Convert 1.000 000 000 000 000 222 044 604 925 037 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

1.000 000 000 000 000 222 044 604 925 037(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: 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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =


1(2)


3. Convert to the binary (base 2) the fractional part: 0.000 000 000 000 000 222 044 604 925 037.

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) 0.000 000 000 000 000 222 044 604 925 037 × 2 = 0 + 0.000 000 000 000 000 444 089 209 850 074;
  • 2) 0.000 000 000 000 000 444 089 209 850 074 × 2 = 0 + 0.000 000 000 000 000 888 178 419 700 148;
  • 3) 0.000 000 000 000 000 888 178 419 700 148 × 2 = 0 + 0.000 000 000 000 001 776 356 839 400 296;
  • 4) 0.000 000 000 000 001 776 356 839 400 296 × 2 = 0 + 0.000 000 000 000 003 552 713 678 800 592;
  • 5) 0.000 000 000 000 003 552 713 678 800 592 × 2 = 0 + 0.000 000 000 000 007 105 427 357 601 184;
  • 6) 0.000 000 000 000 007 105 427 357 601 184 × 2 = 0 + 0.000 000 000 000 014 210 854 715 202 368;
  • 7) 0.000 000 000 000 014 210 854 715 202 368 × 2 = 0 + 0.000 000 000 000 028 421 709 430 404 736;
  • 8) 0.000 000 000 000 028 421 709 430 404 736 × 2 = 0 + 0.000 000 000 000 056 843 418 860 809 472;
  • 9) 0.000 000 000 000 056 843 418 860 809 472 × 2 = 0 + 0.000 000 000 000 113 686 837 721 618 944;
  • 10) 0.000 000 000 000 113 686 837 721 618 944 × 2 = 0 + 0.000 000 000 000 227 373 675 443 237 888;
  • 11) 0.000 000 000 000 227 373 675 443 237 888 × 2 = 0 + 0.000 000 000 000 454 747 350 886 475 776;
  • 12) 0.000 000 000 000 454 747 350 886 475 776 × 2 = 0 + 0.000 000 000 000 909 494 701 772 951 552;
  • 13) 0.000 000 000 000 909 494 701 772 951 552 × 2 = 0 + 0.000 000 000 001 818 989 403 545 903 104;
  • 14) 0.000 000 000 001 818 989 403 545 903 104 × 2 = 0 + 0.000 000 000 003 637 978 807 091 806 208;
  • 15) 0.000 000 000 003 637 978 807 091 806 208 × 2 = 0 + 0.000 000 000 007 275 957 614 183 612 416;
  • 16) 0.000 000 000 007 275 957 614 183 612 416 × 2 = 0 + 0.000 000 000 014 551 915 228 367 224 832;
  • 17) 0.000 000 000 014 551 915 228 367 224 832 × 2 = 0 + 0.000 000 000 029 103 830 456 734 449 664;
  • 18) 0.000 000 000 029 103 830 456 734 449 664 × 2 = 0 + 0.000 000 000 058 207 660 913 468 899 328;
  • 19) 0.000 000 000 058 207 660 913 468 899 328 × 2 = 0 + 0.000 000 000 116 415 321 826 937 798 656;
  • 20) 0.000 000 000 116 415 321 826 937 798 656 × 2 = 0 + 0.000 000 000 232 830 643 653 875 597 312;
  • 21) 0.000 000 000 232 830 643 653 875 597 312 × 2 = 0 + 0.000 000 000 465 661 287 307 751 194 624;
  • 22) 0.000 000 000 465 661 287 307 751 194 624 × 2 = 0 + 0.000 000 000 931 322 574 615 502 389 248;
  • 23) 0.000 000 000 931 322 574 615 502 389 248 × 2 = 0 + 0.000 000 001 862 645 149 231 004 778 496;
  • 24) 0.000 000 001 862 645 149 231 004 778 496 × 2 = 0 + 0.000 000 003 725 290 298 462 009 556 992;
  • 25) 0.000 000 003 725 290 298 462 009 556 992 × 2 = 0 + 0.000 000 007 450 580 596 924 019 113 984;
  • 26) 0.000 000 007 450 580 596 924 019 113 984 × 2 = 0 + 0.000 000 014 901 161 193 848 038 227 968;
  • 27) 0.000 000 014 901 161 193 848 038 227 968 × 2 = 0 + 0.000 000 029 802 322 387 696 076 455 936;
  • 28) 0.000 000 029 802 322 387 696 076 455 936 × 2 = 0 + 0.000 000 059 604 644 775 392 152 911 872;
  • 29) 0.000 000 059 604 644 775 392 152 911 872 × 2 = 0 + 0.000 000 119 209 289 550 784 305 823 744;
  • 30) 0.000 000 119 209 289 550 784 305 823 744 × 2 = 0 + 0.000 000 238 418 579 101 568 611 647 488;
  • 31) 0.000 000 238 418 579 101 568 611 647 488 × 2 = 0 + 0.000 000 476 837 158 203 137 223 294 976;
  • 32) 0.000 000 476 837 158 203 137 223 294 976 × 2 = 0 + 0.000 000 953 674 316 406 274 446 589 952;
  • 33) 0.000 000 953 674 316 406 274 446 589 952 × 2 = 0 + 0.000 001 907 348 632 812 548 893 179 904;
  • 34) 0.000 001 907 348 632 812 548 893 179 904 × 2 = 0 + 0.000 003 814 697 265 625 097 786 359 808;
  • 35) 0.000 003 814 697 265 625 097 786 359 808 × 2 = 0 + 0.000 007 629 394 531 250 195 572 719 616;
  • 36) 0.000 007 629 394 531 250 195 572 719 616 × 2 = 0 + 0.000 015 258 789 062 500 391 145 439 232;
  • 37) 0.000 015 258 789 062 500 391 145 439 232 × 2 = 0 + 0.000 030 517 578 125 000 782 290 878 464;
  • 38) 0.000 030 517 578 125 000 782 290 878 464 × 2 = 0 + 0.000 061 035 156 250 001 564 581 756 928;
  • 39) 0.000 061 035 156 250 001 564 581 756 928 × 2 = 0 + 0.000 122 070 312 500 003 129 163 513 856;
  • 40) 0.000 122 070 312 500 003 129 163 513 856 × 2 = 0 + 0.000 244 140 625 000 006 258 327 027 712;
  • 41) 0.000 244 140 625 000 006 258 327 027 712 × 2 = 0 + 0.000 488 281 250 000 012 516 654 055 424;
  • 42) 0.000 488 281 250 000 012 516 654 055 424 × 2 = 0 + 0.000 976 562 500 000 025 033 308 110 848;
  • 43) 0.000 976 562 500 000 025 033 308 110 848 × 2 = 0 + 0.001 953 125 000 000 050 066 616 221 696;
  • 44) 0.001 953 125 000 000 050 066 616 221 696 × 2 = 0 + 0.003 906 250 000 000 100 133 232 443 392;
  • 45) 0.003 906 250 000 000 100 133 232 443 392 × 2 = 0 + 0.007 812 500 000 000 200 266 464 886 784;
  • 46) 0.007 812 500 000 000 200 266 464 886 784 × 2 = 0 + 0.015 625 000 000 000 400 532 929 773 568;
  • 47) 0.015 625 000 000 000 400 532 929 773 568 × 2 = 0 + 0.031 250 000 000 000 801 065 859 547 136;
  • 48) 0.031 250 000 000 000 801 065 859 547 136 × 2 = 0 + 0.062 500 000 000 001 602 131 719 094 272;
  • 49) 0.062 500 000 000 001 602 131 719 094 272 × 2 = 0 + 0.125 000 000 000 003 204 263 438 188 544;
  • 50) 0.125 000 000 000 003 204 263 438 188 544 × 2 = 0 + 0.250 000 000 000 006 408 526 876 377 088;
  • 51) 0.250 000 000 000 006 408 526 876 377 088 × 2 = 0 + 0.500 000 000 000 012 817 053 752 754 176;
  • 52) 0.500 000 000 000 012 817 053 752 754 176 × 2 = 1 + 0.000 000 000 000 025 634 107 505 508 352;
  • 53) 0.000 000 000 000 025 634 107 505 508 352 × 2 = 0 + 0.000 000 000 000 051 268 215 011 016 704;

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:

0.000 000 000 000 000 222 044 604 925 037(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0(2)


5. Positive number before normalization:

1.000 000 000 000 000 222 044 604 925 037(10) =


1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left so that only one non zero digit remains to the left of it:

1.000 000 000 000 000 222 044 604 925 037(10) =


1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0(2) =


1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0(2) × 20


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): 0


Mantissa (not normalized):
1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


0 + 2(11-1) - 1 =


(0 + 1 023)(10) =


1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 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) =


1023(10) =


011 1111 1111(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, 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 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0 =


0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001


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 1111


Mantissa (52 bits) =
0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001


Number 1.000 000 000 000 000 222 044 604 925 037 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1111 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001

(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
    • 1

      52
  • Mantissa (52 bits):

    • 0

      51
    • 0

      50
    • 0

      49
    • 0

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 0

      43
    • 0

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 0

      34
    • 0

      33
    • 0

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 0

      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
    • 1

      0

More operations of this kind:

1.000 000 000 000 000 222 044 604 925 036 = ? ... 1.000 000 000 000 000 222 044 604 925 038 = ?


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

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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