-0.016 738 891 601 562 496 455 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 455(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
-0.016 738 891 601 562 496 455(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:

|-0.016 738 891 601 562 496 455| = 0.016 738 891 601 562 496 455


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: 0.016 738 891 601 562 496 455.

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) 0.016 738 891 601 562 496 455 × 2 = 0 + 0.033 477 783 203 124 992 91;
  • 2) 0.033 477 783 203 124 992 91 × 2 = 0 + 0.066 955 566 406 249 985 82;
  • 3) 0.066 955 566 406 249 985 82 × 2 = 0 + 0.133 911 132 812 499 971 64;
  • 4) 0.133 911 132 812 499 971 64 × 2 = 0 + 0.267 822 265 624 999 943 28;
  • 5) 0.267 822 265 624 999 943 28 × 2 = 0 + 0.535 644 531 249 999 886 56;
  • 6) 0.535 644 531 249 999 886 56 × 2 = 1 + 0.071 289 062 499 999 773 12;
  • 7) 0.071 289 062 499 999 773 12 × 2 = 0 + 0.142 578 124 999 999 546 24;
  • 8) 0.142 578 124 999 999 546 24 × 2 = 0 + 0.285 156 249 999 999 092 48;
  • 9) 0.285 156 249 999 999 092 48 × 2 = 0 + 0.570 312 499 999 998 184 96;
  • 10) 0.570 312 499 999 998 184 96 × 2 = 1 + 0.140 624 999 999 996 369 92;
  • 11) 0.140 624 999 999 996 369 92 × 2 = 0 + 0.281 249 999 999 992 739 84;
  • 12) 0.281 249 999 999 992 739 84 × 2 = 0 + 0.562 499 999 999 985 479 68;
  • 13) 0.562 499 999 999 985 479 68 × 2 = 1 + 0.124 999 999 999 970 959 36;
  • 14) 0.124 999 999 999 970 959 36 × 2 = 0 + 0.249 999 999 999 941 918 72;
  • 15) 0.249 999 999 999 941 918 72 × 2 = 0 + 0.499 999 999 999 883 837 44;
  • 16) 0.499 999 999 999 883 837 44 × 2 = 0 + 0.999 999 999 999 767 674 88;
  • 17) 0.999 999 999 999 767 674 88 × 2 = 1 + 0.999 999 999 999 535 349 76;
  • 18) 0.999 999 999 999 535 349 76 × 2 = 1 + 0.999 999 999 999 070 699 52;
  • 19) 0.999 999 999 999 070 699 52 × 2 = 1 + 0.999 999 999 998 141 399 04;
  • 20) 0.999 999 999 998 141 399 04 × 2 = 1 + 0.999 999 999 996 282 798 08;
  • 21) 0.999 999 999 996 282 798 08 × 2 = 1 + 0.999 999 999 992 565 596 16;
  • 22) 0.999 999 999 992 565 596 16 × 2 = 1 + 0.999 999 999 985 131 192 32;
  • 23) 0.999 999 999 985 131 192 32 × 2 = 1 + 0.999 999 999 970 262 384 64;
  • 24) 0.999 999 999 970 262 384 64 × 2 = 1 + 0.999 999 999 940 524 769 28;
  • 25) 0.999 999 999 940 524 769 28 × 2 = 1 + 0.999 999 999 881 049 538 56;
  • 26) 0.999 999 999 881 049 538 56 × 2 = 1 + 0.999 999 999 762 099 077 12;
  • 27) 0.999 999 999 762 099 077 12 × 2 = 1 + 0.999 999 999 524 198 154 24;
  • 28) 0.999 999 999 524 198 154 24 × 2 = 1 + 0.999 999 999 048 396 308 48;
  • 29) 0.999 999 999 048 396 308 48 × 2 = 1 + 0.999 999 998 096 792 616 96;
  • 30) 0.999 999 998 096 792 616 96 × 2 = 1 + 0.999 999 996 193 585 233 92;
  • 31) 0.999 999 996 193 585 233 92 × 2 = 1 + 0.999 999 992 387 170 467 84;
  • 32) 0.999 999 992 387 170 467 84 × 2 = 1 + 0.999 999 984 774 340 935 68;
  • 33) 0.999 999 984 774 340 935 68 × 2 = 1 + 0.999 999 969 548 681 871 36;
  • 34) 0.999 999 969 548 681 871 36 × 2 = 1 + 0.999 999 939 097 363 742 72;
  • 35) 0.999 999 939 097 363 742 72 × 2 = 1 + 0.999 999 878 194 727 485 44;
  • 36) 0.999 999 878 194 727 485 44 × 2 = 1 + 0.999 999 756 389 454 970 88;
  • 37) 0.999 999 756 389 454 970 88 × 2 = 1 + 0.999 999 512 778 909 941 76;
  • 38) 0.999 999 512 778 909 941 76 × 2 = 1 + 0.999 999 025 557 819 883 52;
  • 39) 0.999 999 025 557 819 883 52 × 2 = 1 + 0.999 998 051 115 639 767 04;
  • 40) 0.999 998 051 115 639 767 04 × 2 = 1 + 0.999 996 102 231 279 534 08;
  • 41) 0.999 996 102 231 279 534 08 × 2 = 1 + 0.999 992 204 462 559 068 16;
  • 42) 0.999 992 204 462 559 068 16 × 2 = 1 + 0.999 984 408 925 118 136 32;
  • 43) 0.999 984 408 925 118 136 32 × 2 = 1 + 0.999 968 817 850 236 272 64;
  • 44) 0.999 968 817 850 236 272 64 × 2 = 1 + 0.999 937 635 700 472 545 28;
  • 45) 0.999 937 635 700 472 545 28 × 2 = 1 + 0.999 875 271 400 945 090 56;
  • 46) 0.999 875 271 400 945 090 56 × 2 = 1 + 0.999 750 542 801 890 181 12;
  • 47) 0.999 750 542 801 890 181 12 × 2 = 1 + 0.999 501 085 603 780 362 24;
  • 48) 0.999 501 085 603 780 362 24 × 2 = 1 + 0.999 002 171 207 560 724 48;
  • 49) 0.999 002 171 207 560 724 48 × 2 = 1 + 0.998 004 342 415 121 448 96;
  • 50) 0.998 004 342 415 121 448 96 × 2 = 1 + 0.996 008 684 830 242 897 92;
  • 51) 0.996 008 684 830 242 897 92 × 2 = 1 + 0.992 017 369 660 485 795 84;
  • 52) 0.992 017 369 660 485 795 84 × 2 = 1 + 0.984 034 739 320 971 591 68;
  • 53) 0.984 034 739 320 971 591 68 × 2 = 1 + 0.968 069 478 641 943 183 36;
  • 54) 0.968 069 478 641 943 183 36 × 2 = 1 + 0.936 138 957 283 886 366 72;
  • 55) 0.936 138 957 283 886 366 72 × 2 = 1 + 0.872 277 914 567 772 733 44;
  • 56) 0.872 277 914 567 772 733 44 × 2 = 1 + 0.744 555 829 135 545 466 88;
  • 57) 0.744 555 829 135 545 466 88 × 2 = 1 + 0.489 111 658 271 090 933 76;
  • 58) 0.489 111 658 271 090 933 76 × 2 = 0 + 0.978 223 316 542 181 867 52;

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:


0.016 738 891 601 562 496 455(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 455(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:


0.016 738 891 601 562 496 455(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 -0.016 738 891 601 562 496 455 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