Find one counterexample to show that each conjecture is false.

The sum of two numbers is greater than either number.

2 + -3 = -1

Prove the conjecture false by finding a counter example if a number is divisible by 10 then it is divisible by 20

To find a counterexample for the conjecture that the sum of two numbers is always greater than either number individually, we need to find two numbers for which the sum is not greater than either of the numbers.

Let's choose the numbers 3 and 5. If we add them together, we get 3 + 5 = 8.

Now, we compare the sum (8) to each of the numbers individually. Is 8 greater than 3? Yes. Is 8 greater than 5? Yes.

So, in this case, the sum of the two numbers (8) is indeed greater than either number individually. Therefore, these numbers do not provide a counterexample to the conjecture.

Let's try another pair of numbers: -7 and -4. If we add them together, we get -7 + (-4) = -11.

Now, we compare the sum (-11) to each of the numbers individually. Is -11 greater than -7? No. Is -11 greater than -4? No.

Therefore, in this case, the sum of the two numbers (-11) is not greater than either number individually. So, this pair of numbers serves as a counterexample to the conjecture.

Hence, the counterexample to the conjecture that the sum of two numbers is always greater than either number individually is -7 and -4.