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Question Number 113134 by bemath last updated on 11/Sep/20
any one can explain me how to  change decimal number to   biner number. i′m forgot.  example (315)_(10)  = (...)_2   thank you
$$\mathrm{any}\:\mathrm{one}\:\mathrm{can}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\:\mathrm{to} \\ $$$$\mathrm{change}\:\mathrm{decimal}\:\mathrm{number}\:\mathrm{to}\: \\ $$$$\mathrm{biner}\:\mathrm{number}.\:\mathrm{i}'\mathrm{m}\:\mathrm{forgot}. \\ $$$$\mathrm{example}\:\left(\mathrm{315}\right)_{\mathrm{10}} \:=\:\left(…\right)_{\mathrm{2}} \\ $$$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by I want to learn more last updated on 11/Sep/20
                   2                   315             Remainder             −−−−∣−−−−−∣−−−−−−−−                     2           ∣       157          ∣               1                            2           ∣         78           ∣               1                            2           ∣         39           ∣               0                            2           ∣         19           ∣               1                            2           ∣          9             ∣               1                            2           ∣          4             ∣               1                            2           ∣          2             ∣               0                            2           ∣          1             ∣               0                            2           ∣          0             ∣               1         read the remainder from bottom to top        315_(10)    =   100111011_2
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{315}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Remainder} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−\mid−−−−−\mid−−−−−−−− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\mathrm{157}\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{78}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{39}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\mathrm{19}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\: \\ $$$$\mathrm{read}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{from}\:\mathrm{bottom}\:\mathrm{to}\:\mathrm{top} \\ $$$$\:\:\:\:\:\:\mathrm{315}_{\mathrm{10}} \:\:\:=\:\:\:\mathrm{100111011}_{\mathrm{2}} \\ $$
Commented by bemath last updated on 11/Sep/20
thank you friend
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{friend} \\ $$
Answered by MJS_new last updated on 11/Sep/20
find the largest n with 2^n ≤315  ⇒ n=8  (315)_(10) =(1a_7 a_6 a_5 a_4 a_3 a_2 a_1 a_0 )_2   315−2^8 =59  find the largest n with 2^n ≤59  ⇒ n=5  (315)_(10) =(1001a_4 a_3 a_2 a_1 a_0 )_2   59−2^5 =27  ...
$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:{n}\:\mathrm{with}\:\mathrm{2}^{{n}} \leqslant\mathrm{315} \\ $$$$\Rightarrow\:{n}=\mathrm{8} \\ $$$$\left(\mathrm{315}\right)_{\mathrm{10}} =\left(\mathrm{1}\underset{\mathrm{7}} {{a}}\underset{\mathrm{6}} {{a}}\underset{\mathrm{5}} {{a}}\underset{\mathrm{4}} {{a}}\underset{\mathrm{3}} {{a}}\underset{\mathrm{2}} {{a}}\underset{\mathrm{1}} {{a}}\underset{\mathrm{0}} {{a}}\right)_{\mathrm{2}} \\ $$$$\mathrm{315}−\mathrm{2}^{\mathrm{8}} =\mathrm{59} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{largest}\:{n}\:\mathrm{with}\:\mathrm{2}^{{n}} \leqslant\mathrm{59} \\ $$$$\Rightarrow\:{n}=\mathrm{5} \\ $$$$\left(\mathrm{315}\right)_{\mathrm{10}} =\left(\mathrm{1001}\underset{\mathrm{4}} {{a}}\underset{\mathrm{3}} {{a}}\underset{\mathrm{2}} {{a}}\underset{\mathrm{1}} {{a}}\underset{\mathrm{0}} {{a}}\right)_{\mathrm{2}} \\ $$$$\mathrm{59}−\mathrm{2}^{\mathrm{5}} =\mathrm{27} \\ $$$$… \\ $$
Commented by bemath last updated on 11/Sep/20
thank you prof
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$
Answered by mr W last updated on 11/Sep/20
⌊log_2  315⌋=8  315−2^8 =59  ⌊log_2  59⌋=5  59−2^5 =27  ⌊log_2  27⌋=4  27−2^4 =11  ⌊log_2  11⌋=3  11−2^3 =3  ⌊log_2  3⌋=1  3−2^1 =1=2^0   ⇒315=2^8 +2^5 +2^4 +2^3 +2^1 +2^0   =1×2^8 +0×2^7 +0×2^6 +1×2^5 +1×2^4 +1×2^3 +0×2^2 +1×2^1 +1×2^0   =(100111011)_2
$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{315}\rfloor=\mathrm{8} \\ $$$$\mathrm{315}−\mathrm{2}^{\mathrm{8}} =\mathrm{59} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{59}\rfloor=\mathrm{5} \\ $$$$\mathrm{59}−\mathrm{2}^{\mathrm{5}} =\mathrm{27} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{27}\rfloor=\mathrm{4} \\ $$$$\mathrm{27}−\mathrm{2}^{\mathrm{4}} =\mathrm{11} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{11}\rfloor=\mathrm{3} \\ $$$$\mathrm{11}−\mathrm{2}^{\mathrm{3}} =\mathrm{3} \\ $$$$\lfloor\mathrm{log}_{\mathrm{2}} \:\mathrm{3}\rfloor=\mathrm{1} \\ $$$$\mathrm{3}−\mathrm{2}^{\mathrm{1}} =\mathrm{1}=\mathrm{2}^{\mathrm{0}} \\ $$$$\Rightarrow\mathrm{315}=\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{5}} +\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{1}} +\mathrm{2}^{\mathrm{0}} \\ $$$$=\mathrm{1}×\mathrm{2}^{\mathrm{8}} +\mathrm{0}×\mathrm{2}^{\mathrm{7}} +\mathrm{0}×\mathrm{2}^{\mathrm{6}} +\mathrm{1}×\mathrm{2}^{\mathrm{5}} +\mathrm{1}×\mathrm{2}^{\mathrm{4}} +\mathrm{1}×\mathrm{2}^{\mathrm{3}} +\mathrm{0}×\mathrm{2}^{\mathrm{2}} +\mathrm{1}×\mathrm{2}^{\mathrm{1}} +\mathrm{1}×\mathrm{2}^{\mathrm{0}} \\ $$$$=\left(\mathrm{100111011}\right)_{\mathrm{2}} \\ $$
Commented by bemath last updated on 11/Sep/20
thank you prof
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof} \\ $$
Answered by Aziztisffola last updated on 11/Sep/20
Commented by bemath last updated on 11/Sep/20
thank you friend
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{friend} \\ $$

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