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# srdanova math

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### #31 bmk1245

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Posted 28 January 2012 - 11:50 AM

msbiljanica, on 28 January 2012 - 08:28 AM, said:

srcko ab, abc , 55={5,10,15,20,...} ,22222={2,4,6,...,220,222}

Thanks, I guess...

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Posted 28 January 2012 - 12:42 PM

Uhm, ok.

So what does N2M  mean in your math? I am guessing  N2M={N, N+2, N+4,...,M}. If correct, great, so now you have redefined how we denote sets. So, how do you now denote variables? Like the variables N1, N2, M19 etc?

Cheers,

Edited by badeskov, 28 January 2012 - 01:30 PM.

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Posted 28 January 2012 - 12:44 PM

bmk1245, on 28 January 2012 - 11:50 AM, said:

Thanks, I guess...

I rather like it. You know, bmk, d/dx is rather boring and dull. Also, coombacoombla is a nice long and good sounding word, it sooo takes the emphasis over boring numbers in long equations

Cheers,

"Life is not a journey to the grave with the intention to arrive safely in a pretty and well-preserved body, but rather to skid in broadside, thoroughly used, totally worn out, and loudly proclaiming: Wow!! What a ride!". Said to to Dean Karnazes by a running buddy.

### #34 bmk1245

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Posted 28 January 2012 - 04:21 PM

badeskov, on 28 January 2012 - 12:44 PM, said:

I rather like it. You know, bmk, d/dx is rather boring and dull. Also, coombacoombla is a nice long and good sounding word, it sooo takes the emphasis over boring numbers in long equations

Cheers,
Math (as well as other scientific) books would be unliftable

Arguing with fool is like playing chess with pigeon: he will scatter pieces, peck King's crown, crap on bishop, and fly away bragging how he won the game... (heard once, author unknown).
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### #35 msbiljanica

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Posted 29 January 2012 - 08:50 AM

Presupposition-Parts number (a) and mobile number (b ) have a contact , the contact is delete
Process:
P1 4-(.0)2=2
P2 4-(.1)2=¤1(2)1¤ image
P3 4-(.2)2=2
P4 4-(.3)2=¤3(1)1¤
P5 4-(.4)2=¤4(0)2¤

General form
a-(.0)b=c
a-(.1)b=c
...
a-(.d)b=c
[S19]-subtraction
CM-only form a-(s.0/s.0)b=c , others do not know , axiom
_______________________________________________________________________
Presupposition-In the expression a+(.b)c=d , d-(s.0/s.0))11f or d-(s.0/s.0))number (more) from 11f
Process:
P1 3+(.s.0)5=8-(s.0/s.0))118 , 3+(s.0)5>017
P2 5+(.0)5=5-(s.0/s.0))224 , 5+(.0)5>321
...
General form - a+(.b)c=d-(s.0/s.0))11f ,a+(s.b)c>01e
a+(.b)c=d-(s.0/s.0)e , a+(.b)c>f
a+(.b)c=d-(s.0/s.0)efg , a+(.b)c>hij ...
CM-[S20]-know

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Posted 30 January 2012 - 05:29 AM

Sooo, care to elaborate on how you denote variables in your math?

Cheers,

"Life is not a journey to the grave with the intention to arrive safely in a pretty and well-preserved body, but rather to skid in broadside, thoroughly used, totally worn out, and loudly proclaiming: Wow!! What a ride!". Said to to Dean Karnazes by a running buddy.

### #37 msbiljanica

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Posted 30 January 2012 - 10:10 AM

Presupposition-Two ( more) addition (left and right inequalities) can be short to write
Process:
P1 3+(.013)4=y , 3+(.013)4>y ,3+(.013)4<y
P2 8+(.228)5=y , 8+(.228)5>y ,8+(.228)5<y
...
General form - a+(.bcd)e=y ,a+(.bcd)e>y , a+(.bcd)e<y
a+(.bcd_e)f=y , a+(.bcd_e)f>y , a+(.bcd_e)f<y ,...
CM-[S21]-does no know
______________________________________________________
Presupposition-Gap number ( value z) is a variable (same value) with the gaps that is constant
Process:
P1 ¤5(2)0¤,¤4(2)1¤,3(2)2¤,¤2(2)3¤,¤1(2)4¤,¤0(2)5¤ --5¤¤(2)

P2 ¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
¤2(2)0¤,¤1(2)1¤.¤0(2)2¤
¤1(2)0¤,¤0(2)1¤
¤0(2)0¤ --013¤¤(2)

¤2(2)0¤,¤1(2)1¤,¤0(2)2¤
¤3(2)0¤,¤2(2)1¤,¤1(2)2¤,¤0(2)3¤
...
21¤¤(2)

[S22]-variability of z number
CM-[S22]-does no know
_______________________________________-
PDF - http://www.mediafire...jzheej0qu93jjkm

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Posted 30 January 2012 - 11:06 PM

badeskov, on 30 January 2012 - 05:29 AM, said:

Sooo, care to elaborate on how you denote variables in your math?

I guess that is a no then....

Cheers,

"Life is not a journey to the grave with the intention to arrive safely in a pretty and well-preserved body, but rather to skid in broadside, thoroughly used, totally worn out, and loudly proclaiming: Wow!! What a ride!". Said to to Dean Karnazes by a running buddy.

### #39 sepulchrave

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Posted 31 January 2012 - 12:32 AM

badeskov, on 30 January 2012 - 11:06 PM, said:

I guess that is a no then....
Yeah, it seems like msbiljanica is just dumping his ``mathematics manifesto'' into google translate and then throwing it up on the forums, rather than actively explaining or discussing the content of his posts.

I mean since the OP stated that current mathematics were ``limited and sinful'' (emphasis mine) - which I am assuming (or hoping!) was just a poor translation - I wouldn't hold my breath for any extra insight into his existing posts.

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Posted 31 January 2012 - 01:09 AM

sepulchrave, on 31 January 2012 - 12:32 AM, said:

Yeah, it seems like msbiljanica is just dumping his ``mathematics manifesto'' into google translate and then throwing it up on the forums, rather than actively explaining or discussing the content of his posts.

I mean since the OP stated that current mathematics were ``limited and sinful'' (emphasis mine) - which I am assuming (or hoping!) was just a poor translation - I wouldn't hold my breath for any extra insight into his existing posts.

I fear that you are correct in your assessment - not unexpectedly, though. And  I wonder too how current mathematics are sinful. I have heard mathematics being called much, but I think sinful is a first.

Cheers,

"Life is not a journey to the grave with the intention to arrive safely in a pretty and well-preserved body, but rather to skid in broadside, thoroughly used, totally worn out, and loudly proclaiming: Wow!! What a ride!". Said to to Dean Karnazes by a running buddy.

### #41 msbiljanica

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Posted 31 January 2012 - 10:18 AM

Presupposition-Translation of gap number in the variability z ( with constant gap ) can addition
(s.0)
Process:
P1 ¤3(2)3¤+(.z)¤4(2)4=6¤¤(2)+(.z)8¤¤(2)=14¤¤(2)
P2 ¤1(6)1(9)1¤+(.z)¤3(6)2(9)1¤=3¤¤(6)(9)+(.z)6¤¤(6)(9)=9¤¤(6)(9)
...
General form ¤a(b)c¤+(.z)¤d(b)e¤=f¤¤(b )+(.z)g¤¤(b )=h¤¤(b ) ...
CM-[S23]-does no know
__________________________________________
Presupposition-Translation of gap number in the variability z ( with constant gap ) can subtraction
(s.0/s.0)
Process:
P1 ¤3(2)3¤-(.z)¤1(2)1=6¤¤(2)-(.z)2¤¤(2)=4¤¤(2)
P2 ¤3(6)2(9)1¤-(.z)¤1(6)1(9)1¤=6¤¤(6)(9)-(.z)3¤¤(6)(9)=3¤¤(6)(9)
...
General form ¤a(b)c¤-(.z)¤d(b)e¤=f¤¤(b )-(.z)g¤¤(b )=h¤¤(b ) ...
[S24]-z subtraction
CM-[S24]-does no know

### #42 msbiljanica

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Posted 01 February 2012 - 10:38 AM

Presupposition-In the expression a-(.b)c=d , d+(s.0))11 or d+(s.0)number (more) from 11
and d+(.z)11¤¤e or d+(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 4-(.2)2=2+(s.0)11 , 4-(.2)2<31
4-(.3)2=¤1(2)1¤+(.z)11¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)11¤¤(2) , 4-(.3)2<31¤¤(2)
P2 4-(.2)2=2+(s.0)4 , 4-(.2)2<6
4-(.3)2=¤1(2)1¤+(.z)7¤¤(2) , 4-(.3)2<2¤¤(2)+(.z)7¤¤(2), 4-(.3)<9¤¤(2)
...
General form - a-(.b)c=d-(s.0)11 ,a+(s.b)c<g1
a-(.b)c=d+(s.0)g , a+(.b)c<l
a-(.b)c=d+(s.0)kpg , a+(.b)c<hij ...
a-(.b)c=d+(.z)11¤¤e , a-(.b)c<s¤¤e+(.z)11¤¤e , a+(s.b)c<g1¤¤e
a-(.b)c=d+(.z)g¤¤e , a-(.b)c<s¤¤¤e+(.z)g¤¤e , a+(.b)c<l¤¤e
a-(.b)c=d+(.z)kpg¤¤e , a-(.b)c<s¤¤e+(.z)kpg¤¤e , a+(.b)c<hij¤¤e ...
[S25]-left inequality subtraction
CM-[S25]-know , forms without any gaps numbers not known
____________________________________________________________________
Presupposition-In the expression a-(.b)c=d , d-(s.0/s.0))11p or d+(s.0/s.0)number (more) from 11p
and d-(.z)11p¤¤e or d-(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 4-(.2)2=2-(s.0/s.0))112 , 4-(.2)2>011
4-(.3)2=¤1(2)1¤-(.z)211¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)211¤¤(2) , 4-(.3)2>011¤¤(2)
P2 4-(.2)2=2-(s.0/s.0))1 , 4-(.2)2>1
4-(.3)2=¤1(2)1¤-(.z)1¤¤(2) , 4-(.3)2>2¤¤(2)-(.z)1¤¤(2) , 4-(.3)>1¤¤(2)
...
General form - a-(.b)c=d-(s.0/s.0)11p ,a-(s.b)c>g1l
a-(.b)c=d-(s.0/s.0))g , a-(.b)c>l
a-(.b)c=d-(s.0/s.0))ksg , a-(.b)c>hij ...
a-(.b)c=d-(.z)11p¤¤e , a-(.b)c>s¤¤e-(.z)11p¤¤e , a-(s.b)c>g1k¤¤e
a-(.b)c=d-(.z)g¤¤e , a-(.b)c>s¤¤e-(.z)g¤¤e , a-(.b)c>l¤¤e
a-(.b)c=d-(.z)kpg¤¤e , a-(.b)c>s¤¤e-(.z)kpg¤¤e , a-(.b)c>hij¤¤e ...
[S26]-right inequality subtraction
CM-[S26]-know , forms without any gaps numbers not known

### #43 msbiljanica

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Posted 02 February 2012 - 09:39 AM

Presupposition-Two ( more) subtraction (left and right inequalities) can be short to write
Process:
P1 7-(.013)4=y , 7-(.013)4>y ,7-(.013)4<y
P2 8-(.228)5=y , 8-(.228)5>y ,8-(.228)5<y
...
General form a-(.bcd)e=y ,a-(.bcd)e>y , a-(.bcd)e<y
a-(.bcd_e)f=y , a-(.bcd_e)f>y , a-(.bcd_e)f<y ,...
[S27]-function subtraction
CM-[S27]-does no know
_____________________________________
Presupposition-Parts number (a) and mobile number (b ) have a contact , contact remains , the
rest is delete
Process:

P1 4 - (.0)2=2
P2 4 - (.1)2=2 image
P3 4 - (.2)2=2
P4 4 - (.3)2=1
P5 4 - (.4)2=1
General form
a - (.0)b=c
a - (.1)b=c
...
a - (.d)b=c

[S28]-opposite subtraction
CM-[S28]-does no know
-sign for. opposite subtraction (when you download the following PDF you will see how it looks)

### #44 msbiljanica

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Posted 03 February 2012 - 12:27 PM

3.how to solve this current knowledge of mathematics:
along a (20m) ,deleted between 10 m and 15 m (b=5m) , wet get c
20m-(.5m)=¤10m(5m)5m¤
___________________________________________________
Presupposition-In the expression a - (.b)c=d , d+(s.0))11 or d+(s.0)number (more) from 11
and d+(.z)11¤¤e or d+(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 5 - (.0)5=5+(s.0)11 , 5 - (.0)5<61
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)11¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤<2¤¤(2)+(.z)11¤¤(2) , ¤2(2)2¤ - (.1)¤1/1)2¤<31¤¤(2)
P2 5 - (.0)5=5+(s.0)4 , 5 - (.0)5<9
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤+(.z)7¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2)¤<2¤¤(2)+(.z)7¤¤(2), ¤2(2)2¤ - (.1)¤1(1)2¤<9¤¤(2)
...
General form a - (.b)c=d-(s.0)11 ,a - (s.b)c<g1
a - (.b)c=d+(s.0)g , a - (.b)c<l
a - (.b)c=d+(s.0)kpg , a - (.b)c<hij ...
a -(.b)c=d+(.z)11¤¤e , a - (.b)c<s¤¤e+(.z)11¤¤e , a - (s.b)c<g1¤¤e
a - (.b)c=d+(.z)g¤¤e , a - (.b)c<s¤¤¤e+(.z)g¤¤e , a - (.b)c<l¤¤e
a - (.b)c=d+(.z)kpg¤¤e , a - (.b)c<s¤¤e+(.z)kpg¤¤e , a - (.b)c<hij¤¤e ...
[S29]-left inequality opposite subtraction
CM-[S29]-does no know
_________________________________
Presupposition-In the expression a - (.b)c=d , d-(s.0/s.0))11p or d+(s.0/s.0)number (more) from
11p and d-(.z)11p¤¤e or d-(.z) number ( more ) from 11¤¤e , e={(f),(f)(f),(f)(f)(f),...}
Process:
P1 5 - (.0)5=5 -(s.0/s.0))115 , 5 - (.0)5>014
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)211¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)211¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>011¤¤(2)
P2 5 - (.0)5=5-(s.0/s.0))1 , 5 - (.0)5>1
¤2(2)2¤ - (.1)¤1(1)2¤=¤1(2)1¤-(.z)1¤¤(2)
¤2(2)2¤ - (.1)¤1(1)2¤>2¤¤(2)-(.z)1¤¤(2) , ¤2(2)2¤ - (.1)¤1(1)2¤>1¤¤(2)
...
General form a - (.b)c=d-(s.0/s.0)11p ,a - (s.b)c>g1l
a - (.b)c=d-(s.0/s.0))g , a - (.b)c>l
a - (.b)c=d-(s.0/s.0))ksg , a - (.b)c>hij ...
a - (.b)c=d-(.z)11p¤¤e , a - (.b)c>s¤¤e-(.z)11p¤¤e , a - (s.b)c>g1k¤¤e
a - (.b)c=d-(.z)g¤¤e , a - (.b)c>s¤¤e-(.z)g¤¤e , a - (.b)c>l¤¤e
a - (.b)c=d-(.z)kpg¤¤e , a - (.b)c>s¤¤e-(.z)kpg¤¤e , a - (.b)c>hij¤¤e ...
[S30]-right inequality opposite subtraction
CM-[S30]-does no know
__________________________________-
PDF - http://www.mediafire...qle11bvr2yuey9o

### #45 msbiljanica

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Posted 04 February 2012 - 04:34 PM

Presupposition-Two ( more) opposite subtraction (left and right inequalities) can be short to write
Process:
P1 7 - (.013)4=y , 7 - (.013)4>y ,7 - (.013)4<y
P2 8 - (.228)5=y ,  8 - (.228)5>y ,8 - (.228)5<y
...
General form  a - (.bcd)e=y ,a - (.bcd)e>y , a - (.bcd)e<y
a - (.bcd_e)f=y , a - (.bcd_e)f>y , a - (.bcd_e)f<y ,...
[S31]-function opposite subtraction
CM-[S31]-does no know
__________________________________
Presupposition-Two ( more) addition the same number can be written in shorted form
Process:
P1 a+(s.c)a=a×(s.c)2 , a×(s.c)b
P2 a+(s.c)a+(s.c)a=a×(s.c)3 , a×(s.c)b
P3 a+(s.c)a+(s.c)a+(s.c)a=a×(s.c)4 , a×(s.c)b
...
[S32]-multiplication
CM-[S32]-know , axiom
Presupposition-In terms a×(s.c)b , b can be number 0 (1)
Process:
P1 a×(s.c)0
P2 a×(s.c)1
[S32a]-multiplication-amendment

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