-0.016 738 891 601 562 496 548 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 548 8(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
-0.016 738 891 601 562 496 548 8(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:

|-0.016 738 891 601 562 496 548 8| = 0.016 738 891 601 562 496 548 8


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: 0.016 738 891 601 562 496 548 8.

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) 0.016 738 891 601 562 496 548 8 × 2 = 0 + 0.033 477 783 203 124 993 097 6;
  • 2) 0.033 477 783 203 124 993 097 6 × 2 = 0 + 0.066 955 566 406 249 986 195 2;
  • 3) 0.066 955 566 406 249 986 195 2 × 2 = 0 + 0.133 911 132 812 499 972 390 4;
  • 4) 0.133 911 132 812 499 972 390 4 × 2 = 0 + 0.267 822 265 624 999 944 780 8;
  • 5) 0.267 822 265 624 999 944 780 8 × 2 = 0 + 0.535 644 531 249 999 889 561 6;
  • 6) 0.535 644 531 249 999 889 561 6 × 2 = 1 + 0.071 289 062 499 999 779 123 2;
  • 7) 0.071 289 062 499 999 779 123 2 × 2 = 0 + 0.142 578 124 999 999 558 246 4;
  • 8) 0.142 578 124 999 999 558 246 4 × 2 = 0 + 0.285 156 249 999 999 116 492 8;
  • 9) 0.285 156 249 999 999 116 492 8 × 2 = 0 + 0.570 312 499 999 998 232 985 6;
  • 10) 0.570 312 499 999 998 232 985 6 × 2 = 1 + 0.140 624 999 999 996 465 971 2;
  • 11) 0.140 624 999 999 996 465 971 2 × 2 = 0 + 0.281 249 999 999 992 931 942 4;
  • 12) 0.281 249 999 999 992 931 942 4 × 2 = 0 + 0.562 499 999 999 985 863 884 8;
  • 13) 0.562 499 999 999 985 863 884 8 × 2 = 1 + 0.124 999 999 999 971 727 769 6;
  • 14) 0.124 999 999 999 971 727 769 6 × 2 = 0 + 0.249 999 999 999 943 455 539 2;
  • 15) 0.249 999 999 999 943 455 539 2 × 2 = 0 + 0.499 999 999 999 886 911 078 4;
  • 16) 0.499 999 999 999 886 911 078 4 × 2 = 0 + 0.999 999 999 999 773 822 156 8;
  • 17) 0.999 999 999 999 773 822 156 8 × 2 = 1 + 0.999 999 999 999 547 644 313 6;
  • 18) 0.999 999 999 999 547 644 313 6 × 2 = 1 + 0.999 999 999 999 095 288 627 2;
  • 19) 0.999 999 999 999 095 288 627 2 × 2 = 1 + 0.999 999 999 998 190 577 254 4;
  • 20) 0.999 999 999 998 190 577 254 4 × 2 = 1 + 0.999 999 999 996 381 154 508 8;
  • 21) 0.999 999 999 996 381 154 508 8 × 2 = 1 + 0.999 999 999 992 762 309 017 6;
  • 22) 0.999 999 999 992 762 309 017 6 × 2 = 1 + 0.999 999 999 985 524 618 035 2;
  • 23) 0.999 999 999 985 524 618 035 2 × 2 = 1 + 0.999 999 999 971 049 236 070 4;
  • 24) 0.999 999 999 971 049 236 070 4 × 2 = 1 + 0.999 999 999 942 098 472 140 8;
  • 25) 0.999 999 999 942 098 472 140 8 × 2 = 1 + 0.999 999 999 884 196 944 281 6;
  • 26) 0.999 999 999 884 196 944 281 6 × 2 = 1 + 0.999 999 999 768 393 888 563 2;
  • 27) 0.999 999 999 768 393 888 563 2 × 2 = 1 + 0.999 999 999 536 787 777 126 4;
  • 28) 0.999 999 999 536 787 777 126 4 × 2 = 1 + 0.999 999 999 073 575 554 252 8;
  • 29) 0.999 999 999 073 575 554 252 8 × 2 = 1 + 0.999 999 998 147 151 108 505 6;
  • 30) 0.999 999 998 147 151 108 505 6 × 2 = 1 + 0.999 999 996 294 302 217 011 2;
  • 31) 0.999 999 996 294 302 217 011 2 × 2 = 1 + 0.999 999 992 588 604 434 022 4;
  • 32) 0.999 999 992 588 604 434 022 4 × 2 = 1 + 0.999 999 985 177 208 868 044 8;
  • 33) 0.999 999 985 177 208 868 044 8 × 2 = 1 + 0.999 999 970 354 417 736 089 6;
  • 34) 0.999 999 970 354 417 736 089 6 × 2 = 1 + 0.999 999 940 708 835 472 179 2;
  • 35) 0.999 999 940 708 835 472 179 2 × 2 = 1 + 0.999 999 881 417 670 944 358 4;
  • 36) 0.999 999 881 417 670 944 358 4 × 2 = 1 + 0.999 999 762 835 341 888 716 8;
  • 37) 0.999 999 762 835 341 888 716 8 × 2 = 1 + 0.999 999 525 670 683 777 433 6;
  • 38) 0.999 999 525 670 683 777 433 6 × 2 = 1 + 0.999 999 051 341 367 554 867 2;
  • 39) 0.999 999 051 341 367 554 867 2 × 2 = 1 + 0.999 998 102 682 735 109 734 4;
  • 40) 0.999 998 102 682 735 109 734 4 × 2 = 1 + 0.999 996 205 365 470 219 468 8;
  • 41) 0.999 996 205 365 470 219 468 8 × 2 = 1 + 0.999 992 410 730 940 438 937 6;
  • 42) 0.999 992 410 730 940 438 937 6 × 2 = 1 + 0.999 984 821 461 880 877 875 2;
  • 43) 0.999 984 821 461 880 877 875 2 × 2 = 1 + 0.999 969 642 923 761 755 750 4;
  • 44) 0.999 969 642 923 761 755 750 4 × 2 = 1 + 0.999 939 285 847 523 511 500 8;
  • 45) 0.999 939 285 847 523 511 500 8 × 2 = 1 + 0.999 878 571 695 047 023 001 6;
  • 46) 0.999 878 571 695 047 023 001 6 × 2 = 1 + 0.999 757 143 390 094 046 003 2;
  • 47) 0.999 757 143 390 094 046 003 2 × 2 = 1 + 0.999 514 286 780 188 092 006 4;
  • 48) 0.999 514 286 780 188 092 006 4 × 2 = 1 + 0.999 028 573 560 376 184 012 8;
  • 49) 0.999 028 573 560 376 184 012 8 × 2 = 1 + 0.998 057 147 120 752 368 025 6;
  • 50) 0.998 057 147 120 752 368 025 6 × 2 = 1 + 0.996 114 294 241 504 736 051 2;
  • 51) 0.996 114 294 241 504 736 051 2 × 2 = 1 + 0.992 228 588 483 009 472 102 4;
  • 52) 0.992 228 588 483 009 472 102 4 × 2 = 1 + 0.984 457 176 966 018 944 204 8;
  • 53) 0.984 457 176 966 018 944 204 8 × 2 = 1 + 0.968 914 353 932 037 888 409 6;
  • 54) 0.968 914 353 932 037 888 409 6 × 2 = 1 + 0.937 828 707 864 075 776 819 2;
  • 55) 0.937 828 707 864 075 776 819 2 × 2 = 1 + 0.875 657 415 728 151 553 638 4;
  • 56) 0.875 657 415 728 151 553 638 4 × 2 = 1 + 0.751 314 831 456 303 107 276 8;
  • 57) 0.751 314 831 456 303 107 276 8 × 2 = 1 + 0.502 629 662 912 606 214 553 6;
  • 58) 0.502 629 662 912 606 214 553 6 × 2 = 1 + 0.005 259 325 825 212 429 107 2;

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:


0.016 738 891 601 562 496 548 8(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 548 8(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 548 8(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

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


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 548 8 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 1111

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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