QuestionProve by induction that for $n$ numbers $a_1, a_2, \ldots, a_n$ , $|a_1 + \cdots + a_n| \leq |a_1| + \cdots + |a_n|$ . Also, show $|b| < a$ iff $-a < b < a$ .

Studdy Solution

STEP 1

Assumptions1. The triangle inequality states that for any real numbers $x$ and $y$ , we have $|x + y| \leq |x| + |y|$ . . The numbers $a1, a, ..., a_n$ are real numbers.
3. The symbol $|\cdot|$ denotes the absolute value of a number.
4. The symbol $\leq$ denotes less than or equal to.

STEP 2

We will use mathematical induction to prove the inequality. The base case is when $n=1$ . $\left|a_{1}\right| \leq \left|a_{1}\right|$ This is obviously true.

STEP 3

Assume that the inequality holds for some $k \geq1$ , that is $\left|a_{1}+a_{2}+\cdots+a_{k}\right| \leq \left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{k}\right|$

STEP 4

We need to show that the inequality holds for $k+1$ , that is $\left|a_{1}+a_{2}+\cdots+a_{k}+a_{k+1}\right| \leq \left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{k}\right| + \left|a_{k+1}\right|$

STEP 5

We can rewrite the left side of the inequality as $\left|\left(a_{1}+a_{2}+\cdots+a_{k}\right) + a_{k+1}\right|$

STEP 6

By the triangle inequality, we have $\left|\left(a_{1}+a_{2}+\cdots+a_{k}\right) + a_{k+1}\right| \leq \left|\left(a_{1}+a_{2}+\cdots+a_{k}\right)\right| + \left|a_{k+1}\right|$

STEP 7

By the induction hypothesis, we can replace $\left|\left(a_{1}+a_{2}+\cdots+a_{k}\right)\right|$ with $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{k}\right|$ , so we have $\left|\left(a_{1}+a_{2}+\cdots+a_{k}\right) + a_{k+1}\right| \leq \left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{k}\right| + \left|a_{k+1}\right|$

STEP 8

This completes the induction step, and hence the inequality holds for all $n \geq1$ .

STEP 9

For the second part of the problem, we need to show that $|b|<a$ if and only if $-a<b<a$ .

STEP 10

Assume that $|b|<a$ . This means that $-a<b$ and $b<a$ . This is the definition of the absolute value.

STEP 11

Now assume that $-a<b<a$ . This implies that $b<a$ and $-b<a$ . By the definition of absolute value, this means that $|b|<a$ .

STEP 12

Therefore, we have shown that $|b|<a$ if and only if $-a<b<a$ .

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