Wikipedia:Reference desk/Archives/Mathematics/2009 March 28
| Mathematics desk | ||
|---|---|---|
| < March 27 | << Feb | March | Apr >> | March 29 > |
| Welcome to the Wikipedia Mathematics Reference Desk Archives |
|---|
| The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
March 28
[edit]coordinate axes
[edit]how do we define the slopes of the coordinate axes??? or do we simply say that he slope of x-axis is zero and that of the y-axis is infinity?? —Preceding unsigned comment added by 117.197.116.50 (talk) 02:16, 28 March 2009 (UTC)
- Slope is defined as 122.107.207.98 (talk) 03:19, 28 March 2009 (UTC)
- Using that definition, I would say that if the axes are perpendicular (like they usually are), the x-axis has a slope of zero () and the y-axis's slope is undefined (). But can't you have non-perpendicular axes? --Evan ¤ Seeds 03:42, 28 March 2009 (UTC)
- The formulae, m1 * m2 = -1 for perpendicular lines having slopes m1 and m2, does not hold if either m1 or m2 is undefined. I agree with a part of the previous answer. Non-perpendicular axis would not form a basis so it is usually appropriate to deal with perpendicular axis. --PST 05:59, 28 March 2009 (UTC)
- Wouldn't they though? It certainly wouldn't be an orthogonal basis, obviously, but you could still have a basis. Consider the vectors (0, 1) and (1, 1) (let's say and , respectively). They're obviously not orthogonal, but I think they're still a basis of R2. After all, for any vector, . I could easily be wrong though. --Evan ¤ Seeds 08:21, 28 March 2009 (UTC)
- They're a perfectly good basis, yes. Of course, when dealing with an inner product space the natural notion is 'orthonormal basis'. Algebraist 13:36, 28 March 2009 (UTC)
- Wouldn't they though? It certainly wouldn't be an orthogonal basis, obviously, but you could still have a basis. Consider the vectors (0, 1) and (1, 1) (let's say and , respectively). They're obviously not orthogonal, but I think they're still a basis of R2. After all, for any vector, . I could easily be wrong though. --Evan ¤ Seeds 08:21, 28 March 2009 (UTC)
- The formulae, m1 * m2 = -1 for perpendicular lines having slopes m1 and m2, does not hold if either m1 or m2 is undefined. I agree with a part of the previous answer. Non-perpendicular axis would not form a basis so it is usually appropriate to deal with perpendicular axis. --PST 05:59, 28 March 2009 (UTC)
- Using that definition, I would say that if the axes are perpendicular (like they usually are), the x-axis has a slope of zero () and the y-axis's slope is undefined (). But can't you have non-perpendicular axes? --Evan ¤ Seeds 03:42, 28 March 2009 (UTC)
