Question

Prove that if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

Answer 1

It is proved that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

According to the statement

we have to prove that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

And for this proof we use the contradiction

So,

Proof by contradiction is a common proof technique that is based on a very simple principle: something that leads to a contradiction can not be true, and if so, the opposite must be true.

So for this purpose,

Assume a is even, so a = 2k for some integer k. Now let a and b be integers such that a divides b and a + b is odd.

Since a divides b, b = an for integer n, and in turn b = 2nk, which means b is even and hence a + b is also even. But this contradicts our initial assumption, so a must be odd.

So, It is proved that the if a and b are nonzero integers, a divides b, and a b is odd, then a is odd.

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