-0.016 738 891 601 562 531 225 022 567 576 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 531 225 022 567 576(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
-0.016 738 891 601 562 531 225 022 567 576(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. Start with the positive version of the number:

|-0.016 738 891 601 562 531 225 022 567 576| = 0.016 738 891 601 562 531 225 022 567 576


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 531 225 022 567 576.

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) 0.016 738 891 601 562 531 225 022 567 576 × 2 = 0 + 0.033 477 783 203 125 062 450 045 135 152;
  • 2) 0.033 477 783 203 125 062 450 045 135 152 × 2 = 0 + 0.066 955 566 406 250 124 900 090 270 304;
  • 3) 0.066 955 566 406 250 124 900 090 270 304 × 2 = 0 + 0.133 911 132 812 500 249 800 180 540 608;
  • 4) 0.133 911 132 812 500 249 800 180 540 608 × 2 = 0 + 0.267 822 265 625 000 499 600 361 081 216;
  • 5) 0.267 822 265 625 000 499 600 361 081 216 × 2 = 0 + 0.535 644 531 250 000 999 200 722 162 432;
  • 6) 0.535 644 531 250 000 999 200 722 162 432 × 2 = 1 + 0.071 289 062 500 001 998 401 444 324 864;
  • 7) 0.071 289 062 500 001 998 401 444 324 864 × 2 = 0 + 0.142 578 125 000 003 996 802 888 649 728;
  • 8) 0.142 578 125 000 003 996 802 888 649 728 × 2 = 0 + 0.285 156 250 000 007 993 605 777 299 456;
  • 9) 0.285 156 250 000 007 993 605 777 299 456 × 2 = 0 + 0.570 312 500 000 015 987 211 554 598 912;
  • 10) 0.570 312 500 000 015 987 211 554 598 912 × 2 = 1 + 0.140 625 000 000 031 974 423 109 197 824;
  • 11) 0.140 625 000 000 031 974 423 109 197 824 × 2 = 0 + 0.281 250 000 000 063 948 846 218 395 648;
  • 12) 0.281 250 000 000 063 948 846 218 395 648 × 2 = 0 + 0.562 500 000 000 127 897 692 436 791 296;
  • 13) 0.562 500 000 000 127 897 692 436 791 296 × 2 = 1 + 0.125 000 000 000 255 795 384 873 582 592;
  • 14) 0.125 000 000 000 255 795 384 873 582 592 × 2 = 0 + 0.250 000 000 000 511 590 769 747 165 184;
  • 15) 0.250 000 000 000 511 590 769 747 165 184 × 2 = 0 + 0.500 000 000 001 023 181 539 494 330 368;
  • 16) 0.500 000 000 001 023 181 539 494 330 368 × 2 = 1 + 0.000 000 000 002 046 363 078 988 660 736;
  • 17) 0.000 000 000 002 046 363 078 988 660 736 × 2 = 0 + 0.000 000 000 004 092 726 157 977 321 472;
  • 18) 0.000 000 000 004 092 726 157 977 321 472 × 2 = 0 + 0.000 000 000 008 185 452 315 954 642 944;
  • 19) 0.000 000 000 008 185 452 315 954 642 944 × 2 = 0 + 0.000 000 000 016 370 904 631 909 285 888;
  • 20) 0.000 000 000 016 370 904 631 909 285 888 × 2 = 0 + 0.000 000 000 032 741 809 263 818 571 776;
  • 21) 0.000 000 000 032 741 809 263 818 571 776 × 2 = 0 + 0.000 000 000 065 483 618 527 637 143 552;
  • 22) 0.000 000 000 065 483 618 527 637 143 552 × 2 = 0 + 0.000 000 000 130 967 237 055 274 287 104;
  • 23) 0.000 000 000 130 967 237 055 274 287 104 × 2 = 0 + 0.000 000 000 261 934 474 110 548 574 208;
  • 24) 0.000 000 000 261 934 474 110 548 574 208 × 2 = 0 + 0.000 000 000 523 868 948 221 097 148 416;
  • 25) 0.000 000 000 523 868 948 221 097 148 416 × 2 = 0 + 0.000 000 001 047 737 896 442 194 296 832;
  • 26) 0.000 000 001 047 737 896 442 194 296 832 × 2 = 0 + 0.000 000 002 095 475 792 884 388 593 664;
  • 27) 0.000 000 002 095 475 792 884 388 593 664 × 2 = 0 + 0.000 000 004 190 951 585 768 777 187 328;
  • 28) 0.000 000 004 190 951 585 768 777 187 328 × 2 = 0 + 0.000 000 008 381 903 171 537 554 374 656;
  • 29) 0.000 000 008 381 903 171 537 554 374 656 × 2 = 0 + 0.000 000 016 763 806 343 075 108 749 312;
  • 30) 0.000 000 016 763 806 343 075 108 749 312 × 2 = 0 + 0.000 000 033 527 612 686 150 217 498 624;
  • 31) 0.000 000 033 527 612 686 150 217 498 624 × 2 = 0 + 0.000 000 067 055 225 372 300 434 997 248;
  • 32) 0.000 000 067 055 225 372 300 434 997 248 × 2 = 0 + 0.000 000 134 110 450 744 600 869 994 496;
  • 33) 0.000 000 134 110 450 744 600 869 994 496 × 2 = 0 + 0.000 000 268 220 901 489 201 739 988 992;
  • 34) 0.000 000 268 220 901 489 201 739 988 992 × 2 = 0 + 0.000 000 536 441 802 978 403 479 977 984;
  • 35) 0.000 000 536 441 802 978 403 479 977 984 × 2 = 0 + 0.000 001 072 883 605 956 806 959 955 968;
  • 36) 0.000 001 072 883 605 956 806 959 955 968 × 2 = 0 + 0.000 002 145 767 211 913 613 919 911 936;
  • 37) 0.000 002 145 767 211 913 613 919 911 936 × 2 = 0 + 0.000 004 291 534 423 827 227 839 823 872;
  • 38) 0.000 004 291 534 423 827 227 839 823 872 × 2 = 0 + 0.000 008 583 068 847 654 455 679 647 744;
  • 39) 0.000 008 583 068 847 654 455 679 647 744 × 2 = 0 + 0.000 017 166 137 695 308 911 359 295 488;
  • 40) 0.000 017 166 137 695 308 911 359 295 488 × 2 = 0 + 0.000 034 332 275 390 617 822 718 590 976;
  • 41) 0.000 034 332 275 390 617 822 718 590 976 × 2 = 0 + 0.000 068 664 550 781 235 645 437 181 952;
  • 42) 0.000 068 664 550 781 235 645 437 181 952 × 2 = 0 + 0.000 137 329 101 562 471 290 874 363 904;
  • 43) 0.000 137 329 101 562 471 290 874 363 904 × 2 = 0 + 0.000 274 658 203 124 942 581 748 727 808;
  • 44) 0.000 274 658 203 124 942 581 748 727 808 × 2 = 0 + 0.000 549 316 406 249 885 163 497 455 616;
  • 45) 0.000 549 316 406 249 885 163 497 455 616 × 2 = 0 + 0.001 098 632 812 499 770 326 994 911 232;
  • 46) 0.001 098 632 812 499 770 326 994 911 232 × 2 = 0 + 0.002 197 265 624 999 540 653 989 822 464;
  • 47) 0.002 197 265 624 999 540 653 989 822 464 × 2 = 0 + 0.004 394 531 249 999 081 307 979 644 928;
  • 48) 0.004 394 531 249 999 081 307 979 644 928 × 2 = 0 + 0.008 789 062 499 998 162 615 959 289 856;
  • 49) 0.008 789 062 499 998 162 615 959 289 856 × 2 = 0 + 0.017 578 124 999 996 325 231 918 579 712;
  • 50) 0.017 578 124 999 996 325 231 918 579 712 × 2 = 0 + 0.035 156 249 999 992 650 463 837 159 424;
  • 51) 0.035 156 249 999 992 650 463 837 159 424 × 2 = 0 + 0.070 312 499 999 985 300 927 674 318 848;
  • 52) 0.070 312 499 999 985 300 927 674 318 848 × 2 = 0 + 0.140 624 999 999 970 601 855 348 637 696;
  • 53) 0.140 624 999 999 970 601 855 348 637 696 × 2 = 0 + 0.281 249 999 999 941 203 710 697 275 392;
  • 54) 0.281 249 999 999 941 203 710 697 275 392 × 2 = 0 + 0.562 499 999 999 882 407 421 394 550 784;
  • 55) 0.562 499 999 999 882 407 421 394 550 784 × 2 = 1 + 0.124 999 999 999 764 814 842 789 101 568;
  • 56) 0.124 999 999 999 764 814 842 789 101 568 × 2 = 0 + 0.249 999 999 999 529 629 685 578 203 136;
  • 57) 0.249 999 999 999 529 629 685 578 203 136 × 2 = 0 + 0.499 999 999 999 059 259 371 156 406 272;
  • 58) 0.499 999 999 999 059 259 371 156 406 272 × 2 = 0 + 0.999 999 999 998 118 518 742 312 812 544;

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).


5. 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:


0.016 738 891 601 562 531 225 022 567 576(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

6. Positive number before normalization:

0.016 738 891 601 562 531 225 022 567 576(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2)

7. 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:


0.016 738 891 601 562 531 225 022 567 576(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 00(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


9. 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)


10. 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;

11. 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)


12. 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. 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


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

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

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000


Decimal number -0.016 738 891 601 562 531 225 022 567 576 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000

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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