Real Numbers

Proof. Suppose \(e_1,e_2\in\mathbb{R}\) are additive identities. Then, by Axiom 2 \[e_1 = e_1 + e_2\] since \(e_2\) is an additive identity. Likewise, by Axiom 2 \[e_1 + e_2 = e_2\] since \(e_1\) is an additive identity. It follows that \(e_1=e_2\). ◻

Proof. Let \(a\in\mathbb{R}\). Suppose \(b,c\in\mathbb{R}\) are additive inverses of \(a\). Then, by Axiom 2 \[b = b + 0.\] By Axiom 3 \[b + 0 = b + (a + c)\] since \(c\) is an additive inverse of \(a\). By Axiom 1 \[b + (a + c) = (b + a) + c.\] By Axiom 3 \[(b + a) + c = 0 + c\] since \(b\) is an additive inverse of \(a\). By Axiom 2 \[0 + c = c.\] It follows that \(b=c\). ◻

Proof. Suppose \(e_1,e_2\in\mathbb{R}\) are multiplicative identities. Then, by Axiom 6 \[e_1 = e_1 \cdot e_2\] since \(e_2\) is a multiplicative identity. Likewise, by Axiom 6 \[e_1 \cdot e_2 = e_2\] since \(e_1\) is a multiplicative identity. It follows that \(e_1 = e_2\). ◻

Proof. Let \(a\in\mathbb{R}\setminus\{0\}\). Suppose \(b,c\in\mathbb{R}\) are multiplicative inverses of \(a\). Then, by Axiom 6 \[b = b \cdot 1.\] By Axiom 7 \[b \cdot 1 = b \cdot (a \cdot c)\] since \(c\) is a multiplicative inverse of \(a\). By Axiom 5 \[b \cdot (a \cdot c) = (b \cdot a) \cdot c.\] By Axiom 7 \[(b \cdot a) \cdot c = 1 \cdot c\] since \(b\) is a multiplicative inverse of \(a\). By Axiom 6 \[1 \cdot c = c.\] It follows that \(b=c\). ◻

Proof. Since \(<\) and \(=\) are reflexive and transitive, so is \(\leq\). Supose \(a,b\in\mathbb{R}\) with \(a\leq b\) and \(b\leq a\). By Axiom 10, we cannot have \(a = b\) and either of \(a < b\) or \(b < a\) simultaneously, so it must be the case that \(a = b\). Then \(\leq\) is antisymmetric. ◻

Proof. Suppose \(c_1,c_2\in\mathbb{R}\) are supremums of \(A \subseteq\mathbb{R}\). Then \(c_1\) is an upper bound of \(A\). Since \(c_2\) is a supremum of \(A\) \[c_2 \leq c_1.\] Likewise, \(c_2\) is an upper bound of \(A\). Since \(c_1\) is a supremum of \(A\) \[c_1 \leq c_2.\] Since \(\leq\) is antisymmetric, it follows that \(c_1 = c_2\). ◻

Proof. Let \(x,y\in\mathbb{R}\). Without loss of generality, suppose \(x \leq y\). We have four cases to consider.
Case 1: Suppose \(0 \leq x \leq y\). Then \(x+y \geq 0\) and so \[|x+y| = x+y = |x|+|y|.\] Case 2: Suppose \(x \leq y < 0\). Then \(x+y < 0\) and so \[|x+y| = -(x+y) = -x-y = |x|+|y|.\] Case 3: Suppose \(x < 0 \leq y\) with \(|x|\leq |y|\). Then \(x+y \geq 0\) and so \[|x+y| = x+y = -|x|+|y| \leq |x|+|y|.\] Case 4: Suppose \(x < 0 \leq y\) with \(|y| < |x|\). Then \(x+y < 0\) and so \[|x+y| = -(x+y) = -x-y = |x|-|y| \leq |x|+|y|.\] In any case, \(|x+y| \leq |x|+|y|\). ◻

Proof. Let \(a,b\in\mathbb{R}\). By the triangle inequality \[|b| = |a+(b-a)| \leq |a| + |b-a| \quad\text{and}\quad |a| = |b+(a-b)| \leq |b| + |a-b|.\] Then \[|b|-|a| \leq |b-a| \quad\text{and}\quad |a|-|b| \leq |a-b|.\] It follows that \[-|a-b| = -|b-a| \leq |a|-|b| \leq |a-b|\] and so \[||a|-|b|| \leq |a-b|.\] ◻

Proof. Let \(x\in\mathbb{R}\). If \(x\geq 0\), then \(|x|=x\geq 0\). If \(x<0\), then \(|x|=-x>0\). It follows that \(|x|\geq 0\). Moreover, \(|x|=0\) if and only if \(x\geq 0\) and \(x=0\) or \(x<0\) and \(-x=0\). Only one case is possible, from which it follows \(|x|=0\iff x=0\). ◻

Proof. Let \(c,x\in\mathbb{R}\). Without loss of generality, suppose \(c\leq x\). We have three cases to consider.
Case 1: Suppose \(0 \leq c \leq x\). Then \(c \geq 0\), \(x\geq 0\), and \(c\cdot x\geq 0\). It follows that \[|c\cdot x| = c\cdot x = |c|\cdot|x|.\] Case 2: Suppose \(c \leq x < 0\). Then \(c < 0\), \(x < 0\), and \(c\cdot x > 0\). It follows that \[|c\cdot x| = c\cdot x = (-c)\cdot (-x) = |c|\cdot|x|.\] Case 3: Suppose \(c < 0 \leq x\). Then \(c < 0\), \(x\geq 0\), and \(c\cdot x \leq 0\). It follows that \[|c\cdot x| = -(c\cdot x) = (-c)\cdot x = |c|\cdot|x|.\] In any case, \(|c\cdot x| = |c|\cdot|x|\). ◻