# The Proof of the Cancellation Law of Addition

In the preceding posts, I have discussed the Axioms of Real Numbers and the Axioms of Equality. We will use these axioms to prove several basic theorems in Algebra. In this post, we prove the cancellation law of addition. That is, for any numbers a, b, and c,  if a + c = b + c, then a = b.

Theorem

For any real numbers a, b, c,  if a + c = b + c, then a = b.

Proof

a + c = b + c

Adding  –c  to both sides, we have

(a + c) + (-c) = (b + c) + (-c)

Using Axiom 2A, the associativity of addition, we have

a + (c + (-c)) = b + (c + (-c)).

By Axiom 6A, the existence of additive inverse, we get

a + 0 = b + 0.

By Axiom 5A, the existence of additive identity

a = b.

That ends the proof.

# A Proof that Square Root of 2 Plus Square Root of 3 is Irrational

#### Introduction

In the previous post, we have learned that $latex \sqrt{6}$ is irrational.  In this post, we will use that theorem to show that $latex \sqrt{2} + \sqrt{3}$ is also irrational. This proof was explained by Ivan Niven in his book Numbers: Rational and Irrational.

A common argument that might arise is that since $latex \sqrt{2}$ is irrational and $latex \sqrt{3}$ is irrational, then their sum is irrational. Like multiplication, the set of irrational numbers is not closed under addition. The sum of two irrational numbers is not necessarily irrational. For instance, $latex \sqrt{2}$ is irrational, $latex -\sqrt{2}$ is irrational, but $latex \sqrt{2} + (-\sqrt{2}) = 0$ is rational.

#### Theorem

$latex \sqrt{2} + \sqrt{3}$ is irrational.  Continue reading

# Proof that Square Root of 6 is Irrational

#### Introduction

We have learned that $latex \sqrt{3}$ is irrational and thousands of sites on the internet will show you the proof that $latex \sqrt{2}$ is irrational. In this post, we show that $latex \sqrt{6}$ is irrational. This is a prerequisite for the next post which will show that $latex \sqrt{2} + \sqrt{3}$ is irrational.

You will probably argue that since $latex \sqrt{2}$ is irrational and $latex \sqrt{3}$ is irrational, then their product should be irrational. This is not always the case. The product of two irrational numbers is not always irrational.  For instance, $latex \sqrt{2}$ is irrational but $latex (\sqrt{2})(\sqrt{2}) = 2$ is rational. And now, we proceed the the theorem and its proof. Continue reading