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(i) Given that p and q are integers such that $$pq$$ is even use algebra to prove by contradiction that at least one of p or q is even - Edexcel - A-Level Maths Pure - Question 9 - 2022 - Paper 1

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(i) Given that p and q are integers such that $$pq$$ is even use algebra to prove by contradiction that at least one of p or q is even. (ii) Given that x and y ar... show full transcript

Worked Solution & Example Answer:(i) Given that p and q are integers such that $$pq$$ is even use algebra to prove by contradiction that at least one of p or q is even - Edexcel - A-Level Maths Pure - Question 9 - 2022 - Paper 1

Step 1

Given that p and q are integers such that $$pq$$ is even

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Answer

To prove that at least one of p or q is even using a proof by contradiction:

  1. Assume both p and q are odd.
    Let p be represented as p=2m+1p = 2m + 1 and q as q=2n+1q = 2n + 1, where m and n are integers.

  2. Calculate pq:
    pq=(2m+1)(2n+1)=4mn+2m+2n+1pq = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1
    Rearranging gives us
    pq=2(2mn+m+n)+1pq = 2(2mn + m + n) + 1
    This indicates that pq is odd.

  3. Arrive at a contradiction:
    Since we began with the assumption that pq is even, we have reached a contradiction.
    Therefore, at least one of p or q must be even.

Step 2

Given that x and y are integers such that * $x < 0$ * $(x+y)^2 < 9x^2 + y^2$

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Answer

To show that y>4xy > 4x:

  1. Start with the given inequality:
    We have
    (x+y)2<9x2+y2(x+y)^2 < 9x^2 + y^2
    Expanding (x+y)2(x+y)^2 gives
    x2+2xy+y2<9x2+y2x^2 + 2xy + y^2 < 9x^2 + y^2

  2. Simplify the inequality:
    By rearranging, we get
    2xy<8x22xy < 8x^2
    which simplifies to
    xy<4x2xy < 4x^2

  3. Divide by x (note x < 0):
    Since x<0x < 0, dividing both sides by x will reverse the inequality:
    y>4xy > 4x
    Thus, we have shown that under the given conditions, y>4xy > 4x.

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