-0.016 738 891 601 562 496 529 92 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 92(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
-0.016 738 891 601 562 496 529 92(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:

|-0.016 738 891 601 562 496 529 92| = 0.016 738 891 601 562 496 529 92


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: 0.016 738 891 601 562 496 529 92.

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) 0.016 738 891 601 562 496 529 92 × 2 = 0 + 0.033 477 783 203 124 993 059 84;
  • 2) 0.033 477 783 203 124 993 059 84 × 2 = 0 + 0.066 955 566 406 249 986 119 68;
  • 3) 0.066 955 566 406 249 986 119 68 × 2 = 0 + 0.133 911 132 812 499 972 239 36;
  • 4) 0.133 911 132 812 499 972 239 36 × 2 = 0 + 0.267 822 265 624 999 944 478 72;
  • 5) 0.267 822 265 624 999 944 478 72 × 2 = 0 + 0.535 644 531 249 999 888 957 44;
  • 6) 0.535 644 531 249 999 888 957 44 × 2 = 1 + 0.071 289 062 499 999 777 914 88;
  • 7) 0.071 289 062 499 999 777 914 88 × 2 = 0 + 0.142 578 124 999 999 555 829 76;
  • 8) 0.142 578 124 999 999 555 829 76 × 2 = 0 + 0.285 156 249 999 999 111 659 52;
  • 9) 0.285 156 249 999 999 111 659 52 × 2 = 0 + 0.570 312 499 999 998 223 319 04;
  • 10) 0.570 312 499 999 998 223 319 04 × 2 = 1 + 0.140 624 999 999 996 446 638 08;
  • 11) 0.140 624 999 999 996 446 638 08 × 2 = 0 + 0.281 249 999 999 992 893 276 16;
  • 12) 0.281 249 999 999 992 893 276 16 × 2 = 0 + 0.562 499 999 999 985 786 552 32;
  • 13) 0.562 499 999 999 985 786 552 32 × 2 = 1 + 0.124 999 999 999 971 573 104 64;
  • 14) 0.124 999 999 999 971 573 104 64 × 2 = 0 + 0.249 999 999 999 943 146 209 28;
  • 15) 0.249 999 999 999 943 146 209 28 × 2 = 0 + 0.499 999 999 999 886 292 418 56;
  • 16) 0.499 999 999 999 886 292 418 56 × 2 = 0 + 0.999 999 999 999 772 584 837 12;
  • 17) 0.999 999 999 999 772 584 837 12 × 2 = 1 + 0.999 999 999 999 545 169 674 24;
  • 18) 0.999 999 999 999 545 169 674 24 × 2 = 1 + 0.999 999 999 999 090 339 348 48;
  • 19) 0.999 999 999 999 090 339 348 48 × 2 = 1 + 0.999 999 999 998 180 678 696 96;
  • 20) 0.999 999 999 998 180 678 696 96 × 2 = 1 + 0.999 999 999 996 361 357 393 92;
  • 21) 0.999 999 999 996 361 357 393 92 × 2 = 1 + 0.999 999 999 992 722 714 787 84;
  • 22) 0.999 999 999 992 722 714 787 84 × 2 = 1 + 0.999 999 999 985 445 429 575 68;
  • 23) 0.999 999 999 985 445 429 575 68 × 2 = 1 + 0.999 999 999 970 890 859 151 36;
  • 24) 0.999 999 999 970 890 859 151 36 × 2 = 1 + 0.999 999 999 941 781 718 302 72;
  • 25) 0.999 999 999 941 781 718 302 72 × 2 = 1 + 0.999 999 999 883 563 436 605 44;
  • 26) 0.999 999 999 883 563 436 605 44 × 2 = 1 + 0.999 999 999 767 126 873 210 88;
  • 27) 0.999 999 999 767 126 873 210 88 × 2 = 1 + 0.999 999 999 534 253 746 421 76;
  • 28) 0.999 999 999 534 253 746 421 76 × 2 = 1 + 0.999 999 999 068 507 492 843 52;
  • 29) 0.999 999 999 068 507 492 843 52 × 2 = 1 + 0.999 999 998 137 014 985 687 04;
  • 30) 0.999 999 998 137 014 985 687 04 × 2 = 1 + 0.999 999 996 274 029 971 374 08;
  • 31) 0.999 999 996 274 029 971 374 08 × 2 = 1 + 0.999 999 992 548 059 942 748 16;
  • 32) 0.999 999 992 548 059 942 748 16 × 2 = 1 + 0.999 999 985 096 119 885 496 32;
  • 33) 0.999 999 985 096 119 885 496 32 × 2 = 1 + 0.999 999 970 192 239 770 992 64;
  • 34) 0.999 999 970 192 239 770 992 64 × 2 = 1 + 0.999 999 940 384 479 541 985 28;
  • 35) 0.999 999 940 384 479 541 985 28 × 2 = 1 + 0.999 999 880 768 959 083 970 56;
  • 36) 0.999 999 880 768 959 083 970 56 × 2 = 1 + 0.999 999 761 537 918 167 941 12;
  • 37) 0.999 999 761 537 918 167 941 12 × 2 = 1 + 0.999 999 523 075 836 335 882 24;
  • 38) 0.999 999 523 075 836 335 882 24 × 2 = 1 + 0.999 999 046 151 672 671 764 48;
  • 39) 0.999 999 046 151 672 671 764 48 × 2 = 1 + 0.999 998 092 303 345 343 528 96;
  • 40) 0.999 998 092 303 345 343 528 96 × 2 = 1 + 0.999 996 184 606 690 687 057 92;
  • 41) 0.999 996 184 606 690 687 057 92 × 2 = 1 + 0.999 992 369 213 381 374 115 84;
  • 42) 0.999 992 369 213 381 374 115 84 × 2 = 1 + 0.999 984 738 426 762 748 231 68;
  • 43) 0.999 984 738 426 762 748 231 68 × 2 = 1 + 0.999 969 476 853 525 496 463 36;
  • 44) 0.999 969 476 853 525 496 463 36 × 2 = 1 + 0.999 938 953 707 050 992 926 72;
  • 45) 0.999 938 953 707 050 992 926 72 × 2 = 1 + 0.999 877 907 414 101 985 853 44;
  • 46) 0.999 877 907 414 101 985 853 44 × 2 = 1 + 0.999 755 814 828 203 971 706 88;
  • 47) 0.999 755 814 828 203 971 706 88 × 2 = 1 + 0.999 511 629 656 407 943 413 76;
  • 48) 0.999 511 629 656 407 943 413 76 × 2 = 1 + 0.999 023 259 312 815 886 827 52;
  • 49) 0.999 023 259 312 815 886 827 52 × 2 = 1 + 0.998 046 518 625 631 773 655 04;
  • 50) 0.998 046 518 625 631 773 655 04 × 2 = 1 + 0.996 093 037 251 263 547 310 08;
  • 51) 0.996 093 037 251 263 547 310 08 × 2 = 1 + 0.992 186 074 502 527 094 620 16;
  • 52) 0.992 186 074 502 527 094 620 16 × 2 = 1 + 0.984 372 149 005 054 189 240 32;
  • 53) 0.984 372 149 005 054 189 240 32 × 2 = 1 + 0.968 744 298 010 108 378 480 64;
  • 54) 0.968 744 298 010 108 378 480 64 × 2 = 1 + 0.937 488 596 020 216 756 961 28;
  • 55) 0.937 488 596 020 216 756 961 28 × 2 = 1 + 0.874 977 192 040 433 513 922 56;
  • 56) 0.874 977 192 040 433 513 922 56 × 2 = 1 + 0.749 954 384 080 867 027 845 12;
  • 57) 0.749 954 384 080 867 027 845 12 × 2 = 1 + 0.499 908 768 161 734 055 690 24;
  • 58) 0.499 908 768 161 734 055 690 24 × 2 = 0 + 0.999 817 536 323 468 111 380 48;

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:


0.016 738 891 601 562 496 529 92(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 529 92(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:


0.016 738 891 601 562 496 529 92(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 -0.016 738 891 601 562 496 529 92 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