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Proof by contradiction: if a and b are both negative integers, then a+b is always negative

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Answer:

To prove by contradiction that if a and b are both negative integers, then a+b is always negative, we assume the opposite, that is:

a and b are negative integers, but a+b is not negative.

Let's suppose that a and b are negative integers. This means that a and b are less than zero. Therefore, we can express a and b as:

a = -x (where x is a positive integer)

b = -y (where y is a positive integer)

Now, let's consider the sum of a and b:

a + b = (-x) + (-y) = -(x + y)

Since x and y are both positive integers, their sum (x + y) is also a positive integer. Therefore, we have:

a + b = -(x + y)

But since x + y is positive, the opposite of x + y (which is -(x + y)) is negative. Thus, we have proven that if a and b are both negative integers, then a+b is always negative.

Therefore, our assumption that a and b are negative integers, but a+b is not negative, is false. Hence, the statement "if a and b are both negative integers, then a+b is always negative" is true.

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