0.000 000 000 000 284 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 284(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.000 000 000 000 284(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in 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 binary (base 2) the fractional part: 0.000 000 000 000 284.

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) 0.000 000 000 000 284 × 2 = 0 + 0.000 000 000 000 568;
  • 2) 0.000 000 000 000 568 × 2 = 0 + 0.000 000 000 001 136;
  • 3) 0.000 000 000 001 136 × 2 = 0 + 0.000 000 000 002 272;
  • 4) 0.000 000 000 002 272 × 2 = 0 + 0.000 000 000 004 544;
  • 5) 0.000 000 000 004 544 × 2 = 0 + 0.000 000 000 009 088;
  • 6) 0.000 000 000 009 088 × 2 = 0 + 0.000 000 000 018 176;
  • 7) 0.000 000 000 018 176 × 2 = 0 + 0.000 000 000 036 352;
  • 8) 0.000 000 000 036 352 × 2 = 0 + 0.000 000 000 072 704;
  • 9) 0.000 000 000 072 704 × 2 = 0 + 0.000 000 000 145 408;
  • 10) 0.000 000 000 145 408 × 2 = 0 + 0.000 000 000 290 816;
  • 11) 0.000 000 000 290 816 × 2 = 0 + 0.000 000 000 581 632;
  • 12) 0.000 000 000 581 632 × 2 = 0 + 0.000 000 001 163 264;
  • 13) 0.000 000 001 163 264 × 2 = 0 + 0.000 000 002 326 528;
  • 14) 0.000 000 002 326 528 × 2 = 0 + 0.000 000 004 653 056;
  • 15) 0.000 000 004 653 056 × 2 = 0 + 0.000 000 009 306 112;
  • 16) 0.000 000 009 306 112 × 2 = 0 + 0.000 000 018 612 224;
  • 17) 0.000 000 018 612 224 × 2 = 0 + 0.000 000 037 224 448;
  • 18) 0.000 000 037 224 448 × 2 = 0 + 0.000 000 074 448 896;
  • 19) 0.000 000 074 448 896 × 2 = 0 + 0.000 000 148 897 792;
  • 20) 0.000 000 148 897 792 × 2 = 0 + 0.000 000 297 795 584;
  • 21) 0.000 000 297 795 584 × 2 = 0 + 0.000 000 595 591 168;
  • 22) 0.000 000 595 591 168 × 2 = 0 + 0.000 001 191 182 336;
  • 23) 0.000 001 191 182 336 × 2 = 0 + 0.000 002 382 364 672;
  • 24) 0.000 002 382 364 672 × 2 = 0 + 0.000 004 764 729 344;
  • 25) 0.000 004 764 729 344 × 2 = 0 + 0.000 009 529 458 688;
  • 26) 0.000 009 529 458 688 × 2 = 0 + 0.000 019 058 917 376;
  • 27) 0.000 019 058 917 376 × 2 = 0 + 0.000 038 117 834 752;
  • 28) 0.000 038 117 834 752 × 2 = 0 + 0.000 076 235 669 504;
  • 29) 0.000 076 235 669 504 × 2 = 0 + 0.000 152 471 339 008;
  • 30) 0.000 152 471 339 008 × 2 = 0 + 0.000 304 942 678 016;
  • 31) 0.000 304 942 678 016 × 2 = 0 + 0.000 609 885 356 032;
  • 32) 0.000 609 885 356 032 × 2 = 0 + 0.001 219 770 712 064;
  • 33) 0.001 219 770 712 064 × 2 = 0 + 0.002 439 541 424 128;
  • 34) 0.002 439 541 424 128 × 2 = 0 + 0.004 879 082 848 256;
  • 35) 0.004 879 082 848 256 × 2 = 0 + 0.009 758 165 696 512;
  • 36) 0.009 758 165 696 512 × 2 = 0 + 0.019 516 331 393 024;
  • 37) 0.019 516 331 393 024 × 2 = 0 + 0.039 032 662 786 048;
  • 38) 0.039 032 662 786 048 × 2 = 0 + 0.078 065 325 572 096;
  • 39) 0.078 065 325 572 096 × 2 = 0 + 0.156 130 651 144 192;
  • 40) 0.156 130 651 144 192 × 2 = 0 + 0.312 261 302 288 384;
  • 41) 0.312 261 302 288 384 × 2 = 0 + 0.624 522 604 576 768;
  • 42) 0.624 522 604 576 768 × 2 = 1 + 0.249 045 209 153 536;
  • 43) 0.249 045 209 153 536 × 2 = 0 + 0.498 090 418 307 072;
  • 44) 0.498 090 418 307 072 × 2 = 0 + 0.996 180 836 614 144;
  • 45) 0.996 180 836 614 144 × 2 = 1 + 0.992 361 673 228 288;
  • 46) 0.992 361 673 228 288 × 2 = 1 + 0.984 723 346 456 576;
  • 47) 0.984 723 346 456 576 × 2 = 1 + 0.969 446 692 913 152;
  • 48) 0.969 446 692 913 152 × 2 = 1 + 0.938 893 385 826 304;
  • 49) 0.938 893 385 826 304 × 2 = 1 + 0.877 786 771 652 608;
  • 50) 0.877 786 771 652 608 × 2 = 1 + 0.755 573 543 305 216;
  • 51) 0.755 573 543 305 216 × 2 = 1 + 0.511 147 086 610 432;
  • 52) 0.511 147 086 610 432 × 2 = 1 + 0.022 294 173 220 864;
  • 53) 0.022 294 173 220 864 × 2 = 0 + 0.044 588 346 441 728;
  • 54) 0.044 588 346 441 728 × 2 = 0 + 0.089 176 692 883 456;
  • 55) 0.089 176 692 883 456 × 2 = 0 + 0.178 353 385 766 912;
  • 56) 0.178 353 385 766 912 × 2 = 0 + 0.356 706 771 533 824;
  • 57) 0.356 706 771 533 824 × 2 = 0 + 0.713 413 543 067 648;
  • 58) 0.713 413 543 067 648 × 2 = 1 + 0.426 827 086 135 296;
  • 59) 0.426 827 086 135 296 × 2 = 0 + 0.853 654 172 270 592;
  • 60) 0.853 654 172 270 592 × 2 = 1 + 0.707 308 344 541 184;
  • 61) 0.707 308 344 541 184 × 2 = 1 + 0.414 616 689 082 368;
  • 62) 0.414 616 689 082 368 × 2 = 0 + 0.829 233 378 164 736;
  • 63) 0.829 233 378 164 736 × 2 = 1 + 0.658 466 756 329 472;
  • 64) 0.658 466 756 329 472 × 2 = 1 + 0.316 933 512 658 944;
  • 65) 0.316 933 512 658 944 × 2 = 0 + 0.633 867 025 317 888;
  • 66) 0.633 867 025 317 888 × 2 = 1 + 0.267 734 050 635 776;
  • 67) 0.267 734 050 635 776 × 2 = 0 + 0.535 468 101 271 552;
  • 68) 0.535 468 101 271 552 × 2 = 1 + 0.070 936 202 543 104;
  • 69) 0.070 936 202 543 104 × 2 = 0 + 0.141 872 405 086 208;
  • 70) 0.141 872 405 086 208 × 2 = 0 + 0.283 744 810 172 416;
  • 71) 0.283 744 810 172 416 × 2 = 0 + 0.567 489 620 344 832;
  • 72) 0.567 489 620 344 832 × 2 = 1 + 0.134 979 240 689 664;
  • 73) 0.134 979 240 689 664 × 2 = 0 + 0.269 958 481 379 328;
  • 74) 0.269 958 481 379 328 × 2 = 0 + 0.539 916 962 758 656;
  • 75) 0.539 916 962 758 656 × 2 = 1 + 0.079 833 925 517 312;
  • 76) 0.079 833 925 517 312 × 2 = 0 + 0.159 667 851 034 624;
  • 77) 0.159 667 851 034 624 × 2 = 0 + 0.319 335 702 069 248;
  • 78) 0.319 335 702 069 248 × 2 = 0 + 0.638 671 404 138 496;
  • 79) 0.638 671 404 138 496 × 2 = 1 + 0.277 342 808 276 992;
  • 80) 0.277 342 808 276 992 × 2 = 0 + 0.554 685 616 553 984;
  • 81) 0.554 685 616 553 984 × 2 = 1 + 0.109 371 233 107 968;
  • 82) 0.109 371 233 107 968 × 2 = 0 + 0.218 742 466 215 936;
  • 83) 0.218 742 466 215 936 × 2 = 0 + 0.437 484 932 431 872;
  • 84) 0.437 484 932 431 872 × 2 = 0 + 0.874 969 864 863 744;
  • 85) 0.874 969 864 863 744 × 2 = 1 + 0.749 939 729 727 488;
  • 86) 0.749 939 729 727 488 × 2 = 1 + 0.499 879 459 454 976;
  • 87) 0.499 879 459 454 976 × 2 = 0 + 0.999 758 918 909 952;
  • 88) 0.999 758 918 909 952 × 2 = 1 + 0.999 517 837 819 904;
  • 89) 0.999 517 837 819 904 × 2 = 1 + 0.999 035 675 639 808;
  • 90) 0.999 035 675 639 808 × 2 = 1 + 0.998 071 351 279 616;
  • 91) 0.998 071 351 279 616 × 2 = 1 + 0.996 142 702 559 232;
  • 92) 0.996 142 702 559 232 × 2 = 1 + 0.992 285 405 118 464;
  • 93) 0.992 285 405 118 464 × 2 = 1 + 0.984 570 810 236 928;
  • 94) 0.984 570 810 236 928 × 2 = 1 + 0.969 141 620 473 856;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 000 000 284(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111 1111 0000 0101 1011 0101 0001 0010 0010 1000 1101 1111 11(2)

5. Positive number before normalization:

0.000 000 000 000 284(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111 1111 0000 0101 1011 0101 0001 0010 0010 1000 1101 1111 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 42 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 284(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111 1111 0000 0101 1011 0101 0001 0010 0010 1000 1101 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 1111 1111 0000 0101 1011 0101 0001 0010 0010 1000 1101 1111 11(2) × 20 =

1.0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111(2) × 2-42


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): -42

Mantissa (not normalized):
1.0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-42 + 2(11-1) - 1 =

(-42 + 1 023)(10) =

981(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;
  • 981 ÷ 2 = 490 + 1;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

981(10) =

011 1101 0101(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. 0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111 =

0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111


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 1101 0101

Mantissa (52 bits) =
0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111


Decimal number 0.000 000 000 000 284 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1101 0101 - 0011 1111 1100 0001 0110 1101 0100 0100 1000 1010 0011 0111 1111

Share

» Convert decimal 0.000 000 000 000 189 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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