Question

The sum of two even integers is even (do a proof by contradiction).

Answer 1

# Som of two even numbers is even. negate the statement in first let us we get :- # Sum of two even numbers is odd. Now let's prove it! Let 'a' and 'b' two integers. Then definitely da and 26 ore even numbers. This is by the de finition of the even numbers. So, Now we get in da + 2b = c m and an C must be odd number 1 5) p Carb) => c 2(xc Leo Carb)=% Now by de finition 2(x) must be even DUMDes 011, a is an integer and b is allo on integer, i (+ b? is also or integere.

Hence (a+b) is also an integer, and 2 Carb) coill be an even integer . by definition of even numbers, so, Now we have :) Adlar b) C a next se, (where caut b) = x) we have assumed that ie is an odd integer Hence, Even 3 Odd is a contradiction, iie. 2(x) t e 0 2 (0+b) c or do +20 Hence our assumption is wrong i that #SOM of hoo even numbers is odd. Hence, we can now say that as . e is not an add integer then it Most be an even integer, which proves the given statement, os original]

Also, asi 2 carb) El so, Carb) : Cis assumed to troed (Must be also true) but we know that is an odd integer here, and ana odd number when divided by 2 does not B gives an integer resolt. the result is a decimal number when an ' odd number is divided by 2. but (a + b) is an integer valve. Hence, integer I decimal Here is the contradie Hon, com and here our assomption feels, which meanse is not an odd Integer, then . de finitely it is an even integer, also, even integer dèvided by a yeóids an integer value Hence it sott's Hies.

Monce our original statement is true, - Sum of two even numbers is even.

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