![Example 13 - Chapter 14 Class 11 Mathematical Reasoning - Part 2](https://d1avenlh0i1xmr.cloudfront.net/d92c3ed7-9645-475f-8b84-45eeafc5f27f/slide44.jpg)
![Example 13 - Chapter 14 Class 11 Mathematical Reasoning - Part 3](https://d1avenlh0i1xmr.cloudfront.net/08a0ddc5-1fb2-4f1e-b60c-c700533d5054/slide45.jpg)
![Example 13 - Chapter 14 Class 11 Mathematical Reasoning - Part 4](https://d1avenlh0i1xmr.cloudfront.net/c3995357-8f63-48af-b8a6-45685d3ce505/slide46.jpg)
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Proving True - If and only if
Proving True - If and only if
Last updated at April 16, 2024 by Teachoo
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