-0.000 282 005 898 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 898 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
-0.000 282 005 898 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:

|-0.000 282 005 898 2| = 0.000 282 005 898 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: 0.000 282 005 898 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) 0.000 282 005 898 2 × 2 = 0 + 0.000 564 011 796 4;
  • 2) 0.000 564 011 796 4 × 2 = 0 + 0.001 128 023 592 8;
  • 3) 0.001 128 023 592 8 × 2 = 0 + 0.002 256 047 185 6;
  • 4) 0.002 256 047 185 6 × 2 = 0 + 0.004 512 094 371 2;
  • 5) 0.004 512 094 371 2 × 2 = 0 + 0.009 024 188 742 4;
  • 6) 0.009 024 188 742 4 × 2 = 0 + 0.018 048 377 484 8;
  • 7) 0.018 048 377 484 8 × 2 = 0 + 0.036 096 754 969 6;
  • 8) 0.036 096 754 969 6 × 2 = 0 + 0.072 193 509 939 2;
  • 9) 0.072 193 509 939 2 × 2 = 0 + 0.144 387 019 878 4;
  • 10) 0.144 387 019 878 4 × 2 = 0 + 0.288 774 039 756 8;
  • 11) 0.288 774 039 756 8 × 2 = 0 + 0.577 548 079 513 6;
  • 12) 0.577 548 079 513 6 × 2 = 1 + 0.155 096 159 027 2;
  • 13) 0.155 096 159 027 2 × 2 = 0 + 0.310 192 318 054 4;
  • 14) 0.310 192 318 054 4 × 2 = 0 + 0.620 384 636 108 8;
  • 15) 0.620 384 636 108 8 × 2 = 1 + 0.240 769 272 217 6;
  • 16) 0.240 769 272 217 6 × 2 = 0 + 0.481 538 544 435 2;
  • 17) 0.481 538 544 435 2 × 2 = 0 + 0.963 077 088 870 4;
  • 18) 0.963 077 088 870 4 × 2 = 1 + 0.926 154 177 740 8;
  • 19) 0.926 154 177 740 8 × 2 = 1 + 0.852 308 355 481 6;
  • 20) 0.852 308 355 481 6 × 2 = 1 + 0.704 616 710 963 2;
  • 21) 0.704 616 710 963 2 × 2 = 1 + 0.409 233 421 926 4;
  • 22) 0.409 233 421 926 4 × 2 = 0 + 0.818 466 843 852 8;
  • 23) 0.818 466 843 852 8 × 2 = 1 + 0.636 933 687 705 6;
  • 24) 0.636 933 687 705 6 × 2 = 1 + 0.273 867 375 411 2;
  • 25) 0.273 867 375 411 2 × 2 = 0 + 0.547 734 750 822 4;
  • 26) 0.547 734 750 822 4 × 2 = 1 + 0.095 469 501 644 8;
  • 27) 0.095 469 501 644 8 × 2 = 0 + 0.190 939 003 289 6;
  • 28) 0.190 939 003 289 6 × 2 = 0 + 0.381 878 006 579 2;
  • 29) 0.381 878 006 579 2 × 2 = 0 + 0.763 756 013 158 4;
  • 30) 0.763 756 013 158 4 × 2 = 1 + 0.527 512 026 316 8;
  • 31) 0.527 512 026 316 8 × 2 = 1 + 0.055 024 052 633 6;
  • 32) 0.055 024 052 633 6 × 2 = 0 + 0.110 048 105 267 2;
  • 33) 0.110 048 105 267 2 × 2 = 0 + 0.220 096 210 534 4;
  • 34) 0.220 096 210 534 4 × 2 = 0 + 0.440 192 421 068 8;
  • 35) 0.440 192 421 068 8 × 2 = 0 + 0.880 384 842 137 6;
  • 36) 0.880 384 842 137 6 × 2 = 1 + 0.760 769 684 275 2;
  • 37) 0.760 769 684 275 2 × 2 = 1 + 0.521 539 368 550 4;
  • 38) 0.521 539 368 550 4 × 2 = 1 + 0.043 078 737 100 8;
  • 39) 0.043 078 737 100 8 × 2 = 0 + 0.086 157 474 201 6;
  • 40) 0.086 157 474 201 6 × 2 = 0 + 0.172 314 948 403 2;
  • 41) 0.172 314 948 403 2 × 2 = 0 + 0.344 629 896 806 4;
  • 42) 0.344 629 896 806 4 × 2 = 0 + 0.689 259 793 612 8;
  • 43) 0.689 259 793 612 8 × 2 = 1 + 0.378 519 587 225 6;
  • 44) 0.378 519 587 225 6 × 2 = 0 + 0.757 039 174 451 2;
  • 45) 0.757 039 174 451 2 × 2 = 1 + 0.514 078 348 902 4;
  • 46) 0.514 078 348 902 4 × 2 = 1 + 0.028 156 697 804 8;
  • 47) 0.028 156 697 804 8 × 2 = 0 + 0.056 313 395 609 6;
  • 48) 0.056 313 395 609 6 × 2 = 0 + 0.112 626 791 219 2;
  • 49) 0.112 626 791 219 2 × 2 = 0 + 0.225 253 582 438 4;
  • 50) 0.225 253 582 438 4 × 2 = 0 + 0.450 507 164 876 8;
  • 51) 0.450 507 164 876 8 × 2 = 0 + 0.901 014 329 753 6;
  • 52) 0.901 014 329 753 6 × 2 = 1 + 0.802 028 659 507 2;
  • 53) 0.802 028 659 507 2 × 2 = 1 + 0.604 057 319 014 4;
  • 54) 0.604 057 319 014 4 × 2 = 1 + 0.208 114 638 028 8;
  • 55) 0.208 114 638 028 8 × 2 = 0 + 0.416 229 276 057 6;
  • 56) 0.416 229 276 057 6 × 2 = 0 + 0.832 458 552 115 2;
  • 57) 0.832 458 552 115 2 × 2 = 1 + 0.664 917 104 230 4;
  • 58) 0.664 917 104 230 4 × 2 = 1 + 0.329 834 208 460 8;
  • 59) 0.329 834 208 460 8 × 2 = 0 + 0.659 668 416 921 6;
  • 60) 0.659 668 416 921 6 × 2 = 1 + 0.319 336 833 843 2;
  • 61) 0.319 336 833 843 2 × 2 = 0 + 0.638 673 667 686 4;
  • 62) 0.638 673 667 686 4 × 2 = 1 + 0.277 347 335 372 8;
  • 63) 0.277 347 335 372 8 × 2 = 0 + 0.554 694 670 745 6;
  • 64) 0.554 694 670 745 6 × 2 = 1 + 0.109 389 341 491 2;

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:


0.000 282 005 898 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101(2)

6. Positive number before normalization:

0.000 282 005 898 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101(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:


0.000 282 005 898 2(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101(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 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101


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 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101 =

0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101


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 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101


Decimal number -0.000 282 005 898 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1100 0010 1100 0001 1100 1101 0101

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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