Prove the statements in 1-11. In each case use only the...

Prove the statements in 1-11. In each case use only the definitions of the terms and the Assumptions listed on page 161, not any previously established properties of odd and even integers. Follow the directions given in this section for writing proofs of universal statements. 1. For every integer n, if n is odd then 3n + 5 is even. 2. For every integer m, if m is even then 3m + 5 is odd. 3. For every integer n, 2n - 1 is odd. 4. Theorem 4.2.2: The difference of any even integer minus any odd integer is odd. 5. If a and b are any odd integers, then a² + b² is even. 6. If k is any odd integer and m is any even integer, then k² + m² is odd.

Transcript

00:01 In this question we are asked to prove certain statement.

00:05 So in the first statement, we are asked to prove that for every integer n, if n is odd, then 3n plus 5 is even.

00:16 So let's take n as a general odd number, say 2k plus 1.

00:27 If we substitute this value of n in 3n plus 5, we get 3 2k plus 1 plus 5, which is 6k plus 3 plus 5, that is 6k plus 8.

00:47 Now since 6k is individually is an even number and 8 is also in an even number and we know that the sum of even and even is always even.

01:01 Or we can write 6k plus 8 as 2, 3k plus 4.

01:07 And we know that whenever a number is multiplied by 2, it always become even.

01:13 So we can say that for every odd integer n, the value of 3n plus 5 is even.

01:22 Now in the next part, that is in the b part, we are asked to prove for every integer m, if m is even, then 3m plus 5 is odd.

01:33 So let's take m as general even function that is 2k substitute this value of 2k in 3m plus 5 we get 3 2k plus 5 that is 6k plus 5 that is 6k plus 4 plus 1 which will be 2 3k plus 2 plus 1 so it is of the form 2n plus 1 2n plus 1.

02:08 Which is an odd form that is whenever a multiple of 2 whenever in a multiple of 2 we add 1 it becomes odd so we can say that for every odd even integer m the value of 3m plus 5 is always odd now in the third part we are asked for every integer n 2n minus 1 is odd for every integer n 2 n minus 1 is odd for every integer n 2 n minus 1 is odd we have to proof that.

02:45 So let's take n as an even integer first.

02:50 That is let's take n as 2k.

02:54 Whenever we substitute the value of n as 2k into n plus 1 we get 2 into 2k minus 1.

03:01 That is 4k minus 1.

03:03 So whenever a multiple of 4 is subtracted by 1 it becomes odd.

03:10 So for n as an even integer the value of 2n minus 1 is odd.

03:15 Now let's take n as odd number, that is take n as 2k plus 1.

03:21 If we substitute the value of n as 2k plus 1 in the equation 2 in minus 1, we get 2 into 2k plus 1 minus 1 which will be 4k plus 2 minus 1.

03:33 That is 2 into 2k plus 1 minus 1.

03:42 So whenever we subtract 1 from the multiple of 2, always become odd.

03:50 So for every odd integer in the value of 2n minus 1 is odd.

03:57 So from these two conclusions we can say that the value of 2n minus 1 is always odd for every integer n...

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