Wikipedia:Reference desk/Archives/Mathematics/2017 July 14
Appearance
Mathematics desk | ||
---|---|---|
< July 13 | << Jun | July | Aug >> | July 15 > |
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. |
July 14
[edit]Slope–intercept form
[edit]Hey,
Can I wrote a slope intercept form for y=8x+5 like this: y-8x+5 (without equality)?
--141.226.146.224 (talk) 05:48, 14 July 2017 (UTC)
- You can write y-8x=5, and this means the same, but without an equals sign you don't have "verb" in your "sentence" so there is no meaning, it is just an expression, and cannot represent a line. Dbfirs 06:02, 14 July 2017 (UTC)
- thank you. I wasn't sure. --141.226.146.224 (talk) 06:31, 14 July 2017 (UTC)
- Note that it should be , not . -- Meni Rosenfeld (talk) 08:46, 14 July 2017 (UTC)
- ... yes, that's an alternative way to write the equation, but it does need an equals sign. Dbfirs 20:48, 14 July 2017 (UTC)
- Yes, but that's a separate problem. My understanding is that the OP tried to inquire if he can write the expression to refer to the line, with the part being implicit. You explained that this is not a thing and the equality should be included. I explained the unrelated issue that the OP had a calculation error and that is really . -- Meni Rosenfeld (talk) 00:54, 16 July 2017 (UTC)
- ... yes, that's an alternative way to write the equation, but it does need an equals sign. Dbfirs 20:48, 14 July 2017 (UTC)
- Note that it should be , not . -- Meni Rosenfeld (talk) 08:46, 14 July 2017 (UTC)
- thank you. I wasn't sure. --141.226.146.224 (talk) 06:31, 14 July 2017 (UTC)
But the Slope–intercept form is specifically y=mx+b. Bubba73 You talkin' to me? 02:59, 15 July 2017 (UTC)
- ... or y=mx+c in the UK. Dbfirs 08:41, 15 July 2017 (UTC)
Algebraic numbers such as those in casus irreducibilis
[edit]If a cubic equation over the rationals has three real roots, none of which are rational, the only way to express those roots algebraically is in terms of non-real complex numbers. This case is called casus irreducibilis.
What do we know about the set of all real algebraic numbers that are expressible algebraically only in terms of non-real complex expressions? Loraof (talk) 15:30, 14 July 2017 (UTC)
- For example, are "almost all" real algebraic numbers in this category? Loraof (talk) 00:00, 16 July 2017 (UTC)
- Since there are countably many algebraic numbers and the subset expressible in terms of real radical expressions is infinite, and so is its complement, it follows they have the same cardinality.--Jasper Deng (talk) 16:58, 17 July 2017 (UTC)