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Q: How do you use inductive reasoning to prove the product of an odd integer and an even integer will always be even?

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Yes.

Yes, the product of 2 integers are always an integers. ex. -2*3=-6

Conjecture: the sum of two even integers is always an even integer.Suppose M and N are two even integers.M is even and so it is divisible by 2 without remainder. That is to say, there is some integer X, such that M = 2X.Similarly, N = 2Y for some Y.Then M + N = 2X + 2Y=2*(X + Y) due to the distributive property of multiplication over addition of integers.And, by the closure of the set of integers under addition, X+Y is an integer.Therefore, M + N is equal to 2 times an integer and therefore it is an even integer.

yes..always a perfect square A perfect square is the product of an integer by itself. If you multiply a perfect square x² by another perfect square y² you get x²y² = x·x·y·y = x·y·x·y = (x·y)² which is a perfect square. Note that the product of two integers will also be an integer so x·y must be an integer because if x² and y² are perfect squares x must be an integer and y must be an integer and x·y is therefore a product of 2 integers.

an integer plus and integer will always be an integer. We say integers are closed under addition.

Related questions

~apex Inductive reasoning

Inductive reasoning use theories and assumptions to validate observations. It involves reasoning from a specific case or cases to derive a general rule. The result of inductive reasoning are not always certain because it uses conclusion from observations to make generalizations. Inductive reasoning is helpful for extrapolation, prediction, and part to whole arguments.

2

always a negative

Negative

always a negative

No

A negative integer multiplied by a negative integer is always a positive integer product. -x * -y = xy

Yes, by definition, the sum of two integers is always an integer. Likewise, the product and difference of two integers is always an integer.

Always.

Always

Yes.

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