-0.016 738 891 601 562 496 531 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 531 19(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
-0.016 738 891 601 562 496 531 19(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:

|-0.016 738 891 601 562 496 531 19| = 0.016 738 891 601 562 496 531 19


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: 0.016 738 891 601 562 496 531 19.

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) 0.016 738 891 601 562 496 531 19 × 2 = 0 + 0.033 477 783 203 124 993 062 38;
  • 2) 0.033 477 783 203 124 993 062 38 × 2 = 0 + 0.066 955 566 406 249 986 124 76;
  • 3) 0.066 955 566 406 249 986 124 76 × 2 = 0 + 0.133 911 132 812 499 972 249 52;
  • 4) 0.133 911 132 812 499 972 249 52 × 2 = 0 + 0.267 822 265 624 999 944 499 04;
  • 5) 0.267 822 265 624 999 944 499 04 × 2 = 0 + 0.535 644 531 249 999 888 998 08;
  • 6) 0.535 644 531 249 999 888 998 08 × 2 = 1 + 0.071 289 062 499 999 777 996 16;
  • 7) 0.071 289 062 499 999 777 996 16 × 2 = 0 + 0.142 578 124 999 999 555 992 32;
  • 8) 0.142 578 124 999 999 555 992 32 × 2 = 0 + 0.285 156 249 999 999 111 984 64;
  • 9) 0.285 156 249 999 999 111 984 64 × 2 = 0 + 0.570 312 499 999 998 223 969 28;
  • 10) 0.570 312 499 999 998 223 969 28 × 2 = 1 + 0.140 624 999 999 996 447 938 56;
  • 11) 0.140 624 999 999 996 447 938 56 × 2 = 0 + 0.281 249 999 999 992 895 877 12;
  • 12) 0.281 249 999 999 992 895 877 12 × 2 = 0 + 0.562 499 999 999 985 791 754 24;
  • 13) 0.562 499 999 999 985 791 754 24 × 2 = 1 + 0.124 999 999 999 971 583 508 48;
  • 14) 0.124 999 999 999 971 583 508 48 × 2 = 0 + 0.249 999 999 999 943 167 016 96;
  • 15) 0.249 999 999 999 943 167 016 96 × 2 = 0 + 0.499 999 999 999 886 334 033 92;
  • 16) 0.499 999 999 999 886 334 033 92 × 2 = 0 + 0.999 999 999 999 772 668 067 84;
  • 17) 0.999 999 999 999 772 668 067 84 × 2 = 1 + 0.999 999 999 999 545 336 135 68;
  • 18) 0.999 999 999 999 545 336 135 68 × 2 = 1 + 0.999 999 999 999 090 672 271 36;
  • 19) 0.999 999 999 999 090 672 271 36 × 2 = 1 + 0.999 999 999 998 181 344 542 72;
  • 20) 0.999 999 999 998 181 344 542 72 × 2 = 1 + 0.999 999 999 996 362 689 085 44;
  • 21) 0.999 999 999 996 362 689 085 44 × 2 = 1 + 0.999 999 999 992 725 378 170 88;
  • 22) 0.999 999 999 992 725 378 170 88 × 2 = 1 + 0.999 999 999 985 450 756 341 76;
  • 23) 0.999 999 999 985 450 756 341 76 × 2 = 1 + 0.999 999 999 970 901 512 683 52;
  • 24) 0.999 999 999 970 901 512 683 52 × 2 = 1 + 0.999 999 999 941 803 025 367 04;
  • 25) 0.999 999 999 941 803 025 367 04 × 2 = 1 + 0.999 999 999 883 606 050 734 08;
  • 26) 0.999 999 999 883 606 050 734 08 × 2 = 1 + 0.999 999 999 767 212 101 468 16;
  • 27) 0.999 999 999 767 212 101 468 16 × 2 = 1 + 0.999 999 999 534 424 202 936 32;
  • 28) 0.999 999 999 534 424 202 936 32 × 2 = 1 + 0.999 999 999 068 848 405 872 64;
  • 29) 0.999 999 999 068 848 405 872 64 × 2 = 1 + 0.999 999 998 137 696 811 745 28;
  • 30) 0.999 999 998 137 696 811 745 28 × 2 = 1 + 0.999 999 996 275 393 623 490 56;
  • 31) 0.999 999 996 275 393 623 490 56 × 2 = 1 + 0.999 999 992 550 787 246 981 12;
  • 32) 0.999 999 992 550 787 246 981 12 × 2 = 1 + 0.999 999 985 101 574 493 962 24;
  • 33) 0.999 999 985 101 574 493 962 24 × 2 = 1 + 0.999 999 970 203 148 987 924 48;
  • 34) 0.999 999 970 203 148 987 924 48 × 2 = 1 + 0.999 999 940 406 297 975 848 96;
  • 35) 0.999 999 940 406 297 975 848 96 × 2 = 1 + 0.999 999 880 812 595 951 697 92;
  • 36) 0.999 999 880 812 595 951 697 92 × 2 = 1 + 0.999 999 761 625 191 903 395 84;
  • 37) 0.999 999 761 625 191 903 395 84 × 2 = 1 + 0.999 999 523 250 383 806 791 68;
  • 38) 0.999 999 523 250 383 806 791 68 × 2 = 1 + 0.999 999 046 500 767 613 583 36;
  • 39) 0.999 999 046 500 767 613 583 36 × 2 = 1 + 0.999 998 093 001 535 227 166 72;
  • 40) 0.999 998 093 001 535 227 166 72 × 2 = 1 + 0.999 996 186 003 070 454 333 44;
  • 41) 0.999 996 186 003 070 454 333 44 × 2 = 1 + 0.999 992 372 006 140 908 666 88;
  • 42) 0.999 992 372 006 140 908 666 88 × 2 = 1 + 0.999 984 744 012 281 817 333 76;
  • 43) 0.999 984 744 012 281 817 333 76 × 2 = 1 + 0.999 969 488 024 563 634 667 52;
  • 44) 0.999 969 488 024 563 634 667 52 × 2 = 1 + 0.999 938 976 049 127 269 335 04;
  • 45) 0.999 938 976 049 127 269 335 04 × 2 = 1 + 0.999 877 952 098 254 538 670 08;
  • 46) 0.999 877 952 098 254 538 670 08 × 2 = 1 + 0.999 755 904 196 509 077 340 16;
  • 47) 0.999 755 904 196 509 077 340 16 × 2 = 1 + 0.999 511 808 393 018 154 680 32;
  • 48) 0.999 511 808 393 018 154 680 32 × 2 = 1 + 0.999 023 616 786 036 309 360 64;
  • 49) 0.999 023 616 786 036 309 360 64 × 2 = 1 + 0.998 047 233 572 072 618 721 28;
  • 50) 0.998 047 233 572 072 618 721 28 × 2 = 1 + 0.996 094 467 144 145 237 442 56;
  • 51) 0.996 094 467 144 145 237 442 56 × 2 = 1 + 0.992 188 934 288 290 474 885 12;
  • 52) 0.992 188 934 288 290 474 885 12 × 2 = 1 + 0.984 377 868 576 580 949 770 24;
  • 53) 0.984 377 868 576 580 949 770 24 × 2 = 1 + 0.968 755 737 153 161 899 540 48;
  • 54) 0.968 755 737 153 161 899 540 48 × 2 = 1 + 0.937 511 474 306 323 799 080 96;
  • 55) 0.937 511 474 306 323 799 080 96 × 2 = 1 + 0.875 022 948 612 647 598 161 92;
  • 56) 0.875 022 948 612 647 598 161 92 × 2 = 1 + 0.750 045 897 225 295 196 323 84;
  • 57) 0.750 045 897 225 295 196 323 84 × 2 = 1 + 0.500 091 794 450 590 392 647 68;
  • 58) 0.500 091 794 450 590 392 647 68 × 2 = 1 + 0.000 183 588 901 180 785 295 36;

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:


0.016 738 891 601 562 496 531 19(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 531 19(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:


0.016 738 891 601 562 496 531 19(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 -0.016 738 891 601 562 496 531 19 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