-0.000 281 992 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 992 9(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
-0.000 281 992 9(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:

|-0.000 281 992 9| = 0.000 281 992 9


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: 0.000 281 992 9.

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) 0.000 281 992 9 × 2 = 0 + 0.000 563 985 8;
  • 2) 0.000 563 985 8 × 2 = 0 + 0.001 127 971 6;
  • 3) 0.001 127 971 6 × 2 = 0 + 0.002 255 943 2;
  • 4) 0.002 255 943 2 × 2 = 0 + 0.004 511 886 4;
  • 5) 0.004 511 886 4 × 2 = 0 + 0.009 023 772 8;
  • 6) 0.009 023 772 8 × 2 = 0 + 0.018 047 545 6;
  • 7) 0.018 047 545 6 × 2 = 0 + 0.036 095 091 2;
  • 8) 0.036 095 091 2 × 2 = 0 + 0.072 190 182 4;
  • 9) 0.072 190 182 4 × 2 = 0 + 0.144 380 364 8;
  • 10) 0.144 380 364 8 × 2 = 0 + 0.288 760 729 6;
  • 11) 0.288 760 729 6 × 2 = 0 + 0.577 521 459 2;
  • 12) 0.577 521 459 2 × 2 = 1 + 0.155 042 918 4;
  • 13) 0.155 042 918 4 × 2 = 0 + 0.310 085 836 8;
  • 14) 0.310 085 836 8 × 2 = 0 + 0.620 171 673 6;
  • 15) 0.620 171 673 6 × 2 = 1 + 0.240 343 347 2;
  • 16) 0.240 343 347 2 × 2 = 0 + 0.480 686 694 4;
  • 17) 0.480 686 694 4 × 2 = 0 + 0.961 373 388 8;
  • 18) 0.961 373 388 8 × 2 = 1 + 0.922 746 777 6;
  • 19) 0.922 746 777 6 × 2 = 1 + 0.845 493 555 2;
  • 20) 0.845 493 555 2 × 2 = 1 + 0.690 987 110 4;
  • 21) 0.690 987 110 4 × 2 = 1 + 0.381 974 220 8;
  • 22) 0.381 974 220 8 × 2 = 0 + 0.763 948 441 6;
  • 23) 0.763 948 441 6 × 2 = 1 + 0.527 896 883 2;
  • 24) 0.527 896 883 2 × 2 = 1 + 0.055 793 766 4;
  • 25) 0.055 793 766 4 × 2 = 0 + 0.111 587 532 8;
  • 26) 0.111 587 532 8 × 2 = 0 + 0.223 175 065 6;
  • 27) 0.223 175 065 6 × 2 = 0 + 0.446 350 131 2;
  • 28) 0.446 350 131 2 × 2 = 0 + 0.892 700 262 4;
  • 29) 0.892 700 262 4 × 2 = 1 + 0.785 400 524 8;
  • 30) 0.785 400 524 8 × 2 = 1 + 0.570 801 049 6;
  • 31) 0.570 801 049 6 × 2 = 1 + 0.141 602 099 2;
  • 32) 0.141 602 099 2 × 2 = 0 + 0.283 204 198 4;
  • 33) 0.283 204 198 4 × 2 = 0 + 0.566 408 396 8;
  • 34) 0.566 408 396 8 × 2 = 1 + 0.132 816 793 6;
  • 35) 0.132 816 793 6 × 2 = 0 + 0.265 633 587 2;
  • 36) 0.265 633 587 2 × 2 = 0 + 0.531 267 174 4;
  • 37) 0.531 267 174 4 × 2 = 1 + 0.062 534 348 8;
  • 38) 0.062 534 348 8 × 2 = 0 + 0.125 068 697 6;
  • 39) 0.125 068 697 6 × 2 = 0 + 0.250 137 395 2;
  • 40) 0.250 137 395 2 × 2 = 0 + 0.500 274 790 4;
  • 41) 0.500 274 790 4 × 2 = 1 + 0.000 549 580 8;
  • 42) 0.000 549 580 8 × 2 = 0 + 0.001 099 161 6;
  • 43) 0.001 099 161 6 × 2 = 0 + 0.002 198 323 2;
  • 44) 0.002 198 323 2 × 2 = 0 + 0.004 396 646 4;
  • 45) 0.004 396 646 4 × 2 = 0 + 0.008 793 292 8;
  • 46) 0.008 793 292 8 × 2 = 0 + 0.017 586 585 6;
  • 47) 0.017 586 585 6 × 2 = 0 + 0.035 173 171 2;
  • 48) 0.035 173 171 2 × 2 = 0 + 0.070 346 342 4;
  • 49) 0.070 346 342 4 × 2 = 0 + 0.140 692 684 8;
  • 50) 0.140 692 684 8 × 2 = 0 + 0.281 385 369 6;
  • 51) 0.281 385 369 6 × 2 = 0 + 0.562 770 739 2;
  • 52) 0.562 770 739 2 × 2 = 1 + 0.125 541 478 4;
  • 53) 0.125 541 478 4 × 2 = 0 + 0.251 082 956 8;
  • 54) 0.251 082 956 8 × 2 = 0 + 0.502 165 913 6;
  • 55) 0.502 165 913 6 × 2 = 1 + 0.004 331 827 2;
  • 56) 0.004 331 827 2 × 2 = 0 + 0.008 663 654 4;
  • 57) 0.008 663 654 4 × 2 = 0 + 0.017 327 308 8;
  • 58) 0.017 327 308 8 × 2 = 0 + 0.034 654 617 6;
  • 59) 0.034 654 617 6 × 2 = 0 + 0.069 309 235 2;
  • 60) 0.069 309 235 2 × 2 = 0 + 0.138 618 470 4;
  • 61) 0.138 618 470 4 × 2 = 0 + 0.277 236 940 8;
  • 62) 0.277 236 940 8 × 2 = 0 + 0.554 473 881 6;
  • 63) 0.554 473 881 6 × 2 = 1 + 0.108 947 763 2;
  • 64) 0.108 947 763 2 × 2 = 0 + 0.217 895 526 4;

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:


0.000 281 992 9(10) =

0.0000 0000 0001 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010(2)

6. Positive number before normalization:

0.000 281 992 9(10) =

0.0000 0000 0001 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 281 992 9(10) =

0.0000 0000 0001 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010(2) =

0.0000 0000 0001 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010(2) × 20 =

1.0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010 =

0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010


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 0011

Mantissa (52 bits) =
0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010


Decimal number -0.000 281 992 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0000 1110 0100 1000 1000 0000 0001 0010 0000 0010

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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