Question

1. Prove the following statements by the method of contradiction.
a) If n^2 is an even number, then n is an even number.​

Answer 1

Answer:

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Yes it is afcource that if a number is even then it's square as well as cube will be even or if a number is odd then the square as well as the cube if the number is odd ✓

so in the case of n is even then the square of it that is n² will be even

Let, us now proof it

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Let us take any even number which is 2

So n= 2

Now square it we get

$= >n = 2 \\ \\ squaring \:(n) \\ \\ = > n ^{2} = 2^{2} \\ \\ = > n ^{2} = 4$

So from the above calculation we find the the value of n ² is 4 ✓ that is when we square any even number we get even no and let us also now proof if we square any odd number we get another odd no ✓

Let us take any odd number which is 3 so n= 3

$= > n = 3 \\ \\ squaring \: n \\ \\ = > n ^{2} = 3 ^{2} \\ \\ = > n ^{2} = 9$

So here we can see the when we have squared the odd number 3 we get another odd number 9 ✓

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Hope it's help u ❣️

Answer 2

Answer:

proof: we have to prove that ''if n is odd then $n^{2}$ is odd'' lets assume that n is odd ,n=2k+1 where k an element of real numbers

$n^{2}$=(2k+1)²

   =4$k^{2}$+4k+1

   =(4$k^{2}$+4k)+1

   =2(2$k^{2}$+2k)+1

let (2$k^{2}$+2k) be m

   =2m+1

therefore n^{2} is odd by contraposition

Step-by-step explanation: