-0.016 738 891 601 562 496 520 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 520 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
-0.016 738 891 601 562 496 520 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:

|-0.016 738 891 601 562 496 520 2| = 0.016 738 891 601 562 496 520 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: 0.016 738 891 601 562 496 520 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) 0.016 738 891 601 562 496 520 2 × 2 = 0 + 0.033 477 783 203 124 993 040 4;
  • 2) 0.033 477 783 203 124 993 040 4 × 2 = 0 + 0.066 955 566 406 249 986 080 8;
  • 3) 0.066 955 566 406 249 986 080 8 × 2 = 0 + 0.133 911 132 812 499 972 161 6;
  • 4) 0.133 911 132 812 499 972 161 6 × 2 = 0 + 0.267 822 265 624 999 944 323 2;
  • 5) 0.267 822 265 624 999 944 323 2 × 2 = 0 + 0.535 644 531 249 999 888 646 4;
  • 6) 0.535 644 531 249 999 888 646 4 × 2 = 1 + 0.071 289 062 499 999 777 292 8;
  • 7) 0.071 289 062 499 999 777 292 8 × 2 = 0 + 0.142 578 124 999 999 554 585 6;
  • 8) 0.142 578 124 999 999 554 585 6 × 2 = 0 + 0.285 156 249 999 999 109 171 2;
  • 9) 0.285 156 249 999 999 109 171 2 × 2 = 0 + 0.570 312 499 999 998 218 342 4;
  • 10) 0.570 312 499 999 998 218 342 4 × 2 = 1 + 0.140 624 999 999 996 436 684 8;
  • 11) 0.140 624 999 999 996 436 684 8 × 2 = 0 + 0.281 249 999 999 992 873 369 6;
  • 12) 0.281 249 999 999 992 873 369 6 × 2 = 0 + 0.562 499 999 999 985 746 739 2;
  • 13) 0.562 499 999 999 985 746 739 2 × 2 = 1 + 0.124 999 999 999 971 493 478 4;
  • 14) 0.124 999 999 999 971 493 478 4 × 2 = 0 + 0.249 999 999 999 942 986 956 8;
  • 15) 0.249 999 999 999 942 986 956 8 × 2 = 0 + 0.499 999 999 999 885 973 913 6;
  • 16) 0.499 999 999 999 885 973 913 6 × 2 = 0 + 0.999 999 999 999 771 947 827 2;
  • 17) 0.999 999 999 999 771 947 827 2 × 2 = 1 + 0.999 999 999 999 543 895 654 4;
  • 18) 0.999 999 999 999 543 895 654 4 × 2 = 1 + 0.999 999 999 999 087 791 308 8;
  • 19) 0.999 999 999 999 087 791 308 8 × 2 = 1 + 0.999 999 999 998 175 582 617 6;
  • 20) 0.999 999 999 998 175 582 617 6 × 2 = 1 + 0.999 999 999 996 351 165 235 2;
  • 21) 0.999 999 999 996 351 165 235 2 × 2 = 1 + 0.999 999 999 992 702 330 470 4;
  • 22) 0.999 999 999 992 702 330 470 4 × 2 = 1 + 0.999 999 999 985 404 660 940 8;
  • 23) 0.999 999 999 985 404 660 940 8 × 2 = 1 + 0.999 999 999 970 809 321 881 6;
  • 24) 0.999 999 999 970 809 321 881 6 × 2 = 1 + 0.999 999 999 941 618 643 763 2;
  • 25) 0.999 999 999 941 618 643 763 2 × 2 = 1 + 0.999 999 999 883 237 287 526 4;
  • 26) 0.999 999 999 883 237 287 526 4 × 2 = 1 + 0.999 999 999 766 474 575 052 8;
  • 27) 0.999 999 999 766 474 575 052 8 × 2 = 1 + 0.999 999 999 532 949 150 105 6;
  • 28) 0.999 999 999 532 949 150 105 6 × 2 = 1 + 0.999 999 999 065 898 300 211 2;
  • 29) 0.999 999 999 065 898 300 211 2 × 2 = 1 + 0.999 999 998 131 796 600 422 4;
  • 30) 0.999 999 998 131 796 600 422 4 × 2 = 1 + 0.999 999 996 263 593 200 844 8;
  • 31) 0.999 999 996 263 593 200 844 8 × 2 = 1 + 0.999 999 992 527 186 401 689 6;
  • 32) 0.999 999 992 527 186 401 689 6 × 2 = 1 + 0.999 999 985 054 372 803 379 2;
  • 33) 0.999 999 985 054 372 803 379 2 × 2 = 1 + 0.999 999 970 108 745 606 758 4;
  • 34) 0.999 999 970 108 745 606 758 4 × 2 = 1 + 0.999 999 940 217 491 213 516 8;
  • 35) 0.999 999 940 217 491 213 516 8 × 2 = 1 + 0.999 999 880 434 982 427 033 6;
  • 36) 0.999 999 880 434 982 427 033 6 × 2 = 1 + 0.999 999 760 869 964 854 067 2;
  • 37) 0.999 999 760 869 964 854 067 2 × 2 = 1 + 0.999 999 521 739 929 708 134 4;
  • 38) 0.999 999 521 739 929 708 134 4 × 2 = 1 + 0.999 999 043 479 859 416 268 8;
  • 39) 0.999 999 043 479 859 416 268 8 × 2 = 1 + 0.999 998 086 959 718 832 537 6;
  • 40) 0.999 998 086 959 718 832 537 6 × 2 = 1 + 0.999 996 173 919 437 665 075 2;
  • 41) 0.999 996 173 919 437 665 075 2 × 2 = 1 + 0.999 992 347 838 875 330 150 4;
  • 42) 0.999 992 347 838 875 330 150 4 × 2 = 1 + 0.999 984 695 677 750 660 300 8;
  • 43) 0.999 984 695 677 750 660 300 8 × 2 = 1 + 0.999 969 391 355 501 320 601 6;
  • 44) 0.999 969 391 355 501 320 601 6 × 2 = 1 + 0.999 938 782 711 002 641 203 2;
  • 45) 0.999 938 782 711 002 641 203 2 × 2 = 1 + 0.999 877 565 422 005 282 406 4;
  • 46) 0.999 877 565 422 005 282 406 4 × 2 = 1 + 0.999 755 130 844 010 564 812 8;
  • 47) 0.999 755 130 844 010 564 812 8 × 2 = 1 + 0.999 510 261 688 021 129 625 6;
  • 48) 0.999 510 261 688 021 129 625 6 × 2 = 1 + 0.999 020 523 376 042 259 251 2;
  • 49) 0.999 020 523 376 042 259 251 2 × 2 = 1 + 0.998 041 046 752 084 518 502 4;
  • 50) 0.998 041 046 752 084 518 502 4 × 2 = 1 + 0.996 082 093 504 169 037 004 8;
  • 51) 0.996 082 093 504 169 037 004 8 × 2 = 1 + 0.992 164 187 008 338 074 009 6;
  • 52) 0.992 164 187 008 338 074 009 6 × 2 = 1 + 0.984 328 374 016 676 148 019 2;
  • 53) 0.984 328 374 016 676 148 019 2 × 2 = 1 + 0.968 656 748 033 352 296 038 4;
  • 54) 0.968 656 748 033 352 296 038 4 × 2 = 1 + 0.937 313 496 066 704 592 076 8;
  • 55) 0.937 313 496 066 704 592 076 8 × 2 = 1 + 0.874 626 992 133 409 184 153 6;
  • 56) 0.874 626 992 133 409 184 153 6 × 2 = 1 + 0.749 253 984 266 818 368 307 2;
  • 57) 0.749 253 984 266 818 368 307 2 × 2 = 1 + 0.498 507 968 533 636 736 614 4;
  • 58) 0.498 507 968 533 636 736 614 4 × 2 = 0 + 0.997 015 937 067 273 473 228 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:


0.016 738 891 601 562 496 520 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 520 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:


0.016 738 891 601 562 496 520 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 -0.016 738 891 601 562 496 520 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