It is not immediately obvious that | a - b | < c implies a - c < b, so you need to show it.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?
Given ∣a−b∣<c & b<d prove a<c+d.How can I prove inequalities of the form:
If |a - b| < c and b < d then a < c + d?
Sorry but I don't accept your proof for the following reason.Given ∣a−b∣<c & b<d prove a<c+d.
⇒−c<a−b<c
⇒b−c<a<c+b
⇒a<c+b<c+d
⇒a<c+d
1) The post is in the Pre-Algebra forum.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.