Convert 10 000 000 111 110 010 100 000 100 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

10 000 000 111 110 010 100 000 100(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. 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;
  • 10 000 000 111 110 010 100 000 100 ÷ 2 = 5 000 000 055 555 005 050 000 050 + 0;
  • 5 000 000 055 555 005 050 000 050 ÷ 2 = 2 500 000 027 777 502 525 000 025 + 0;
  • 2 500 000 027 777 502 525 000 025 ÷ 2 = 1 250 000 013 888 751 262 500 012 + 1;
  • 1 250 000 013 888 751 262 500 012 ÷ 2 = 625 000 006 944 375 631 250 006 + 0;
  • 625 000 006 944 375 631 250 006 ÷ 2 = 312 500 003 472 187 815 625 003 + 0;
  • 312 500 003 472 187 815 625 003 ÷ 2 = 156 250 001 736 093 907 812 501 + 1;
  • 156 250 001 736 093 907 812 501 ÷ 2 = 78 125 000 868 046 953 906 250 + 1;
  • 78 125 000 868 046 953 906 250 ÷ 2 = 39 062 500 434 023 476 953 125 + 0;
  • 39 062 500 434 023 476 953 125 ÷ 2 = 19 531 250 217 011 738 476 562 + 1;
  • 19 531 250 217 011 738 476 562 ÷ 2 = 9 765 625 108 505 869 238 281 + 0;
  • 9 765 625 108 505 869 238 281 ÷ 2 = 4 882 812 554 252 934 619 140 + 1;
  • 4 882 812 554 252 934 619 140 ÷ 2 = 2 441 406 277 126 467 309 570 + 0;
  • 2 441 406 277 126 467 309 570 ÷ 2 = 1 220 703 138 563 233 654 785 + 0;
  • 1 220 703 138 563 233 654 785 ÷ 2 = 610 351 569 281 616 827 392 + 1;
  • 610 351 569 281 616 827 392 ÷ 2 = 305 175 784 640 808 413 696 + 0;
  • 305 175 784 640 808 413 696 ÷ 2 = 152 587 892 320 404 206 848 + 0;
  • 152 587 892 320 404 206 848 ÷ 2 = 76 293 946 160 202 103 424 + 0;
  • 76 293 946 160 202 103 424 ÷ 2 = 38 146 973 080 101 051 712 + 0;
  • 38 146 973 080 101 051 712 ÷ 2 = 19 073 486 540 050 525 856 + 0;
  • 19 073 486 540 050 525 856 ÷ 2 = 9 536 743 270 025 262 928 + 0;
  • 9 536 743 270 025 262 928 ÷ 2 = 4 768 371 635 012 631 464 + 0;
  • 4 768 371 635 012 631 464 ÷ 2 = 2 384 185 817 506 315 732 + 0;
  • 2 384 185 817 506 315 732 ÷ 2 = 1 192 092 908 753 157 866 + 0;
  • 1 192 092 908 753 157 866 ÷ 2 = 596 046 454 376 578 933 + 0;
  • 596 046 454 376 578 933 ÷ 2 = 298 023 227 188 289 466 + 1;
  • 298 023 227 188 289 466 ÷ 2 = 149 011 613 594 144 733 + 0;
  • 149 011 613 594 144 733 ÷ 2 = 74 505 806 797 072 366 + 1;
  • 74 505 806 797 072 366 ÷ 2 = 37 252 903 398 536 183 + 0;
  • 37 252 903 398 536 183 ÷ 2 = 18 626 451 699 268 091 + 1;
  • 18 626 451 699 268 091 ÷ 2 = 9 313 225 849 634 045 + 1;
  • 9 313 225 849 634 045 ÷ 2 = 4 656 612 924 817 022 + 1;
  • 4 656 612 924 817 022 ÷ 2 = 2 328 306 462 408 511 + 0;
  • 2 328 306 462 408 511 ÷ 2 = 1 164 153 231 204 255 + 1;
  • 1 164 153 231 204 255 ÷ 2 = 582 076 615 602 127 + 1;
  • 582 076 615 602 127 ÷ 2 = 291 038 307 801 063 + 1;
  • 291 038 307 801 063 ÷ 2 = 145 519 153 900 531 + 1;
  • 145 519 153 900 531 ÷ 2 = 72 759 576 950 265 + 1;
  • 72 759 576 950 265 ÷ 2 = 36 379 788 475 132 + 1;
  • 36 379 788 475 132 ÷ 2 = 18 189 894 237 566 + 0;
  • 18 189 894 237 566 ÷ 2 = 9 094 947 118 783 + 0;
  • 9 094 947 118 783 ÷ 2 = 4 547 473 559 391 + 1;
  • 4 547 473 559 391 ÷ 2 = 2 273 736 779 695 + 1;
  • 2 273 736 779 695 ÷ 2 = 1 136 868 389 847 + 1;
  • 1 136 868 389 847 ÷ 2 = 568 434 194 923 + 1;
  • 568 434 194 923 ÷ 2 = 284 217 097 461 + 1;
  • 284 217 097 461 ÷ 2 = 142 108 548 730 + 1;
  • 142 108 548 730 ÷ 2 = 71 054 274 365 + 0;
  • 71 054 274 365 ÷ 2 = 35 527 137 182 + 1;
  • 35 527 137 182 ÷ 2 = 17 763 568 591 + 0;
  • 17 763 568 591 ÷ 2 = 8 881 784 295 + 1;
  • 8 881 784 295 ÷ 2 = 4 440 892 147 + 1;
  • 4 440 892 147 ÷ 2 = 2 220 446 073 + 1;
  • 2 220 446 073 ÷ 2 = 1 110 223 036 + 1;
  • 1 110 223 036 ÷ 2 = 555 111 518 + 0;
  • 555 111 518 ÷ 2 = 277 555 759 + 0;
  • 277 555 759 ÷ 2 = 138 777 879 + 1;
  • 138 777 879 ÷ 2 = 69 388 939 + 1;
  • 69 388 939 ÷ 2 = 34 694 469 + 1;
  • 34 694 469 ÷ 2 = 17 347 234 + 1;
  • 17 347 234 ÷ 2 = 8 673 617 + 0;
  • 8 673 617 ÷ 2 = 4 336 808 + 1;
  • 4 336 808 ÷ 2 = 2 168 404 + 0;
  • 2 168 404 ÷ 2 = 1 084 202 + 0;
  • 1 084 202 ÷ 2 = 542 101 + 0;
  • 542 101 ÷ 2 = 271 050 + 1;
  • 271 050 ÷ 2 = 135 525 + 0;
  • 135 525 ÷ 2 = 67 762 + 1;
  • 67 762 ÷ 2 = 33 881 + 0;
  • 33 881 ÷ 2 = 16 940 + 1;
  • 16 940 ÷ 2 = 8 470 + 0;
  • 8 470 ÷ 2 = 4 235 + 0;
  • 4 235 ÷ 2 = 2 117 + 1;
  • 2 117 ÷ 2 = 1 058 + 1;
  • 1 058 ÷ 2 = 529 + 0;
  • 529 ÷ 2 = 264 + 1;
  • 264 ÷ 2 = 132 + 0;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

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

10 000 000 111 110 010 100 000 100(10) =


1000 0100 0101 1001 0101 0001 0111 1001 1110 1011 1111 0011 1111 0111 0101 0000 0000 0010 0101 0110 0100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 83 positions to the left so that only one non zero digit remains to the left of it:

10 000 000 111 110 010 100 000 100(10) =


1000 0100 0101 1001 0101 0001 0111 1001 1110 1011 1111 0011 1111 0111 0101 0000 0000 0010 0101 0110 0100(2) =


1000 0100 0101 1001 0101 0001 0111 1001 1110 1011 1111 0011 1111 0111 0101 0000 0000 0010 0101 0110 0100(2) × 20 =


1.0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110 1110 1010 0000 0000 0100 1010 1100 100(2) × 283


4. 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): 83


Mantissa (not normalized):
1.0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110 1110 1010 0000 0000 0100 1010 1100 100


5. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


83 + 2(11-1) - 1 =


(83 + 1 023)(10) =


1 106(10)


6. 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;
  • 1 106 ÷ 2 = 553 + 0;
  • 553 ÷ 2 = 276 + 1;
  • 276 ÷ 2 = 138 + 0;
  • 138 ÷ 2 = 69 + 0;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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

Exponent (adjusted) =


1106(10) =


100 0101 0010(2)


8. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).

Mantissa (normalized) =


1. 0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110 111 0101 0000 0000 0010 0101 0110 0100 =


0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110


9. 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) =
100 0101 0010


Mantissa (52 bits) =
0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110


Number 10 000 000 111 110 010 100 000 100 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0101 0010 - 0000 1000 1011 0010 1010 0010 1111 0011 1101 0111 1110 0111 1110

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 1

      58
    • 0

      57
    • 1

      56
    • 0

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 1

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 1

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 1

      25
    • 1

      24
    • 0

      23
    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 1

      9
    • 0

      8
    • 0

      7
    • 1

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 0

      0

More operations of this kind:

10 000 000 111 110 010 100 000 099 = ? ... 10 000 000 111 110 010 100 000 101 = ?


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

10 000 000 111 110 010 100 000 100 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:47 UTC (GMT)
26 103 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
41.687 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
103.876 52 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
-160.207 113 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
0.707 106 781 186 547 524 400 2 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
1 038 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
1.666 662 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
1.123 8 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
108.447 026 621 334 160 267 906 554 508 954 286 56 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
3 215 454 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
-2.6 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 UTC (GMT)
642 to 64 bit double precision IEEE 754 binary floating point = ? Apr 18 07:46 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