How to prove inequalities

[CatMeow]

How can I prove inequalities of the form:

If |a - b| < c and b < d then a < c + d?

 

[CatMeow]

I did it like this

From | a - b | < c

a - c < b

And since b < d

a - c < d

Therefore, a < c + d.

Is this correct?

 

[JeffM]

Not quite in my opinion.

$$0 \le (a - b) \text { and } |a - b| < c \implies a - b < c \implies\\ (a - c) < b.$$
I follow that. But if

$$(a - b) < 0 \text { and } |a - b| < c \implies b - a < c \implies b < a + c, \text { NOT } a - c < b.$$

 

[Dr.Peterson]

I did it like this

From | a - b | < c

a - c < b

And since b < d

a - c < d

Therefore, a < c + d.

Is this correct?

It is not immediately obvious that | a - b | < c implies a - c < b, so you need to show it.

If | a - b | < c (and, by implication, c>0), then -c < a-b < c, so b-c < a < b+c, which does imply a-c < b.

Otherwise, what you say is correct.

 

[Steven G]

| a - b | < c implies that -c<a-b<c. So a-b<c and a-c<b. Continue ...

 

[pka]

How can I prove inequalities of the form:
If |a - b| < c and b < d then a < c + d?

Given $|a-b|<c~\&~b<d$ prove $a<c+d$.
$\Rightarrow\;-c<a-b<c$
$\Rightarrow\;b-c<a<c+b$
$\Rightarrow\;a<c+b<c+d$
$\Rightarrow\;a<c+d$

 

[Steven G]

Given $|a-b|<c~\&~b<d$ prove $a<c+d$.
$\Rightarrow\;-c<a-b<c$
$\Rightarrow\;b-c<a<c+b$
$\Rightarrow\;a<c+b<c+d$
$\Rightarrow\;a<c+d$

Sorry but I don't accept your proof for the following reason.
You are clearly assuming that if x<y<z, then x<z. As you know, this needs to be shown.

 

[pka]

Sorry but I don't accept your proof for the following reason.
You are clearly assuming that if x<y<z, then x<z. As you know, this needs to be shown.

1) The post is in the Pre-Algebra forum.
2) The less-than, $<\,$, relation is transitive.
___ $\therefore\;a<c~\&~c<d \Rightarrow\;a<d$