Question

For all real numbers a and t |a+b| < |a| +|b|

Answer 1

proof to do.

Step-by-step explanation:

To prove that for all a & b in R, | a + b | <= | a | + | b |.

There are six cases.

1) a = 0.

LHS = | 0+b| = | b |

RHS = 0 + | b | = LHS.

2 ) b = 0.

LHS = | a + 0 | = | a |

RHS = | a | + 0 = LHS.

3) a & b are both positive.

LHS = | a + b | = | a | + | b | = RHS.

4 ) a & b are both negative. Let a = -a1, and b = -a2, where a1 & a2 are both positive. |a | = | a1 |

LHS = | - a1 - a2 | = | a1 + a2 | = | a1 | + | a2 |

= a1 + a2 = |a| + |b| = RHS

5) a is positive, and b is negative. b = -a2, where a2 > 0. | b | = | a2 | = a2.

LHS = | a + b | = | a - a2 | < a

RHS = |a | + | b | = a + a2 > a.

So LHS < RHS

6) a is negative and b is positive. It's similar to the case 5. LHS is < b. RHS is more than b. So LHS < RHS.

So the proof is complete.

Answer 2

The inequality \(|a + b| < |a| + |b|\) holds true for all real numbers \(a\) and \(b\). This inequality essentially states that the absolute value of the sum of two numbers is always less than or equal to the sum of the absolute values of those numbers.

This property is a consequence of the triangle inequality for real numbers. The triangle inequality states that for any real numbers \(a\) and \(b\), the sum of the absolute values of \(a\) and \(b\) is greater than or equal to the absolute value of their sum:

\[|a| + |b| \geq |a + b|\]

So, the inequality \(|a + b| < |a| + |b|\) is true for all real numbers \(a\) and \(b\).