# Example 13 - Chapter 14 Class 11 Mathematical Reasoning

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 13 Check whether the following statement is true or not. If x, y ∈ Z are such that x and y are odd, then xy is odd. Statement : If x & y ∈ Z are such that x & y are odd, then xy is odd Let p : x , y ∈ Z such . that x and y are odd q : xy is odd That given statement is of the form if p ⇒ q We check the validity of the given statement Statement : If x & y ∈ Z are such that x & y are odd, then xy is odd Method 1 :- Direct Method By assuming that p is true, prove that q must be true. i.e. Assuming x & y are odd , prove that xy is odd Since x & y are odd Let x = 2m + 1 When m, n ∈ Z & y = 2n + 1 Calculating xy = (2m + 1 ) (2n + 1) = 2m (2n + 1 ) + 1 (2n + 1 ) = (2m) (2n) + 2m + 2n + 1 = 4mn + 2m + 2n + 1 = 2 (mn + m + n) + 1 This shows xy is odd Hence q is true. Therefore the given statement is true Method 2 :- Contrapositive Method Statement : If x & y ∈ Z are such that x & y are odd, then xy is odd By assuming that q is false, prove that p must be false. Let xy be not odd , prove that x and y are not odd Since xy is not odd ⇒ xy is even. ⇒ This is possible only if either x or y is even Let us take an example

Chapter 14 Class 11 Mathematical Reasoning

Concept wise

- Statements
- Writing negation of statements
- Negation - Checking if true or not
- Finding Compound Statement
- Words 'And' & 'Or'
- Inclusive and exclusive or
- Quantifiers
- Contrapositive and converse
- If then
- If and only if
- Proving not true/false (by giving counter examples)
- Proving True - If and only if
- Proving True - By Contrapositive
- Proving True - By Contradiction
- Necessary and sufficient condition

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.