Convert 0.900 000 005 960 464 54 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.900 000 005 960 464 54(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.

  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =


0(2)


3. Convert to the binary (base 2) the fractional part: 0.900 000 005 960 464 54.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.900 000 005 960 464 54 × 2 = 1 + 0.800 000 011 920 929 08;
  • 2) 0.800 000 011 920 929 08 × 2 = 1 + 0.600 000 023 841 858 16;
  • 3) 0.600 000 023 841 858 16 × 2 = 1 + 0.200 000 047 683 716 32;
  • 4) 0.200 000 047 683 716 32 × 2 = 0 + 0.400 000 095 367 432 64;
  • 5) 0.400 000 095 367 432 64 × 2 = 0 + 0.800 000 190 734 865 28;
  • 6) 0.800 000 190 734 865 28 × 2 = 1 + 0.600 000 381 469 730 56;
  • 7) 0.600 000 381 469 730 56 × 2 = 1 + 0.200 000 762 939 461 12;
  • 8) 0.200 000 762 939 461 12 × 2 = 0 + 0.400 001 525 878 922 24;
  • 9) 0.400 001 525 878 922 24 × 2 = 0 + 0.800 003 051 757 844 48;
  • 10) 0.800 003 051 757 844 48 × 2 = 1 + 0.600 006 103 515 688 96;
  • 11) 0.600 006 103 515 688 96 × 2 = 1 + 0.200 012 207 031 377 92;
  • 12) 0.200 012 207 031 377 92 × 2 = 0 + 0.400 024 414 062 755 84;
  • 13) 0.400 024 414 062 755 84 × 2 = 0 + 0.800 048 828 125 511 68;
  • 14) 0.800 048 828 125 511 68 × 2 = 1 + 0.600 097 656 251 023 36;
  • 15) 0.600 097 656 251 023 36 × 2 = 1 + 0.200 195 312 502 046 72;
  • 16) 0.200 195 312 502 046 72 × 2 = 0 + 0.400 390 625 004 093 44;
  • 17) 0.400 390 625 004 093 44 × 2 = 0 + 0.800 781 250 008 186 88;
  • 18) 0.800 781 250 008 186 88 × 2 = 1 + 0.601 562 500 016 373 76;
  • 19) 0.601 562 500 016 373 76 × 2 = 1 + 0.203 125 000 032 747 52;
  • 20) 0.203 125 000 032 747 52 × 2 = 0 + 0.406 250 000 065 495 04;
  • 21) 0.406 250 000 065 495 04 × 2 = 0 + 0.812 500 000 130 990 08;
  • 22) 0.812 500 000 130 990 08 × 2 = 1 + 0.625 000 000 261 980 16;
  • 23) 0.625 000 000 261 980 16 × 2 = 1 + 0.250 000 000 523 960 32;
  • 24) 0.250 000 000 523 960 32 × 2 = 0 + 0.500 000 001 047 920 64;
  • 25) 0.500 000 001 047 920 64 × 2 = 1 + 0.000 000 002 095 841 28;
  • 26) 0.000 000 002 095 841 28 × 2 = 0 + 0.000 000 004 191 682 56;
  • 27) 0.000 000 004 191 682 56 × 2 = 0 + 0.000 000 008 383 365 12;
  • 28) 0.000 000 008 383 365 12 × 2 = 0 + 0.000 000 016 766 730 24;
  • 29) 0.000 000 016 766 730 24 × 2 = 0 + 0.000 000 033 533 460 48;
  • 30) 0.000 000 033 533 460 48 × 2 = 0 + 0.000 000 067 066 920 96;
  • 31) 0.000 000 067 066 920 96 × 2 = 0 + 0.000 000 134 133 841 92;
  • 32) 0.000 000 134 133 841 92 × 2 = 0 + 0.000 000 268 267 683 84;
  • 33) 0.000 000 268 267 683 84 × 2 = 0 + 0.000 000 536 535 367 68;
  • 34) 0.000 000 536 535 367 68 × 2 = 0 + 0.000 001 073 070 735 36;
  • 35) 0.000 001 073 070 735 36 × 2 = 0 + 0.000 002 146 141 470 72;
  • 36) 0.000 002 146 141 470 72 × 2 = 0 + 0.000 004 292 282 941 44;
  • 37) 0.000 004 292 282 941 44 × 2 = 0 + 0.000 008 584 565 882 88;
  • 38) 0.000 008 584 565 882 88 × 2 = 0 + 0.000 017 169 131 765 76;
  • 39) 0.000 017 169 131 765 76 × 2 = 0 + 0.000 034 338 263 531 52;
  • 40) 0.000 034 338 263 531 52 × 2 = 0 + 0.000 068 676 527 063 04;
  • 41) 0.000 068 676 527 063 04 × 2 = 0 + 0.000 137 353 054 126 08;
  • 42) 0.000 137 353 054 126 08 × 2 = 0 + 0.000 274 706 108 252 16;
  • 43) 0.000 274 706 108 252 16 × 2 = 0 + 0.000 549 412 216 504 32;
  • 44) 0.000 549 412 216 504 32 × 2 = 0 + 0.001 098 824 433 008 64;
  • 45) 0.001 098 824 433 008 64 × 2 = 0 + 0.002 197 648 866 017 28;
  • 46) 0.002 197 648 866 017 28 × 2 = 0 + 0.004 395 297 732 034 56;
  • 47) 0.004 395 297 732 034 56 × 2 = 0 + 0.008 790 595 464 069 12;
  • 48) 0.008 790 595 464 069 12 × 2 = 0 + 0.017 581 190 928 138 24;
  • 49) 0.017 581 190 928 138 24 × 2 = 0 + 0.035 162 381 856 276 48;
  • 50) 0.035 162 381 856 276 48 × 2 = 0 + 0.070 324 763 712 552 96;
  • 51) 0.070 324 763 712 552 96 × 2 = 0 + 0.140 649 527 425 105 92;
  • 52) 0.140 649 527 425 105 92 × 2 = 0 + 0.281 299 054 850 211 84;
  • 53) 0.281 299 054 850 211 84 × 2 = 0 + 0.562 598 109 700 423 68;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.900 000 005 960 464 54(10) =


0.1110 0110 0110 0110 0110 0110 1000 0000 0000 0000 0000 0000 0000 0(2)


5. Positive number before normalization:

0.900 000 005 960 464 54(10) =


0.1110 0110 0110 0110 0110 0110 1000 0000 0000 0000 0000 0000 0000 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right so that only one non zero digit remains to the left of it:

0.900 000 005 960 464 54(10) =


0.1110 0110 0110 0110 0110 0110 1000 0000 0000 0000 0000 0000 0000 0(2) =


0.1110 0110 0110 0110 0110 0110 1000 0000 0000 0000 0000 0000 0000 0(2) × 20 =


1.1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000(2) × 2-1


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -1


Mantissa (not normalized):
1.1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


-1 + 2(11-1) - 1 =


(-1 + 1 023)(10) =


1 022(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


1022(10) =


011 1111 1110(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).

Mantissa (normalized) =


1. 1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000 =


1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
011 1111 1110


Mantissa (52 bits) =
1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000


Number 0.900 000 005 960 464 54 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1110 - 1100 1100 1100 1100 1100 1101 0000 0000 0000 0000 0000 0000 0000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 1

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 1

      51
    • 1

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 1

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 0

      32
    • 1

      31
    • 1

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 0

      23
    • 0

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 0

      17
    • 0

      16
    • 0

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 0

      10
    • 0

      9
    • 0

      8
    • 0

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

0.900 000 005 960 464 53 = ? ... 0.900 000 005 960 464 55 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.900 000 005 960 464 54 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
-1 396 696 350 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
9 876 543 210 009 876.009 234 567 810 987 234 567 092 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
-14 748 395 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
11 001 100 001 110 101 111 001 101 100 101 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
4.280 003 5 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
41.022 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
21.92 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
-3 211 267 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
162.761 715 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
44.124 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
12.066 406 28 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
1.323 242 342 342 348 to 64 bit double precision IEEE 754 binary floating point = ? May 12 08:38 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =


    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100