-0.016 738 891 601 562 496 541 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 541 6(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
-0.016 738 891 601 562 496 541 6(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:

|-0.016 738 891 601 562 496 541 6| = 0.016 738 891 601 562 496 541 6


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: 0.016 738 891 601 562 496 541 6.

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) 0.016 738 891 601 562 496 541 6 × 2 = 0 + 0.033 477 783 203 124 993 083 2;
  • 2) 0.033 477 783 203 124 993 083 2 × 2 = 0 + 0.066 955 566 406 249 986 166 4;
  • 3) 0.066 955 566 406 249 986 166 4 × 2 = 0 + 0.133 911 132 812 499 972 332 8;
  • 4) 0.133 911 132 812 499 972 332 8 × 2 = 0 + 0.267 822 265 624 999 944 665 6;
  • 5) 0.267 822 265 624 999 944 665 6 × 2 = 0 + 0.535 644 531 249 999 889 331 2;
  • 6) 0.535 644 531 249 999 889 331 2 × 2 = 1 + 0.071 289 062 499 999 778 662 4;
  • 7) 0.071 289 062 499 999 778 662 4 × 2 = 0 + 0.142 578 124 999 999 557 324 8;
  • 8) 0.142 578 124 999 999 557 324 8 × 2 = 0 + 0.285 156 249 999 999 114 649 6;
  • 9) 0.285 156 249 999 999 114 649 6 × 2 = 0 + 0.570 312 499 999 998 229 299 2;
  • 10) 0.570 312 499 999 998 229 299 2 × 2 = 1 + 0.140 624 999 999 996 458 598 4;
  • 11) 0.140 624 999 999 996 458 598 4 × 2 = 0 + 0.281 249 999 999 992 917 196 8;
  • 12) 0.281 249 999 999 992 917 196 8 × 2 = 0 + 0.562 499 999 999 985 834 393 6;
  • 13) 0.562 499 999 999 985 834 393 6 × 2 = 1 + 0.124 999 999 999 971 668 787 2;
  • 14) 0.124 999 999 999 971 668 787 2 × 2 = 0 + 0.249 999 999 999 943 337 574 4;
  • 15) 0.249 999 999 999 943 337 574 4 × 2 = 0 + 0.499 999 999 999 886 675 148 8;
  • 16) 0.499 999 999 999 886 675 148 8 × 2 = 0 + 0.999 999 999 999 773 350 297 6;
  • 17) 0.999 999 999 999 773 350 297 6 × 2 = 1 + 0.999 999 999 999 546 700 595 2;
  • 18) 0.999 999 999 999 546 700 595 2 × 2 = 1 + 0.999 999 999 999 093 401 190 4;
  • 19) 0.999 999 999 999 093 401 190 4 × 2 = 1 + 0.999 999 999 998 186 802 380 8;
  • 20) 0.999 999 999 998 186 802 380 8 × 2 = 1 + 0.999 999 999 996 373 604 761 6;
  • 21) 0.999 999 999 996 373 604 761 6 × 2 = 1 + 0.999 999 999 992 747 209 523 2;
  • 22) 0.999 999 999 992 747 209 523 2 × 2 = 1 + 0.999 999 999 985 494 419 046 4;
  • 23) 0.999 999 999 985 494 419 046 4 × 2 = 1 + 0.999 999 999 970 988 838 092 8;
  • 24) 0.999 999 999 970 988 838 092 8 × 2 = 1 + 0.999 999 999 941 977 676 185 6;
  • 25) 0.999 999 999 941 977 676 185 6 × 2 = 1 + 0.999 999 999 883 955 352 371 2;
  • 26) 0.999 999 999 883 955 352 371 2 × 2 = 1 + 0.999 999 999 767 910 704 742 4;
  • 27) 0.999 999 999 767 910 704 742 4 × 2 = 1 + 0.999 999 999 535 821 409 484 8;
  • 28) 0.999 999 999 535 821 409 484 8 × 2 = 1 + 0.999 999 999 071 642 818 969 6;
  • 29) 0.999 999 999 071 642 818 969 6 × 2 = 1 + 0.999 999 998 143 285 637 939 2;
  • 30) 0.999 999 998 143 285 637 939 2 × 2 = 1 + 0.999 999 996 286 571 275 878 4;
  • 31) 0.999 999 996 286 571 275 878 4 × 2 = 1 + 0.999 999 992 573 142 551 756 8;
  • 32) 0.999 999 992 573 142 551 756 8 × 2 = 1 + 0.999 999 985 146 285 103 513 6;
  • 33) 0.999 999 985 146 285 103 513 6 × 2 = 1 + 0.999 999 970 292 570 207 027 2;
  • 34) 0.999 999 970 292 570 207 027 2 × 2 = 1 + 0.999 999 940 585 140 414 054 4;
  • 35) 0.999 999 940 585 140 414 054 4 × 2 = 1 + 0.999 999 881 170 280 828 108 8;
  • 36) 0.999 999 881 170 280 828 108 8 × 2 = 1 + 0.999 999 762 340 561 656 217 6;
  • 37) 0.999 999 762 340 561 656 217 6 × 2 = 1 + 0.999 999 524 681 123 312 435 2;
  • 38) 0.999 999 524 681 123 312 435 2 × 2 = 1 + 0.999 999 049 362 246 624 870 4;
  • 39) 0.999 999 049 362 246 624 870 4 × 2 = 1 + 0.999 998 098 724 493 249 740 8;
  • 40) 0.999 998 098 724 493 249 740 8 × 2 = 1 + 0.999 996 197 448 986 499 481 6;
  • 41) 0.999 996 197 448 986 499 481 6 × 2 = 1 + 0.999 992 394 897 972 998 963 2;
  • 42) 0.999 992 394 897 972 998 963 2 × 2 = 1 + 0.999 984 789 795 945 997 926 4;
  • 43) 0.999 984 789 795 945 997 926 4 × 2 = 1 + 0.999 969 579 591 891 995 852 8;
  • 44) 0.999 969 579 591 891 995 852 8 × 2 = 1 + 0.999 939 159 183 783 991 705 6;
  • 45) 0.999 939 159 183 783 991 705 6 × 2 = 1 + 0.999 878 318 367 567 983 411 2;
  • 46) 0.999 878 318 367 567 983 411 2 × 2 = 1 + 0.999 756 636 735 135 966 822 4;
  • 47) 0.999 756 636 735 135 966 822 4 × 2 = 1 + 0.999 513 273 470 271 933 644 8;
  • 48) 0.999 513 273 470 271 933 644 8 × 2 = 1 + 0.999 026 546 940 543 867 289 6;
  • 49) 0.999 026 546 940 543 867 289 6 × 2 = 1 + 0.998 053 093 881 087 734 579 2;
  • 50) 0.998 053 093 881 087 734 579 2 × 2 = 1 + 0.996 106 187 762 175 469 158 4;
  • 51) 0.996 106 187 762 175 469 158 4 × 2 = 1 + 0.992 212 375 524 350 938 316 8;
  • 52) 0.992 212 375 524 350 938 316 8 × 2 = 1 + 0.984 424 751 048 701 876 633 6;
  • 53) 0.984 424 751 048 701 876 633 6 × 2 = 1 + 0.968 849 502 097 403 753 267 2;
  • 54) 0.968 849 502 097 403 753 267 2 × 2 = 1 + 0.937 699 004 194 807 506 534 4;
  • 55) 0.937 699 004 194 807 506 534 4 × 2 = 1 + 0.875 398 008 389 615 013 068 8;
  • 56) 0.875 398 008 389 615 013 068 8 × 2 = 1 + 0.750 796 016 779 230 026 137 6;
  • 57) 0.750 796 016 779 230 026 137 6 × 2 = 1 + 0.501 592 033 558 460 052 275 2;
  • 58) 0.501 592 033 558 460 052 275 2 × 2 = 1 + 0.003 184 067 116 920 104 550 4;

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:


0.016 738 891 601 562 496 541 6(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 541 6(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:


0.016 738 891 601 562 496 541 6(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 -0.016 738 891 601 562 496 541 6 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