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ganeshie8 Group Title

\(11_{11}^2 = ?_{11}\)

  • 2 years ago
  • 2 years ago

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  1. mathslover Group Title
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    What is this? @ganeshie8 can u explain this to me?

    • 2 years ago
  2. Zekarias Group Title
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    \[131_{11}\]

    • 2 years ago
  3. ganeshie8 Group Title
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    that looks correct... actually i thought of asking as we move left in base 11, 11^2 becomes 100 ? right ?

    • 2 years ago
  4. UnkleRhaukus Group Title
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    \[121_{11}\]

    • 2 years ago
  5. ganeshie8 Group Title
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    @mathslover subscript means its base 11

    • 2 years ago
  6. UnkleRhaukus Group Title
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    @Zekarias check that

    • 2 years ago
  7. Zekarias Group Title
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    You made algebra mistake I think @UnkleRhaukus

    • 2 years ago
  8. UnkleRhaukus Group Title
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    121+22+1=144

    • 2 years ago
  9. sauravshakya Group Title
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    IS A_11=10_10 ??

    • 2 years ago
  10. UnkleRhaukus Group Title
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    yes @sauravshakya

    • 2 years ago
  11. sauravshakya Group Title
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    Well then the answer must be less than 121

    • 2 years ago
  12. ganeshie8 Group Title
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    im getting 121 too.. but the q i thought of asking was different

    • 2 years ago
  13. UnkleRhaukus Group Title
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    \[11^2_{11}=(11+1)^2=12^2=144\] \[144=121+22+1=121_{11}\]

    • 2 years ago
  14. sauravshakya Group Title
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    |dw:1349083712778:dw|

    • 2 years ago
  15. UnkleRhaukus Group Title
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    http://openstudy.com/study#/updates/50694b4ce4b0e3061a1d953f

    • 2 years ago
  16. sauravshakya Group Title
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    |dw:1349083784134:dw|

    • 2 years ago
  17. ganeshie8 Group Title
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    \(11_{10}^2\) = \(100_{11}\) \(11_{10}^3\) = \(1000_{11}\) \(11_{10}^4\) = \(10000_{11}\)

    • 2 years ago
  18. ganeshie8 Group Title
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    does that mean the positional value changes as smooth as it changes in base 10 ? therez no advantage of base 10... i use to think we use base 10 coz the progression of positional value is smooth in base 10

    • 2 years ago
  19. UnkleRhaukus Group Title
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    im not sure what you mean by "positional value changes as smooth"

    • 2 years ago
  20. ganeshie8 Group Title
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    in base 10, the position value increases in powers of \(10_{10}\), like 10, 100, 1000, 100000 .... . same is happening wid base 11 also, its changing as powers of \(11_{10}\) : 10, 100, 1000, 100000... only if i think in base 11, i see its changing smoothly... else, in base 10 i see it changing as, 11, 121, 1331, 14641...

    • 2 years ago
  21. sauravshakya Group Title
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    Since there are 9 digits only....... So I think base is better

    • 2 years ago
  22. sauravshakya Group Title
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    I mean base 10

    • 2 years ago
  23. ganeshie8 Group Title
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    as long as i think in one base system only, i dont see any problem. all systems look equal to me nw. @sauravshakya 9 digits or 10 digits hw does it matter... u mean 9/10 is easy to remember for us... and since, around 5 digits would be too small, and again, around 20 would be too many to remember... so 10 digits look fairly good to remember so we stuck wid base 10... . looks good rationale to me :)

    • 2 years ago
  24. sauravshakya Group Title
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    I mean it is easy to do calculation using base 10..... like division and multiplication

    • 2 years ago
  25. sauravshakya Group Title
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    This surely takes time: A^2_11 = ?_11

    • 2 years ago
  26. ganeshie8 Group Title
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    |dw:1349086304417:dw|

    • 2 years ago
  27. ganeshie8 Group Title
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    surely it doesnt look easy..but thats becoz we didnt learn multiplication tables in base 11, we learned them in base 10. but if we lived in base 11, we would have done it in a snap. i dont think we would have missed anything.. . but im really not sure.. im just inclining towards thinking like this, not fully sure yet .. .

    • 2 years ago
  28. sauravshakya Group Title
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    Agreed.

    • 2 years ago
  29. UnkleRhaukus Group Title
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    |dw:1349086853953:dw|

    • 2 years ago
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