0.019 932 208 134 231 138 919 244 739 42 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.019 932 208 134 231 138 919 244 739 42(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
0.019 932 208 134 231 138 919 244 739 42(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: 0.019 932 208 134 231 138 919 244 739 42.

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) 0.019 932 208 134 231 138 919 244 739 42 × 2 = 0 + 0.039 864 416 268 462 277 838 489 478 84;
  • 2) 0.039 864 416 268 462 277 838 489 478 84 × 2 = 0 + 0.079 728 832 536 924 555 676 978 957 68;
  • 3) 0.079 728 832 536 924 555 676 978 957 68 × 2 = 0 + 0.159 457 665 073 849 111 353 957 915 36;
  • 4) 0.159 457 665 073 849 111 353 957 915 36 × 2 = 0 + 0.318 915 330 147 698 222 707 915 830 72;
  • 5) 0.318 915 330 147 698 222 707 915 830 72 × 2 = 0 + 0.637 830 660 295 396 445 415 831 661 44;
  • 6) 0.637 830 660 295 396 445 415 831 661 44 × 2 = 1 + 0.275 661 320 590 792 890 831 663 322 88;
  • 7) 0.275 661 320 590 792 890 831 663 322 88 × 2 = 0 + 0.551 322 641 181 585 781 663 326 645 76;
  • 8) 0.551 322 641 181 585 781 663 326 645 76 × 2 = 1 + 0.102 645 282 363 171 563 326 653 291 52;
  • 9) 0.102 645 282 363 171 563 326 653 291 52 × 2 = 0 + 0.205 290 564 726 343 126 653 306 583 04;
  • 10) 0.205 290 564 726 343 126 653 306 583 04 × 2 = 0 + 0.410 581 129 452 686 253 306 613 166 08;
  • 11) 0.410 581 129 452 686 253 306 613 166 08 × 2 = 0 + 0.821 162 258 905 372 506 613 226 332 16;
  • 12) 0.821 162 258 905 372 506 613 226 332 16 × 2 = 1 + 0.642 324 517 810 745 013 226 452 664 32;
  • 13) 0.642 324 517 810 745 013 226 452 664 32 × 2 = 1 + 0.284 649 035 621 490 026 452 905 328 64;
  • 14) 0.284 649 035 621 490 026 452 905 328 64 × 2 = 0 + 0.569 298 071 242 980 052 905 810 657 28;
  • 15) 0.569 298 071 242 980 052 905 810 657 28 × 2 = 1 + 0.138 596 142 485 960 105 811 621 314 56;
  • 16) 0.138 596 142 485 960 105 811 621 314 56 × 2 = 0 + 0.277 192 284 971 920 211 623 242 629 12;
  • 17) 0.277 192 284 971 920 211 623 242 629 12 × 2 = 0 + 0.554 384 569 943 840 423 246 485 258 24;
  • 18) 0.554 384 569 943 840 423 246 485 258 24 × 2 = 1 + 0.108 769 139 887 680 846 492 970 516 48;
  • 19) 0.108 769 139 887 680 846 492 970 516 48 × 2 = 0 + 0.217 538 279 775 361 692 985 941 032 96;
  • 20) 0.217 538 279 775 361 692 985 941 032 96 × 2 = 0 + 0.435 076 559 550 723 385 971 882 065 92;
  • 21) 0.435 076 559 550 723 385 971 882 065 92 × 2 = 0 + 0.870 153 119 101 446 771 943 764 131 84;
  • 22) 0.870 153 119 101 446 771 943 764 131 84 × 2 = 1 + 0.740 306 238 202 893 543 887 528 263 68;
  • 23) 0.740 306 238 202 893 543 887 528 263 68 × 2 = 1 + 0.480 612 476 405 787 087 775 056 527 36;
  • 24) 0.480 612 476 405 787 087 775 056 527 36 × 2 = 0 + 0.961 224 952 811 574 175 550 113 054 72;
  • 25) 0.961 224 952 811 574 175 550 113 054 72 × 2 = 1 + 0.922 449 905 623 148 351 100 226 109 44;
  • 26) 0.922 449 905 623 148 351 100 226 109 44 × 2 = 1 + 0.844 899 811 246 296 702 200 452 218 88;
  • 27) 0.844 899 811 246 296 702 200 452 218 88 × 2 = 1 + 0.689 799 622 492 593 404 400 904 437 76;
  • 28) 0.689 799 622 492 593 404 400 904 437 76 × 2 = 1 + 0.379 599 244 985 186 808 801 808 875 52;
  • 29) 0.379 599 244 985 186 808 801 808 875 52 × 2 = 0 + 0.759 198 489 970 373 617 603 617 751 04;
  • 30) 0.759 198 489 970 373 617 603 617 751 04 × 2 = 1 + 0.518 396 979 940 747 235 207 235 502 08;
  • 31) 0.518 396 979 940 747 235 207 235 502 08 × 2 = 1 + 0.036 793 959 881 494 470 414 471 004 16;
  • 32) 0.036 793 959 881 494 470 414 471 004 16 × 2 = 0 + 0.073 587 919 762 988 940 828 942 008 32;
  • 33) 0.073 587 919 762 988 940 828 942 008 32 × 2 = 0 + 0.147 175 839 525 977 881 657 884 016 64;
  • 34) 0.147 175 839 525 977 881 657 884 016 64 × 2 = 0 + 0.294 351 679 051 955 763 315 768 033 28;
  • 35) 0.294 351 679 051 955 763 315 768 033 28 × 2 = 0 + 0.588 703 358 103 911 526 631 536 066 56;
  • 36) 0.588 703 358 103 911 526 631 536 066 56 × 2 = 1 + 0.177 406 716 207 823 053 263 072 133 12;
  • 37) 0.177 406 716 207 823 053 263 072 133 12 × 2 = 0 + 0.354 813 432 415 646 106 526 144 266 24;
  • 38) 0.354 813 432 415 646 106 526 144 266 24 × 2 = 0 + 0.709 626 864 831 292 213 052 288 532 48;
  • 39) 0.709 626 864 831 292 213 052 288 532 48 × 2 = 1 + 0.419 253 729 662 584 426 104 577 064 96;
  • 40) 0.419 253 729 662 584 426 104 577 064 96 × 2 = 0 + 0.838 507 459 325 168 852 209 154 129 92;
  • 41) 0.838 507 459 325 168 852 209 154 129 92 × 2 = 1 + 0.677 014 918 650 337 704 418 308 259 84;
  • 42) 0.677 014 918 650 337 704 418 308 259 84 × 2 = 1 + 0.354 029 837 300 675 408 836 616 519 68;
  • 43) 0.354 029 837 300 675 408 836 616 519 68 × 2 = 0 + 0.708 059 674 601 350 817 673 233 039 36;
  • 44) 0.708 059 674 601 350 817 673 233 039 36 × 2 = 1 + 0.416 119 349 202 701 635 346 466 078 72;
  • 45) 0.416 119 349 202 701 635 346 466 078 72 × 2 = 0 + 0.832 238 698 405 403 270 692 932 157 44;
  • 46) 0.832 238 698 405 403 270 692 932 157 44 × 2 = 1 + 0.664 477 396 810 806 541 385 864 314 88;
  • 47) 0.664 477 396 810 806 541 385 864 314 88 × 2 = 1 + 0.328 954 793 621 613 082 771 728 629 76;
  • 48) 0.328 954 793 621 613 082 771 728 629 76 × 2 = 0 + 0.657 909 587 243 226 165 543 457 259 52;
  • 49) 0.657 909 587 243 226 165 543 457 259 52 × 2 = 1 + 0.315 819 174 486 452 331 086 914 519 04;
  • 50) 0.315 819 174 486 452 331 086 914 519 04 × 2 = 0 + 0.631 638 348 972 904 662 173 829 038 08;
  • 51) 0.631 638 348 972 904 662 173 829 038 08 × 2 = 1 + 0.263 276 697 945 809 324 347 658 076 16;
  • 52) 0.263 276 697 945 809 324 347 658 076 16 × 2 = 0 + 0.526 553 395 891 618 648 695 316 152 32;
  • 53) 0.526 553 395 891 618 648 695 316 152 32 × 2 = 1 + 0.053 106 791 783 237 297 390 632 304 64;
  • 54) 0.053 106 791 783 237 297 390 632 304 64 × 2 = 0 + 0.106 213 583 566 474 594 781 264 609 28;
  • 55) 0.106 213 583 566 474 594 781 264 609 28 × 2 = 0 + 0.212 427 167 132 949 189 562 529 218 56;
  • 56) 0.212 427 167 132 949 189 562 529 218 56 × 2 = 0 + 0.424 854 334 265 898 379 125 058 437 12;
  • 57) 0.424 854 334 265 898 379 125 058 437 12 × 2 = 0 + 0.849 708 668 531 796 758 250 116 874 24;
  • 58) 0.849 708 668 531 796 758 250 116 874 24 × 2 = 1 + 0.699 417 337 063 593 516 500 233 748 48;

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:


0.019 932 208 134 231 138 919 244 739 42(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2)

5. Positive number before normalization:

0.019 932 208 134 231 138 919 244 739 42(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.019 932 208 134 231 138 919 244 739 42(10) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2) =

0.0000 0101 0001 1010 0100 0110 1111 0110 0001 0010 1101 0110 1010 1000 01(2) × 20 =

1.0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001(2) × 2-6


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): -6

Mantissa (not normalized):
1.0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1017(10) =

011 1111 1001(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. 0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001 =

0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 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 1001

Mantissa (52 bits) =
0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001


Decimal number 0.019 932 208 134 231 138 919 244 739 42 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1001 - 0100 0110 1001 0001 1011 1101 1000 0100 1011 0101 1010 1010 0001

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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