Consider the following conjecture: the sum of two even numbers will be even.
A.) give three examples supporting the conjecture. (I gave 2+2=4 4+4=8 10+10=20)
B.) Prove the conjecture( this is the part I need help with)
proof.
a+b = sum of even numbers
2(a/2)+2(b/2)= where a/2 and b/2 can be even or odd, or both.
2 ((a/2+b/2) ) has to be an even number, because it is evenly divislble by 2 (note the leading 2).
Therefore, the sum of two even numbers is even.
To prove the conjecture that the sum of two even numbers will always be even, we need to demonstrate that any pair of even numbers, when added together, will result in an even number outcome.
Let's proceed with the proof:
Assuming two even numbers, we can represent them as follows:
Let a and b be two even numbers, with a = 2m and b = 2n, where m and n are integers.
Now, let's add these two numbers together:
a + b = 2m + 2n.
We can factor out the common factor of 2, which gives us:
a + b = 2(m + n).
Since (m + n) is an integer (as the sum of two integers is always an integer), we can rewrite the equation as:
a + b = 2k, where k = (m + n).
We have expressed the sum (a + b) as 2 multiplied by another integer k. Therefore, a + b is a multiple of 2 which implies it is an even number.
Hence, the proof shows that the sum of any two even numbers will always be an even number.
To prove the conjecture that the sum of two even numbers will be even, we need to show that for any two even numbers, their sum will always be even.
A.) We can start by giving three examples that support the conjecture:
1) Let's consider the even numbers 2 and 4. Adding them together, we get 2 + 4 = 6. Since 6 is an even number (divisible by 2 without leaving a remainder), this example supports the conjecture.
2) Next, let's take the even numbers 6 and 8. Adding them together, we obtain 6 + 8 = 14. Again, 14 is divisible by 2 without a remainder, confirming the conjecture.
3) Lastly, consider the even numbers 12 and 16. Their sum is 12 + 16 = 28, which is also divisible by 2 without any remainder. This example further supports the conjecture.
B.) Now, let's prove the conjecture more formally:
To prove this conjecture, we need to use the definition of an even number. An even number, by definition, is an integer that can be divided by 2 without leaving a remainder. Let's assume that we have two even numbers, x and y.
Since x and y are even, we can express them as:
x = 2a (where a is an integer)
y = 2b (where b is an integer)
Now, let's find the sum of these two even numbers:
x + y = 2a + 2b
Using the distributive property, we can factor out the common factor of 2:
x + y = 2(a + b)
Here, (a + b) is also an integer, which means the sum x + y can be expressed as a multiple of 2. In other words, the sum of two even numbers is divisible by 2 without leaving a remainder, which verifies that the sum is indeed even.
Therefore, we have proven the conjecture: the sum of two even numbers will always be even.