# Proof that -(a + b) = (-a) + (-b)

This is the second to the last post on proofs of the Basic Algebra Theorems. In this post, we prove that -(a + b) = (-a) + (-b).

Theorem

For any a and b,  -(a + b) = (-a) + (-b).

Proof

– (a + b) = (-1)(a + b) by Corollary 1 of this theorem on the product of positive and negative numbers.

By the Distributive Property (Axiom 4), we have (-1)(a + b) = (-1)(a) + (-1)(b).

(-1)(a) + (-1)(b) = (-a) + (-b) again, by Corollary 1 of this theorem.

This ends the proof.

# Proof that the Product of Any Number and Zero is Zero

We continue the series of proofs of the basic theorems in Algebra.  In this post, we use the Axioms of Real Numbers to prove that the product of any number and zero is zero. You may go back the the preceding link to review the axioms stated below.

Theorem

For any number a, 0a = 0.

Proof

We know that 0 + 0 = 0. Multiplying both sides by a, we have

(0 + 0)a = 0a.

By the Distributive Property (Axiom 4), we get

0a + 0a = 0a.

Since by Axiom 5A, any number has an Additive Identity (recall c = c + 0 for any c), we can add 0, the right hand side of the equation giving us

0a + 0a = 0a + 0.

By the Cancellation Law, we eliminate one 0a on both sides of the equation resulting to

0a = 0

which is what we want to prove.

# The Sum of the First n Positive Integers

According to an anecdote, the math genius Carl Friedrich Gauss, added all the numbers from 1 through 100 when he was at a very young age. The sum of the first positive 100 integers is 5050 and Gauss was able to give this sum in less than a minute. True or not, how can we get the sum of all these numbers with admirable speed? Well, it’s not really that hard.

If you want to be a great mathematician someday like Gauss, you must learn to explore problems and look for patterns. Notice that if you add the first $latex 100$ integers, the sum of the largest and the smallest integer is $latex 101$. Also, the sum of the second largest and the second smallest is $latex 101$. This also goes with the third largest and the third smallest and so on. Since there are the first $latex 100$ numbers can be divided into $latex \frac{100}{2}=50$ pairs, each of which has a sum of $latex 101$, the sum of all the numbers is $latex 50(101) = 5050$. We will use this strategy to generalize. We explore the sum of the first positive n integers. Continue reading