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How does this conjecture be true can you prove this? watch

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    Okay, I am doing this maths homework in Core 1 Proofing unit
    The question asks me to decide if this conjecture is telling the truth.
    IF it's true i have to prove it by suitable method and name the method
    IF it isn't true, i have to come up with a counterexample
    and the questions are as follows

    1.The sum of any three consecutive integers is divisible by 6
    2. An easy way to remember 7times 8 is that 56=7*8 and the number 5,6,7 and 8are consecutive. There is exactly one other multification of two single-digit numbers with the same pattern
    3. x^2>x => x.1
    4. ABCD is a parallelogram. Sides AB and DC are parallel; sides AD and BC are parallel. AC=BD
    5. A triangle with sides (x^2+1), (x^2-1), 2x is right angled.
    6. The circle with equation x^2+y^2=36 passes through exactly four points for which both the x and y coordinates are integers.
    7. The value of (n^2+n+41) is a prime number for all positive integer values of n.
    8.A number is divisible by 9 if the sum of its digits is divisible by 9.
    9. n is prime, then n^2+n+1 is also a prime



    thank you!
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    (Original post by ucjs2)
    Okay, I am doing this maths homework in Core 1 Proofing unit
    The question asks me to decide if this conjecture is telling the truth.
    IF it's true i have to prove it by suitable method and name the method
    IF it isn't true, i have to come up with a counterexample
    and the questions are as follows

    1.The sum of any three consecutive integers is divisible by 6
    2. An easy way to remember 7times 8 is that 56=7*8 and the number 5,6,7 and 8are consecutive. There is exactly one other multification of two single-digit numbers with the same pattern
    3. x^2>x => x.1
    4. ABCD is a parallelogram. Sides AB and DC are parallel; sides AD and BC are parallel. AC=BD
    5. A triangle with sides (x^2+1), (x^2-1), 2x is right angled.
    6. The circle with equation x^2+y^2=36 passes through exactly four points for which both the x and y coordinates are integers.
    7. The value of (n^2+n+41) is a prime number for all positive integer values of n.
    8.A number is divisible by 9 if the sum of its digits is divisible by 9.
    9. n is prime, then n^2+n+1 is also a prime



    thank you!
    1) A number is divisible by 6 if it is divisible by 2 and by 3. Think about the divisibility of 3 consecutive numbers by 2 and 3.

    2) Write it in a general form.  10n + n+1 = (n+2)(n+3). Solve this.
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    (Original post by ucjs2)
    Okay, I am doing this maths homework in Core 1 Proofing unit
    The question asks me to decide if this conjecture is telling the truth.
    IF it's true i have to prove it by suitable method and name the method
    IF it isn't true, i have to come up with a counterexample
    and the questions are as follows

    1.The sum of any three consecutive integers is divisible by 6
    2. An easy way to remember 7times 8 is that 56=7*8 and the number 5,6,7 and 8are consecutive. There is exactly one other multification of two single-digit numbers with the same pattern
    3. x^2>x => x.1
    4. ABCD is a parallelogram. Sides AB and DC are parallel; sides AD and BC are parallel. AC=BD
    5. A triangle with sides (x^2+1), (x^2-1), 2x is right angled.
    6. The circle with equation x^2+y^2=36 passes through exactly four points for which both the x and y coordinates are integers.
    7. The value of (n^2+n+41) is a prime number for all positive integer values of n.
    8.A number is divisible by 9 if the sum of its digits is divisible by 9.
    9. n is prime, then n^2+n+1 is also a prime



    thank you!
    3. x^2 > x \iff x^2 - x > 0 \iff x(x-1) > 0 \Leftarrow x>1
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    (Original post by ucjs2)
    6. The circle with equation x^2+y^2=36 passes through exactly four points for which both the x and y coordinates are integers.
    7. The value of (n^2+n+41) is a prime number for all positive integer values of n.
    6. Notice the radius is an integer, then you can see four integer points generated by x=0 and y=0 - how can you say these are the only four points that are integers?

    7. Use a nice large counter-example.
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    (Original post by ucjs2)
    Okay, I am doing this maths homework in Core 1 Proofing unit
    The question asks me to decide if this conjecture is telling the truth.
    IF it's true i have to prove it by suitable method and name the method
    IF it isn't true, i have to come up with a counterexample
    and the questions are as follows

    1.The sum of any three consecutive integers is divisible by 6
    2. An easy way to remember 7times 8 is that 56=7*8 and the number 5,6,7 and 8are consecutive. There is exactly one other multification of two single-digit numbers with the same pattern
    3. x^2>x => x.1
    4. ABCD is a parallelogram. Sides AB and DC are parallel; sides AD and BC are parallel. AC=BD
    5. A triangle with sides (x^2+1), (x^2-1), 2x is right angled.
    6. The circle with equation x^2+y^2=36 passes through exactly four points for which both the x and y coordinates are integers.
    7. The value of (n^2+n+41) is a prime number for all positive integer values of n.
    8.A number is divisible by 9 if the sum of its digits is divisible by 9.
    9. n is prime, then n^2+n+1 is also a prime



    thank you!
    5: (x^2+1)^2 - (x^2-1)^2 = 4x^2 Find x, if it solves for a real value it is right angled; otherwise it's not.
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    (Original post by Andy98)
    5: (x^2+1)^2 - (x^2-1)^2 = 4x^2 Find x, if it solves for a real value it is right angled; otherwise it's not.
    You don't really need to find x do you? You can just look at the discriminant?
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    (Original post by Zacken)
    You don't really need to find x do you? You can just look at the discriminant?
    Well you can do, but I just can't find it in me to use an equation but not solve it if it is solvable.
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    (Original post by Andy98)
    Well you can do, but I just can't find it in me to use an equation but not solve it if it is solvable.
    Fair enough. :lol:
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    (Original post by Zacken)
    Fair enough. :lol:
    Yeah, I'm weird like that :teehee:
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    (Original post by Andy98)
    5: (x^2+1)^2 - (x^2-1)^2 = 4x^2 Find x, if it solves for a real value it is right angled; otherwise it's not.
    It's an identity.
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    (Original post by Zacken)
    3. x^2 > x \iff x^2 - x > 0 \iff x(x-1) > 0 \implies x>1
    Apropos your questions on another thread, please note that your final implication here is false. (Think about the graph of y=x(x-1))
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    (Original post by atsruser)
    Apropos your questions on another thread, please note that your final implication here is false. (Think about the graph of y=x(x-1))
    Yes - I asked that question because I was fairly sure what I said here wasn't correct, I meant to back and change it to x(x-1) \Leftarrow x>1, I'll do that now!
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    (Original post by Andy98)
    5: (x^2+1)^2 - (x^2-1)^2 = 4x^2 Find x, if it solves for a real value it is right angled; otherwise it's not.
    This is not a valid answer IMHO.

    (It would be a valid answer if the question was "There exists x such that the triangle with sides x^2+1, x^2-1 and 2x is right angled", but as it stands the question wants you to show it for all x).
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    (Original post by DFranklin)
    This is not a valid answer IMHO.

    (It would be a valid answer if the question was "There exists x such that the triangle with sides x^2+1, x^2-1 and 2x is right angled", but as it stands the question wants you to show it for all x).
    Off the top of my head, would you use the cosine rule and solve for \cos \alpha and show that \alpha = \frac{\pi}{2}?
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    (Original post by Zacken)
    Off the top of my head, would you use the cosine rule and solve for \cos \alpha and show that \alpha = \frac{\pi}{2}?
    Just use Pythagoras. It's a 1-liner.
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    i) is false... if you add -2, -1, 0 you get -3

    :top2:

    no thank you
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    (Original post by DFranklin)
    Just use Pythagoras. It's a 1-liner.
    Isn't that what Andy did? Or were your remarks directed towards his "find" when he should have said "prove that this identity holds"?
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    (Original post by Zacken)
    Isn't that what Andy did? Or were your remarks directed towards his "find" when he should have said "prove that this identity holds"?
    At the point where he said "if it solves for a real value" (emphasis mine) it was clear he wasn't looking to show it held for general values of x.
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    (Original post by DFranklin)
    At the point where he said "if it solves for a real value" (emphasis mine) it was clear he wasn't looking to show it held for general values of x.
    Makes sense. thanks for clearing it up.
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    (Original post by DFranklin)
    At the point where he said "if it solves for a real value" (emphasis mine) it was clear he wasn't looking to show it held for general values of x.
    Right

    I was just meaning if you can solve it without having to play around with i

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