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Theorem: "n and m are even numbers = n × m is an even number." Proof: Any even number a can be presented in the form a = 2k, where k is some integer number, therefore, let us write n and m in the form = 2k1, m = 2k2, where k1, k2 are integers. The product is m x n = (2kj) × (2k2) = 2 × (2k,k2). Since k1, k2 are integer numbers, the twi the product of them will also be some integer number k3 = 2k,k2. Thus, m x n = 2k3, where k3 is an integer, therefore the product of the integers n and m is even.

Theorem: "n and m are even numbers = n × m is an even number." Proof: Any even number a can be presented in the form a = 2k, where k is some integer number, therefore, let us write n and m in the form = 2k1, m = 2k2, where k1, k2 are integers. The product is m x n = (2kj) × (2k2) = 2 × (2k,k2). Since k1, k2 are integer numbers, the twi the product of them will also be some integer number k3 = 2k,k2. Thus, m x n = 2k3, where k3 is an integer, therefore the product of the integers n and m is even.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Theorem: "n and m are even numbers =nxm is an even number." Proof: Any even number a can be presented in the form a = 2k, where k is some integer number, therefore, let us write n and m in the form n = 2k1, m = 2k2, where k1, k2 are integers. The product is m x n = (2kj) × (2k2) = 2 × (2k,k2). Since k1, k2 are integer numbers, the twice the product of them will also be some integer number k3 = 2kkɔ. Thus, m x n = 2k3, where k3 is an integer, therefore the product of the integers n and m is even. The reasoning used here is O equivalence proof O backward reasoning O reductio ad absurdum. O forward reasoning
Transcribed Image Text:Theorem: "n and m are even numbers =nxm is an even number." Proof: Any even number a can be presented in the form a = 2k, where k is some integer number, therefore, let us write n and m in the form n = 2k1, m = 2k2, where k1, k2 are integers. The product is m x n = (2kj) × (2k2) = 2 × (2k,k2). Since k1, k2 are integer numbers, the twice the product of them will also be some integer number k3 = 2kkɔ. Thus, m x n = 2k3, where k3 is an integer, therefore the product of the integers n and m is even. The reasoning used here is O equivalence proof O backward reasoning O reductio ad absurdum. O forward reasoning
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