 3, 6n-2 = 2(3n-1) so is divisible by three so all primes must be 6+/-1. Not all are though as counter examples of 35 and 25 show. QED


    Posted from TSR Mobile
    Thats the one I was thinking of!!
    Offline

    2
    ReputationRep:
    Name:  ImageUploadedByStudent Room1433995895.816991.jpg
Views: 210
Size:  121.0 KB How do I do the last part to this question? I figured out its the area of the trapezium minus 25/3 but how do I do the trapezium part? ANS: 30 2/3


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    (Original post by Supergirlxxxxxx)
    Name:  ImageUploadedByStudent Room1433995895.816991.jpg
Views: 210
Size:  121.0 KB How do I do the last part to this question? I figured out its the area of the trapezium minus 25/3 but how do I do the trapezium part? ANS: 30 2/3


    Posted from TSR Mobile
    Don't think of it as a trapezium. Work out the area of the triangle bounded by the line y=x, the line x=11 and the x axis. Then work out the little triangle bounded by the line y=x and the line y=3 (Where point p lies).

    Then subtract the integrated component and the little triangle from the big triangle.

    This should give you 92/3 which is the same as 30 2/3.
    Offline

    2
    ReputationRep:
    (Original post by Computer Geek)
    Don't think of it as a trapezium. Work out the area of the triangle bounded by the line y=x, the line x=11 and the x axis. Then work out the little triangle bounded by the line y=x and the line y=3 (Where point p lies).

    Then subtract the integrated component and the little triangle from the big triangle.

    This should give you 92/3 which is the same as 30 2/3.
    Oh that makes so much sense thankyou!! Also I don't quite understand when it asked me to intergrate that curve between certain limits, why does that represent the area above the curve? I think when you integrate your finding the area below the curve?



    Posted from TSR Mobile
    Offline

    2
    ReputationRep:
    the trig identities like sec^2x=1+tan^2x and cosec^2x=1+cot^2x, and the double angle formulae, are C4 topics right? So we shouldn't need them tomorrow??
    I've been neglecting C3 revision to the point where even basic integration is a bit of a struggle right now lol, gonna have to revise like crazy :eek:
    Offline

    9
    ReputationRep:
    (Original post by tingirl)
    the trig identities like sec^2x=1+tan^2x and cosec^2x=1+cot^2x, and the double angle formulae, are C4 topics right? So we shouldn't need them tomorrow??
    I've been neglecting C3 revision to the point where even basic integration is a bit of a struggle right now lol, gonna have to revise like crazy :eek:
    Yeah you do need to know them for c3/c4 I think.

    Just remember \displaystyle \sin^2(\theta)+\cos^2(\theta)=1 and you can get both of them by dividing both sides by either \displaystyle \sin^2(\theta) or \displaystyle \cos^2(\theta).
    Offline

    2
    ReputationRep:
    (Original post by poorform)
    Yeah you do need to know them for c3/c4 I think.

    Just remember \displaystyle \sin^2(\theta)+\cos^2(\theta)=1 and you can get both of them by dividing both sides by either \displaystyle \sin^2(\theta) or \displaystyle \cos^2(\theta).
    Ok! Thanks, that's really helpful!! How about parametric equations - they're just a C4 topic right?
    Offline

    9
    ReputationRep:
    (Original post by tingirl)
    Ok! Thanks, that's really helpful!! How about parametric equations - they're just a C4 topic right?
    Yes I believe so.

    I think this is also helpful.

    http://www.mathshelper.co.uk/MEI%20C...on%20Sheet.pdf

    Use it to help if you are stuck on past papers.
    Offline

    2
    ReputationRep:
    (Original post by poorform)
    Yes I believe so.

    I think this is also helpful.

    http://www.mathshelper.co.uk/MEI%20C...on%20Sheet.pdf

    Use it to help if you are stuck on past papers.
    that looks fab, thanks a lot!!
    Offline

    2
    ReputationRep:
    Name:  image.jpg
Views: 150
Size:  92.0 KB
    Please can someone explain this to me. (See attachments)

    For question three I understand that the first answer is 3-2x=4x

    However, for the second answer I put -(3-2x)=4x so -3+2x=4x so -3=2x so x=-1.5, however according the the mark scheme I get no method marks for this even though I think it is completely correct? They put (3-2x)=-4x, so how come you get a mark for putting a minus on the left but not the right, makes no sense???
    Attached Images
     
    Offline

    3
    ReputationRep:
    (Original post by Connorbwfc)
    Name:  image.jpg
Views: 150
Size:  92.0 KB
    Please can someone explain this to me. (See attachments)

    For question three I understand that the first answer is 3-2x=4x

    However, for the second answer I put -(3-2x)=4x so -3+2x=4x so -3=2x so x=-1.5, however according the the mark scheme I get no method marks for this even though I think it is completely correct? They put (3-2x)=-4x, so how come you get a mark for putting a minus on the left but not the right, makes no sense???
    You would still get all the marks.

    -(3-2x) = 4x is the same as 3-2x=-4x
    Offline

    18
    ReputationRep:
    is irrational

    How would I go about proving that. I can do it for square of 2 but not 3.
    Offline

    3
    ReputationRep:
    Some notes:

    -Odd functions: f(-x) = -f(x) and these functions show rotational symmetry about the origin, order 2
    -Even functions: f(x) = f(-x) and these functions show symmetry across the y-axis
    -Reflecting f(x) across y=x gives f^-1(x) (the inverse function)
    -The domain of f(x) is the range of f^-1(x), and vice versa
    -If the gradient of f(x) at (x,y) is a, then the gradient of f^-1(x) at (y,x) is 1/a

    -Differentiating: sin(x) --> cos(x) --> -sin(x) --> -cos(x) --> sin(x)...
    -Differentiating sin(ax) gives acos(ax)

    -Integrating e^x gives e^x
    -Integrating f'(x)/f(x) gives ln(f(x))
    -integrating e^ax gives (1/a)e^ax
    -Integrating sin(ax) gives -(1/a)cos(ax)
    -Integration by parts formula: uv - (Integral of)vdu
    -Indefinite integrations introduce +c as an unknown value
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: April 20, 2016

2,681

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.