-0.016 738 891 601 562 494 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 53(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
-0.016 738 891 601 562 494 53(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:

|-0.016 738 891 601 562 494 53| = 0.016 738 891 601 562 494 53


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: 0.016 738 891 601 562 494 53.

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) 0.016 738 891 601 562 494 53 × 2 = 0 + 0.033 477 783 203 124 989 06;
  • 2) 0.033 477 783 203 124 989 06 × 2 = 0 + 0.066 955 566 406 249 978 12;
  • 3) 0.066 955 566 406 249 978 12 × 2 = 0 + 0.133 911 132 812 499 956 24;
  • 4) 0.133 911 132 812 499 956 24 × 2 = 0 + 0.267 822 265 624 999 912 48;
  • 5) 0.267 822 265 624 999 912 48 × 2 = 0 + 0.535 644 531 249 999 824 96;
  • 6) 0.535 644 531 249 999 824 96 × 2 = 1 + 0.071 289 062 499 999 649 92;
  • 7) 0.071 289 062 499 999 649 92 × 2 = 0 + 0.142 578 124 999 999 299 84;
  • 8) 0.142 578 124 999 999 299 84 × 2 = 0 + 0.285 156 249 999 998 599 68;
  • 9) 0.285 156 249 999 998 599 68 × 2 = 0 + 0.570 312 499 999 997 199 36;
  • 10) 0.570 312 499 999 997 199 36 × 2 = 1 + 0.140 624 999 999 994 398 72;
  • 11) 0.140 624 999 999 994 398 72 × 2 = 0 + 0.281 249 999 999 988 797 44;
  • 12) 0.281 249 999 999 988 797 44 × 2 = 0 + 0.562 499 999 999 977 594 88;
  • 13) 0.562 499 999 999 977 594 88 × 2 = 1 + 0.124 999 999 999 955 189 76;
  • 14) 0.124 999 999 999 955 189 76 × 2 = 0 + 0.249 999 999 999 910 379 52;
  • 15) 0.249 999 999 999 910 379 52 × 2 = 0 + 0.499 999 999 999 820 759 04;
  • 16) 0.499 999 999 999 820 759 04 × 2 = 0 + 0.999 999 999 999 641 518 08;
  • 17) 0.999 999 999 999 641 518 08 × 2 = 1 + 0.999 999 999 999 283 036 16;
  • 18) 0.999 999 999 999 283 036 16 × 2 = 1 + 0.999 999 999 998 566 072 32;
  • 19) 0.999 999 999 998 566 072 32 × 2 = 1 + 0.999 999 999 997 132 144 64;
  • 20) 0.999 999 999 997 132 144 64 × 2 = 1 + 0.999 999 999 994 264 289 28;
  • 21) 0.999 999 999 994 264 289 28 × 2 = 1 + 0.999 999 999 988 528 578 56;
  • 22) 0.999 999 999 988 528 578 56 × 2 = 1 + 0.999 999 999 977 057 157 12;
  • 23) 0.999 999 999 977 057 157 12 × 2 = 1 + 0.999 999 999 954 114 314 24;
  • 24) 0.999 999 999 954 114 314 24 × 2 = 1 + 0.999 999 999 908 228 628 48;
  • 25) 0.999 999 999 908 228 628 48 × 2 = 1 + 0.999 999 999 816 457 256 96;
  • 26) 0.999 999 999 816 457 256 96 × 2 = 1 + 0.999 999 999 632 914 513 92;
  • 27) 0.999 999 999 632 914 513 92 × 2 = 1 + 0.999 999 999 265 829 027 84;
  • 28) 0.999 999 999 265 829 027 84 × 2 = 1 + 0.999 999 998 531 658 055 68;
  • 29) 0.999 999 998 531 658 055 68 × 2 = 1 + 0.999 999 997 063 316 111 36;
  • 30) 0.999 999 997 063 316 111 36 × 2 = 1 + 0.999 999 994 126 632 222 72;
  • 31) 0.999 999 994 126 632 222 72 × 2 = 1 + 0.999 999 988 253 264 445 44;
  • 32) 0.999 999 988 253 264 445 44 × 2 = 1 + 0.999 999 976 506 528 890 88;
  • 33) 0.999 999 976 506 528 890 88 × 2 = 1 + 0.999 999 953 013 057 781 76;
  • 34) 0.999 999 953 013 057 781 76 × 2 = 1 + 0.999 999 906 026 115 563 52;
  • 35) 0.999 999 906 026 115 563 52 × 2 = 1 + 0.999 999 812 052 231 127 04;
  • 36) 0.999 999 812 052 231 127 04 × 2 = 1 + 0.999 999 624 104 462 254 08;
  • 37) 0.999 999 624 104 462 254 08 × 2 = 1 + 0.999 999 248 208 924 508 16;
  • 38) 0.999 999 248 208 924 508 16 × 2 = 1 + 0.999 998 496 417 849 016 32;
  • 39) 0.999 998 496 417 849 016 32 × 2 = 1 + 0.999 996 992 835 698 032 64;
  • 40) 0.999 996 992 835 698 032 64 × 2 = 1 + 0.999 993 985 671 396 065 28;
  • 41) 0.999 993 985 671 396 065 28 × 2 = 1 + 0.999 987 971 342 792 130 56;
  • 42) 0.999 987 971 342 792 130 56 × 2 = 1 + 0.999 975 942 685 584 261 12;
  • 43) 0.999 975 942 685 584 261 12 × 2 = 1 + 0.999 951 885 371 168 522 24;
  • 44) 0.999 951 885 371 168 522 24 × 2 = 1 + 0.999 903 770 742 337 044 48;
  • 45) 0.999 903 770 742 337 044 48 × 2 = 1 + 0.999 807 541 484 674 088 96;
  • 46) 0.999 807 541 484 674 088 96 × 2 = 1 + 0.999 615 082 969 348 177 92;
  • 47) 0.999 615 082 969 348 177 92 × 2 = 1 + 0.999 230 165 938 696 355 84;
  • 48) 0.999 230 165 938 696 355 84 × 2 = 1 + 0.998 460 331 877 392 711 68;
  • 49) 0.998 460 331 877 392 711 68 × 2 = 1 + 0.996 920 663 754 785 423 36;
  • 50) 0.996 920 663 754 785 423 36 × 2 = 1 + 0.993 841 327 509 570 846 72;
  • 51) 0.993 841 327 509 570 846 72 × 2 = 1 + 0.987 682 655 019 141 693 44;
  • 52) 0.987 682 655 019 141 693 44 × 2 = 1 + 0.975 365 310 038 283 386 88;
  • 53) 0.975 365 310 038 283 386 88 × 2 = 1 + 0.950 730 620 076 566 773 76;
  • 54) 0.950 730 620 076 566 773 76 × 2 = 1 + 0.901 461 240 153 133 547 52;
  • 55) 0.901 461 240 153 133 547 52 × 2 = 1 + 0.802 922 480 306 267 095 04;
  • 56) 0.802 922 480 306 267 095 04 × 2 = 1 + 0.605 844 960 612 534 190 08;
  • 57) 0.605 844 960 612 534 190 08 × 2 = 1 + 0.211 689 921 225 068 380 16;
  • 58) 0.211 689 921 225 068 380 16 × 2 = 0 + 0.423 379 842 450 136 760 32;

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:


0.016 738 891 601 562 494 53(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 494 53(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:


0.016 738 891 601 562 494 53(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 -0.016 738 891 601 562 494 53 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