-0.016 738 891 601 562 496 523 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 523 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
-0.016 738 891 601 562 496 523 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:

|-0.016 738 891 601 562 496 523 2| = 0.016 738 891 601 562 496 523 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: 0.016 738 891 601 562 496 523 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) 0.016 738 891 601 562 496 523 2 × 2 = 0 + 0.033 477 783 203 124 993 046 4;
  • 2) 0.033 477 783 203 124 993 046 4 × 2 = 0 + 0.066 955 566 406 249 986 092 8;
  • 3) 0.066 955 566 406 249 986 092 8 × 2 = 0 + 0.133 911 132 812 499 972 185 6;
  • 4) 0.133 911 132 812 499 972 185 6 × 2 = 0 + 0.267 822 265 624 999 944 371 2;
  • 5) 0.267 822 265 624 999 944 371 2 × 2 = 0 + 0.535 644 531 249 999 888 742 4;
  • 6) 0.535 644 531 249 999 888 742 4 × 2 = 1 + 0.071 289 062 499 999 777 484 8;
  • 7) 0.071 289 062 499 999 777 484 8 × 2 = 0 + 0.142 578 124 999 999 554 969 6;
  • 8) 0.142 578 124 999 999 554 969 6 × 2 = 0 + 0.285 156 249 999 999 109 939 2;
  • 9) 0.285 156 249 999 999 109 939 2 × 2 = 0 + 0.570 312 499 999 998 219 878 4;
  • 10) 0.570 312 499 999 998 219 878 4 × 2 = 1 + 0.140 624 999 999 996 439 756 8;
  • 11) 0.140 624 999 999 996 439 756 8 × 2 = 0 + 0.281 249 999 999 992 879 513 6;
  • 12) 0.281 249 999 999 992 879 513 6 × 2 = 0 + 0.562 499 999 999 985 759 027 2;
  • 13) 0.562 499 999 999 985 759 027 2 × 2 = 1 + 0.124 999 999 999 971 518 054 4;
  • 14) 0.124 999 999 999 971 518 054 4 × 2 = 0 + 0.249 999 999 999 943 036 108 8;
  • 15) 0.249 999 999 999 943 036 108 8 × 2 = 0 + 0.499 999 999 999 886 072 217 6;
  • 16) 0.499 999 999 999 886 072 217 6 × 2 = 0 + 0.999 999 999 999 772 144 435 2;
  • 17) 0.999 999 999 999 772 144 435 2 × 2 = 1 + 0.999 999 999 999 544 288 870 4;
  • 18) 0.999 999 999 999 544 288 870 4 × 2 = 1 + 0.999 999 999 999 088 577 740 8;
  • 19) 0.999 999 999 999 088 577 740 8 × 2 = 1 + 0.999 999 999 998 177 155 481 6;
  • 20) 0.999 999 999 998 177 155 481 6 × 2 = 1 + 0.999 999 999 996 354 310 963 2;
  • 21) 0.999 999 999 996 354 310 963 2 × 2 = 1 + 0.999 999 999 992 708 621 926 4;
  • 22) 0.999 999 999 992 708 621 926 4 × 2 = 1 + 0.999 999 999 985 417 243 852 8;
  • 23) 0.999 999 999 985 417 243 852 8 × 2 = 1 + 0.999 999 999 970 834 487 705 6;
  • 24) 0.999 999 999 970 834 487 705 6 × 2 = 1 + 0.999 999 999 941 668 975 411 2;
  • 25) 0.999 999 999 941 668 975 411 2 × 2 = 1 + 0.999 999 999 883 337 950 822 4;
  • 26) 0.999 999 999 883 337 950 822 4 × 2 = 1 + 0.999 999 999 766 675 901 644 8;
  • 27) 0.999 999 999 766 675 901 644 8 × 2 = 1 + 0.999 999 999 533 351 803 289 6;
  • 28) 0.999 999 999 533 351 803 289 6 × 2 = 1 + 0.999 999 999 066 703 606 579 2;
  • 29) 0.999 999 999 066 703 606 579 2 × 2 = 1 + 0.999 999 998 133 407 213 158 4;
  • 30) 0.999 999 998 133 407 213 158 4 × 2 = 1 + 0.999 999 996 266 814 426 316 8;
  • 31) 0.999 999 996 266 814 426 316 8 × 2 = 1 + 0.999 999 992 533 628 852 633 6;
  • 32) 0.999 999 992 533 628 852 633 6 × 2 = 1 + 0.999 999 985 067 257 705 267 2;
  • 33) 0.999 999 985 067 257 705 267 2 × 2 = 1 + 0.999 999 970 134 515 410 534 4;
  • 34) 0.999 999 970 134 515 410 534 4 × 2 = 1 + 0.999 999 940 269 030 821 068 8;
  • 35) 0.999 999 940 269 030 821 068 8 × 2 = 1 + 0.999 999 880 538 061 642 137 6;
  • 36) 0.999 999 880 538 061 642 137 6 × 2 = 1 + 0.999 999 761 076 123 284 275 2;
  • 37) 0.999 999 761 076 123 284 275 2 × 2 = 1 + 0.999 999 522 152 246 568 550 4;
  • 38) 0.999 999 522 152 246 568 550 4 × 2 = 1 + 0.999 999 044 304 493 137 100 8;
  • 39) 0.999 999 044 304 493 137 100 8 × 2 = 1 + 0.999 998 088 608 986 274 201 6;
  • 40) 0.999 998 088 608 986 274 201 6 × 2 = 1 + 0.999 996 177 217 972 548 403 2;
  • 41) 0.999 996 177 217 972 548 403 2 × 2 = 1 + 0.999 992 354 435 945 096 806 4;
  • 42) 0.999 992 354 435 945 096 806 4 × 2 = 1 + 0.999 984 708 871 890 193 612 8;
  • 43) 0.999 984 708 871 890 193 612 8 × 2 = 1 + 0.999 969 417 743 780 387 225 6;
  • 44) 0.999 969 417 743 780 387 225 6 × 2 = 1 + 0.999 938 835 487 560 774 451 2;
  • 45) 0.999 938 835 487 560 774 451 2 × 2 = 1 + 0.999 877 670 975 121 548 902 4;
  • 46) 0.999 877 670 975 121 548 902 4 × 2 = 1 + 0.999 755 341 950 243 097 804 8;
  • 47) 0.999 755 341 950 243 097 804 8 × 2 = 1 + 0.999 510 683 900 486 195 609 6;
  • 48) 0.999 510 683 900 486 195 609 6 × 2 = 1 + 0.999 021 367 800 972 391 219 2;
  • 49) 0.999 021 367 800 972 391 219 2 × 2 = 1 + 0.998 042 735 601 944 782 438 4;
  • 50) 0.998 042 735 601 944 782 438 4 × 2 = 1 + 0.996 085 471 203 889 564 876 8;
  • 51) 0.996 085 471 203 889 564 876 8 × 2 = 1 + 0.992 170 942 407 779 129 753 6;
  • 52) 0.992 170 942 407 779 129 753 6 × 2 = 1 + 0.984 341 884 815 558 259 507 2;
  • 53) 0.984 341 884 815 558 259 507 2 × 2 = 1 + 0.968 683 769 631 116 519 014 4;
  • 54) 0.968 683 769 631 116 519 014 4 × 2 = 1 + 0.937 367 539 262 233 038 028 8;
  • 55) 0.937 367 539 262 233 038 028 8 × 2 = 1 + 0.874 735 078 524 466 076 057 6;
  • 56) 0.874 735 078 524 466 076 057 6 × 2 = 1 + 0.749 470 157 048 932 152 115 2;
  • 57) 0.749 470 157 048 932 152 115 2 × 2 = 1 + 0.498 940 314 097 864 304 230 4;
  • 58) 0.498 940 314 097 864 304 230 4 × 2 = 0 + 0.997 880 628 195 728 608 460 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:


0.016 738 891 601 562 496 523 2(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 523 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:


0.016 738 891 601 562 496 523 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 -0.016 738 891 601 562 496 523 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