Question #3881

If a < x < b and a < y < b,& show that |x-y|< b-a. Interpret this geometrically.

Expert's answer

Assume x<y. Then we have a<x<y<b. Then

|x-y|=-(x-y)=y-x=b-x+(y-b)<(b-x), because y<b and thus (y-b)<0. Then

|x-y|>b-x=b-a+(a-x)<(b-a), because a<x and thus (a-x)>0.

So we have got |x-y|<(b-a) for the case when x<y.

In the case when x≥y we can just notice that |x-y|=|y-x| and we can get the same situation as in the previous case. So |x-y|=|y-x|<b-a just like in case 1.

So we’ve proved that if a<x<b and a<y<b then |x-y|<(b-a).

What does it mean geometrically? |x-y| is the distance between points x and y. At the same time (b-a) is the distance between points b and a. As you can see the distance between points x and y must be less than the distance between a and b, becausex and y are inside the section between a and b. Look at the picture bellow:

<img src="../../..//assignments/uploaded_files/static/f0/c8/f0c8ccfc4b5f0019ba2dd1697882fffb.bmp" title="" alt="">

|x-y|=-(x-y)=y-x=b-x+(y-b)<(b-x), because y<b and thus (y-b)<0. Then

|x-y|>b-x=b-a+(a-x)<(b-a), because a<x and thus (a-x)>0.

So we have got |x-y|<(b-a) for the case when x<y.

In the case when x≥y we can just notice that |x-y|=|y-x| and we can get the same situation as in the previous case. So |x-y|=|y-x|<b-a just like in case 1.

So we’ve proved that if a<x<b and a<y<b then |x-y|<(b-a).

What does it mean geometrically? |x-y| is the distance between points x and y. At the same time (b-a) is the distance between points b and a. As you can see the distance between points x and y must be less than the distance between a and b, becausex and y are inside the section between a and b. Look at the picture bellow:

<img src="../../..//assignments/uploaded_files/static/f0/c8/f0c8ccfc4b5f0019ba2dd1697882fffb.bmp" title="" alt="">

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