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Gennadij [26K]
3 years ago
11

Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ

er x and for every integer y, if x is odd and y is odd then x y is even. (c) For every integer x, if x is odd then x 3 is odd.
Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

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Christine has always been weak in mathematics. Based on her performance prior to the final exam in Calculus, there is a 65% chan
amm1812

Answer:

0.49

Step-by-step explanation:

We apply conditional probability to solve this question.

We have Events X and Events Y

P(X) = Probability Christine fails the course = ??? Unknown

P(Y) = Probability Christine finds a tutor

= 0.35

P(Y') = Probability Christine does not find a tutor = 1 - 0.35 = 0.75

P(X|Y) = Decrease in her probability of failing = 0.35

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