By Randall R. Holmes

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**Extra info for Introduction to Advanced Mathematics**

**Sample text**

Assume that n = 1. Then n2 + n − 6 = (1)2 + (1) − 6 = −4 < 0. Discussion: We write the symbol (⇒) to indicate to the reader that we are proving that the first statement implies the second statement. 1 applies and this is what we use. Similarly, we write (⇐) to indicate that we are proving that the second statement implies the first statement. The statement [ P if and only if Q ] is the statement that P and Q are equivalent, meaning that they are both true or both false. For instance, in the preceding example, both P and Q are true in the case n = 1, while in the case n = 2 they are both false.

Here is the method for proving a general if-then statement using a string of implications. Statement: Proof : P ⇒ Q. ) We have P ⇒ · · · ⇒ Q. 2 Example Let m ∈ Z. Prove, using a string of implications: 2m ∈ {n ∈ Z | n + 1 ≥ 6} =⇒ m > 2. Proof We have 2m ∈ {n ∈ Z | n + 1 ≥ 6} =⇒ =⇒ (2m) + 1 ≥ 6 5 m ≥ > 2. 2 Discussion: For the first implication, we use what it means for 2m to be in the indicated set, namely, when 2m plays the role of the n a true statement results. 3 Example for some n ∈ Z+ . Proof Let r ∈ R.

2 Discussion: For the first implication, we use what it means for 2m to be in the indicated set, namely, when 2m plays the role of the n a true statement results. 3 Example for some n ∈ Z+ . Proof Let r ∈ R. Assume that 2r − 4 ∈ [5, ∞). Put n = 4 and note that n ∈ Z+ . Since 2r − 4 ∈ [5, ∞), we have 2r − 4 ≥ 5, so r ≥ 9/2 > 4 = n. Discussion: The statement is a for-every statement, so we begin with [ Let r ∈ R ] . Then, the statement about this fixed element r is of the form [ If P , then Q ] , so we begin by assuming P .