91Ó°ÊÓ

The proof is by contradiction. Suppose \(\exists a \in \mathbb{R}\) such that \(a^{2}=-1\) (note the symbol "I" means "there exists," the symbol \(\mathbb{R}\) denotes the real numbers, and the expression " \(a \in \mathbb{R}^{"}\) means that \(a\) is contained in \(\mathbb{R}\), that is, \(a\) is a real number). There are two cases: either (i) \(a \geq 0\) or (ii) \(a<0\). In Case (i), then \(a^{2}=a \cdot a=(\) nonnegative \() \cdot(\) nonnegative \() \geqq 0,\) which contradicts the supposition. (ii), then \(a^{2}=a \cdot a=(\) negative \() \cdot\) (negative) \(>0\), which contradicts In Case the supposition. By contradiction, it follows that -1 has no real square root. You may note that in the streamlined case, we reduced the number of cases from three to two. That's because we noticed that we really could combine the "positive" and the "zero" case into a single case.

(a) Write down the complex number with real part 0 and imaginary part \(7 .\) (b) Write down a complex number whose real part is the negative of its imaginary part. (c) Write down a complex number that is also a real number.

Use substitution to prove the following statement: if \(3 \mid n\) and \(4 \mid m,\) then \(12 \mid m n\) (the notation \(-3 \mid n^{\prime \prime}\) means that 3 divides \(n\) ). ("Hint \(^{*}\) ) 0

(a) Suppose that \(a, b, c\) are integers and \((a / b)^{2}=c .\) Suppose further that \(a\) and \(b\) have no common factors except 1: that is, any integer \(x>1\) which divides \(b\) doesn't divide \(a\). Prove by contradiction that \(b=1\). (b) Generalize part (a): Suppose that \(a, b, c\) are integers and \((a / b)^{n}=c,\) where \(n\) is a positive integer. If \(a\) and \(b\) have no common factors, prove by contradiction that \(b=1\). (c) Use part (b) to prove the following: Let \(a\) and \(n\) be integers, both greater than \(1 .\) Let \(x\) be a real nth root of \(a .\) If \(x\) is not an integer, then \(x\) is irrational.