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Mind Map - Coggle Diagram

- - - - A = A number ends in a zero, B = A number is a multiple of 10
      - A = A number can be divided by 2, B = A number is even
      - A = A number is a multiple of 3, B = The square of a number is a multiple of 3
      - A implies B and B implies A
    - - A implies B
      - A = A number ends in 0, B = a number is a multiple of 5
      - A = A number is more than 9, B = A number is more than 6
      - A = A number is 5, B = The square of a number is 25
  - - - Find one example of where the Statement is false
      - The difference between any two square numbers is odd
        
        (3)^2 = 9, (5)^2 = 25, 25 - 9 = 16 and 16 is even, proving the statement false
    - - Assume the converse of the statement to be true, then show that is leads to a contradiction
      - Prove that root(2) is irrational
        
        First assume that root(2) is rational meaning it can be written in the form a/b where a and be are co-prime this can be shown by root(2) = a/b, from here we times both sides by b getting us broot(2) = a, then we can square both sides getting 2(b)^2=(a)^2, because 2 can be taken out of the left side is shows that (a)^2 is even, if (a)^2 is even then a must also be even. We can now write a as 2c, putting 2c back into the equation gets us 2(b)^2 = (2c)2 the right hand side can be expanded to give 2(b)^2 = 4(c)^2, from here both sides can be divided by 2 to get (b)^2 = 2(c)^2, now we can take a multiple of 2 out of the right side meaning that the left side must also be even, if (b)^2 is even then b must also be even. However, at the start it was stated that a and b were co-prime but now we have just shown that they both hare a common factor of 2, thus showing that root(2) cannot be rational and therefor must be rational
    - - Using algebra reduce the statement to an equation you know to be true/ false
      - Prove that if n is an even number then (n)^2 + n is also even
        
        if n is even then n=2k where k is half of n, we can then put 2k into (n)^2 + n getting is (2k)^2 + 2k which simplifies down to 4(k^2) + 2k from there we can take out a 2 giving us 2(2(k)^2 + k), if you can take a multiple of 2 out of a number then number must be even
    - - Test every possibility for the statement until you find one where the statement is false
      - Show that (n)^2-1 is multiple of 3 if n is not multiple of 3
        
        for n to not be a multiple of 3 it can either be one above or two above a multiple of 3, if it where 3 above it would be back to a multiple of 3. so we can enter the only 2 possible values into the formula and check both of their results: if we use n = 3k+1, (3k+1)^2-1 which simplifies down to 9(k)^2+6k+1-1 which cancels down to 9(k)^2+6k from here you can take out a multiple of 3 making 3(3(k)^2+2k), if you can take out a multiple of 3 from the number then the number must also be a multiple of 3. For the second possibility where n is 2 above a multiple of 3 we are going to use n = 3k+2, (3k+2)^2-1 which expands to form 9(k)^2+12k+4-1 which cancels down to make 9(k)^2+12k+3 from here we can take out a multiple of 3 to make 3(3(k)^2+4k+1), if you can take out a multiple of three from a number then the number must be a multiple of 3. Now we have tried both possibilities for numbers that aren't a multiple of three and shown that (n)^2-1 is a multiple of 3 if n is not a multiple of 3.
- - - - is approximately
        
        1-1/2(x)^2
    - - is approximately
        
        x
    - - Is approximately
        
        x
  - - - Radians to degrees
        
        multiply by 180/pi
      - Degrees to radians
        
        multiply by pi/180
    - - Area of Sector
        
        1/2(Θ(r)^2)
      - Area of Segment
        
        1/2(r)^2(Θ-sin(Θ))
      - Length of Arc
        
        Θr