-0.016 738 891 601 562 496 530 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 2(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
-0.016 738 891 601 562 496 530 2(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:

|-0.016 738 891 601 562 496 530 2| = 0.016 738 891 601 562 496 530 2


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: 0.016 738 891 601 562 496 530 2.

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) 0.016 738 891 601 562 496 530 2 × 2 = 0 + 0.033 477 783 203 124 993 060 4;
  • 2) 0.033 477 783 203 124 993 060 4 × 2 = 0 + 0.066 955 566 406 249 986 120 8;
  • 3) 0.066 955 566 406 249 986 120 8 × 2 = 0 + 0.133 911 132 812 499 972 241 6;
  • 4) 0.133 911 132 812 499 972 241 6 × 2 = 0 + 0.267 822 265 624 999 944 483 2;
  • 5) 0.267 822 265 624 999 944 483 2 × 2 = 0 + 0.535 644 531 249 999 888 966 4;
  • 6) 0.535 644 531 249 999 888 966 4 × 2 = 1 + 0.071 289 062 499 999 777 932 8;
  • 7) 0.071 289 062 499 999 777 932 8 × 2 = 0 + 0.142 578 124 999 999 555 865 6;
  • 8) 0.142 578 124 999 999 555 865 6 × 2 = 0 + 0.285 156 249 999 999 111 731 2;
  • 9) 0.285 156 249 999 999 111 731 2 × 2 = 0 + 0.570 312 499 999 998 223 462 4;
  • 10) 0.570 312 499 999 998 223 462 4 × 2 = 1 + 0.140 624 999 999 996 446 924 8;
  • 11) 0.140 624 999 999 996 446 924 8 × 2 = 0 + 0.281 249 999 999 992 893 849 6;
  • 12) 0.281 249 999 999 992 893 849 6 × 2 = 0 + 0.562 499 999 999 985 787 699 2;
  • 13) 0.562 499 999 999 985 787 699 2 × 2 = 1 + 0.124 999 999 999 971 575 398 4;
  • 14) 0.124 999 999 999 971 575 398 4 × 2 = 0 + 0.249 999 999 999 943 150 796 8;
  • 15) 0.249 999 999 999 943 150 796 8 × 2 = 0 + 0.499 999 999 999 886 301 593 6;
  • 16) 0.499 999 999 999 886 301 593 6 × 2 = 0 + 0.999 999 999 999 772 603 187 2;
  • 17) 0.999 999 999 999 772 603 187 2 × 2 = 1 + 0.999 999 999 999 545 206 374 4;
  • 18) 0.999 999 999 999 545 206 374 4 × 2 = 1 + 0.999 999 999 999 090 412 748 8;
  • 19) 0.999 999 999 999 090 412 748 8 × 2 = 1 + 0.999 999 999 998 180 825 497 6;
  • 20) 0.999 999 999 998 180 825 497 6 × 2 = 1 + 0.999 999 999 996 361 650 995 2;
  • 21) 0.999 999 999 996 361 650 995 2 × 2 = 1 + 0.999 999 999 992 723 301 990 4;
  • 22) 0.999 999 999 992 723 301 990 4 × 2 = 1 + 0.999 999 999 985 446 603 980 8;
  • 23) 0.999 999 999 985 446 603 980 8 × 2 = 1 + 0.999 999 999 970 893 207 961 6;
  • 24) 0.999 999 999 970 893 207 961 6 × 2 = 1 + 0.999 999 999 941 786 415 923 2;
  • 25) 0.999 999 999 941 786 415 923 2 × 2 = 1 + 0.999 999 999 883 572 831 846 4;
  • 26) 0.999 999 999 883 572 831 846 4 × 2 = 1 + 0.999 999 999 767 145 663 692 8;
  • 27) 0.999 999 999 767 145 663 692 8 × 2 = 1 + 0.999 999 999 534 291 327 385 6;
  • 28) 0.999 999 999 534 291 327 385 6 × 2 = 1 + 0.999 999 999 068 582 654 771 2;
  • 29) 0.999 999 999 068 582 654 771 2 × 2 = 1 + 0.999 999 998 137 165 309 542 4;
  • 30) 0.999 999 998 137 165 309 542 4 × 2 = 1 + 0.999 999 996 274 330 619 084 8;
  • 31) 0.999 999 996 274 330 619 084 8 × 2 = 1 + 0.999 999 992 548 661 238 169 6;
  • 32) 0.999 999 992 548 661 238 169 6 × 2 = 1 + 0.999 999 985 097 322 476 339 2;
  • 33) 0.999 999 985 097 322 476 339 2 × 2 = 1 + 0.999 999 970 194 644 952 678 4;
  • 34) 0.999 999 970 194 644 952 678 4 × 2 = 1 + 0.999 999 940 389 289 905 356 8;
  • 35) 0.999 999 940 389 289 905 356 8 × 2 = 1 + 0.999 999 880 778 579 810 713 6;
  • 36) 0.999 999 880 778 579 810 713 6 × 2 = 1 + 0.999 999 761 557 159 621 427 2;
  • 37) 0.999 999 761 557 159 621 427 2 × 2 = 1 + 0.999 999 523 114 319 242 854 4;
  • 38) 0.999 999 523 114 319 242 854 4 × 2 = 1 + 0.999 999 046 228 638 485 708 8;
  • 39) 0.999 999 046 228 638 485 708 8 × 2 = 1 + 0.999 998 092 457 276 971 417 6;
  • 40) 0.999 998 092 457 276 971 417 6 × 2 = 1 + 0.999 996 184 914 553 942 835 2;
  • 41) 0.999 996 184 914 553 942 835 2 × 2 = 1 + 0.999 992 369 829 107 885 670 4;
  • 42) 0.999 992 369 829 107 885 670 4 × 2 = 1 + 0.999 984 739 658 215 771 340 8;
  • 43) 0.999 984 739 658 215 771 340 8 × 2 = 1 + 0.999 969 479 316 431 542 681 6;
  • 44) 0.999 969 479 316 431 542 681 6 × 2 = 1 + 0.999 938 958 632 863 085 363 2;
  • 45) 0.999 938 958 632 863 085 363 2 × 2 = 1 + 0.999 877 917 265 726 170 726 4;
  • 46) 0.999 877 917 265 726 170 726 4 × 2 = 1 + 0.999 755 834 531 452 341 452 8;
  • 47) 0.999 755 834 531 452 341 452 8 × 2 = 1 + 0.999 511 669 062 904 682 905 6;
  • 48) 0.999 511 669 062 904 682 905 6 × 2 = 1 + 0.999 023 338 125 809 365 811 2;
  • 49) 0.999 023 338 125 809 365 811 2 × 2 = 1 + 0.998 046 676 251 618 731 622 4;
  • 50) 0.998 046 676 251 618 731 622 4 × 2 = 1 + 0.996 093 352 503 237 463 244 8;
  • 51) 0.996 093 352 503 237 463 244 8 × 2 = 1 + 0.992 186 705 006 474 926 489 6;
  • 52) 0.992 186 705 006 474 926 489 6 × 2 = 1 + 0.984 373 410 012 949 852 979 2;
  • 53) 0.984 373 410 012 949 852 979 2 × 2 = 1 + 0.968 746 820 025 899 705 958 4;
  • 54) 0.968 746 820 025 899 705 958 4 × 2 = 1 + 0.937 493 640 051 799 411 916 8;
  • 55) 0.937 493 640 051 799 411 916 8 × 2 = 1 + 0.874 987 280 103 598 823 833 6;
  • 56) 0.874 987 280 103 598 823 833 6 × 2 = 1 + 0.749 974 560 207 197 647 667 2;
  • 57) 0.749 974 560 207 197 647 667 2 × 2 = 1 + 0.499 949 120 414 395 295 334 4;
  • 58) 0.499 949 120 414 395 295 334 4 × 2 = 0 + 0.999 898 240 828 790 590 668 8;

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:


0.016 738 891 601 562 496 530 2(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 2(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 2(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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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