Timeline for Can a right triangle have odd-length legs and even-length hypotenuse?
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2016 at 3:05 | audit | First posts | |||
| Dec 16, 2016 at 3:05 | |||||
| Nov 30, 2016 at 13:32 | comment | added | Vidyanshu Mishra | Yes, @CarstenS but I just edited my question as S.G. suggested me. There is no difference in these two. | |
| Nov 30, 2016 at 13:28 | comment | added | Carsten S | Btw, I do not find “1 is not a multiple of 2” to be more evident than “2 is not a multiple of 4”. | |
| Nov 30, 2016 at 13:20 | history | edited | Vidyanshu Mishra | CC BY-SA 3.0 |
bring it to people's notice
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| Nov 30, 2016 at 11:09 | comment | added | Vidyanshu Mishra | YEs, I agree @ypercubeᵀᴹ and then saying that since k,m,n are integers so it can't be true. | |
| Nov 30, 2016 at 10:26 | comment | added | ypercubeᵀᴹ | Another way is to write it as $2(-k^2-m^2-k-m+n^2) = 1$ | |
| Nov 30, 2016 at 1:19 | history | edited | Vidyanshu Mishra | CC BY-SA 3.0 |
Improved
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| Nov 30, 2016 at 1:18 | comment | added | Vidyanshu Mishra | Yes @TonyK I was just saying same thing in two ways. Thanks for suggestion | |
| Nov 30, 2016 at 0:20 | comment | added | TonyK | It's a bit confusing to say "you were wrong OR there does not exist such a triangle." Logically, what is true is "you were wrong AND there does not exist such a triangle"! I know that by "OR", you mean "to put it another way"; but in a mathematical context that is not so clear. A better word is THEREFORE. | |
| Nov 29, 2016 at 18:41 | comment | added | Vidyanshu Mishra | I have taken your suggestion into account as it will improve the answer.Thanks @S.G. | |
| Nov 29, 2016 at 18:39 | history | edited | Vidyanshu Mishra | CC BY-SA 3.0 |
added necessary info, took the comment into consideration.
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| Nov 29, 2016 at 17:46 | history | edited | Vidyanshu Mishra | CC BY-SA 3.0 |
added 1 character in body
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| Nov 29, 2016 at 17:36 | comment | added | S. G. | I prefer dividing the equation by 2, which yields an odd integer on the left equalling an even integer on the right, violating parity. I think that's about as intuitive as equating a half-integer to an integer. | |
| Nov 29, 2016 at 17:32 | comment | added | Peter - Reinstate Monica | I find the contradiction even more obvious if the equation is divided by 4. The left side then clearly is not an integer (given all variables are integers, the bracketed sum is one, to which we add 1/2), while the right side clearly is. I know, it's the same statement modulo 4, but still. | |
| S Nov 29, 2016 at 15:49 | history | suggested | costrom | CC BY-SA 3.0 |
Moved exponent outside of parens
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| Nov 29, 2016 at 15:45 | review | Suggested edits | |||
| S Nov 29, 2016 at 15:49 | |||||
| Nov 29, 2016 at 15:29 | history | edited | Vidyanshu Mishra | CC BY-SA 3.0 |
removed some errors
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| Nov 29, 2016 at 15:25 | history | edited | Dominik | CC BY-SA 3.0 |
Removed brackets
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| Nov 29, 2016 at 15:23 | history | answered | Vidyanshu Mishra | CC BY-SA 3.0 |