-0.000 282 005 916 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 916 6(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
-0.000 282 005 916 6(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:

|-0.000 282 005 916 6| = 0.000 282 005 916 6


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: 0.000 282 005 916 6.

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) 0.000 282 005 916 6 × 2 = 0 + 0.000 564 011 833 2;
  • 2) 0.000 564 011 833 2 × 2 = 0 + 0.001 128 023 666 4;
  • 3) 0.001 128 023 666 4 × 2 = 0 + 0.002 256 047 332 8;
  • 4) 0.002 256 047 332 8 × 2 = 0 + 0.004 512 094 665 6;
  • 5) 0.004 512 094 665 6 × 2 = 0 + 0.009 024 189 331 2;
  • 6) 0.009 024 189 331 2 × 2 = 0 + 0.018 048 378 662 4;
  • 7) 0.018 048 378 662 4 × 2 = 0 + 0.036 096 757 324 8;
  • 8) 0.036 096 757 324 8 × 2 = 0 + 0.072 193 514 649 6;
  • 9) 0.072 193 514 649 6 × 2 = 0 + 0.144 387 029 299 2;
  • 10) 0.144 387 029 299 2 × 2 = 0 + 0.288 774 058 598 4;
  • 11) 0.288 774 058 598 4 × 2 = 0 + 0.577 548 117 196 8;
  • 12) 0.577 548 117 196 8 × 2 = 1 + 0.155 096 234 393 6;
  • 13) 0.155 096 234 393 6 × 2 = 0 + 0.310 192 468 787 2;
  • 14) 0.310 192 468 787 2 × 2 = 0 + 0.620 384 937 574 4;
  • 15) 0.620 384 937 574 4 × 2 = 1 + 0.240 769 875 148 8;
  • 16) 0.240 769 875 148 8 × 2 = 0 + 0.481 539 750 297 6;
  • 17) 0.481 539 750 297 6 × 2 = 0 + 0.963 079 500 595 2;
  • 18) 0.963 079 500 595 2 × 2 = 1 + 0.926 159 001 190 4;
  • 19) 0.926 159 001 190 4 × 2 = 1 + 0.852 318 002 380 8;
  • 20) 0.852 318 002 380 8 × 2 = 1 + 0.704 636 004 761 6;
  • 21) 0.704 636 004 761 6 × 2 = 1 + 0.409 272 009 523 2;
  • 22) 0.409 272 009 523 2 × 2 = 0 + 0.818 544 019 046 4;
  • 23) 0.818 544 019 046 4 × 2 = 1 + 0.637 088 038 092 8;
  • 24) 0.637 088 038 092 8 × 2 = 1 + 0.274 176 076 185 6;
  • 25) 0.274 176 076 185 6 × 2 = 0 + 0.548 352 152 371 2;
  • 26) 0.548 352 152 371 2 × 2 = 1 + 0.096 704 304 742 4;
  • 27) 0.096 704 304 742 4 × 2 = 0 + 0.193 408 609 484 8;
  • 28) 0.193 408 609 484 8 × 2 = 0 + 0.386 817 218 969 6;
  • 29) 0.386 817 218 969 6 × 2 = 0 + 0.773 634 437 939 2;
  • 30) 0.773 634 437 939 2 × 2 = 1 + 0.547 268 875 878 4;
  • 31) 0.547 268 875 878 4 × 2 = 1 + 0.094 537 751 756 8;
  • 32) 0.094 537 751 756 8 × 2 = 0 + 0.189 075 503 513 6;
  • 33) 0.189 075 503 513 6 × 2 = 0 + 0.378 151 007 027 2;
  • 34) 0.378 151 007 027 2 × 2 = 0 + 0.756 302 014 054 4;
  • 35) 0.756 302 014 054 4 × 2 = 1 + 0.512 604 028 108 8;
  • 36) 0.512 604 028 108 8 × 2 = 1 + 0.025 208 056 217 6;
  • 37) 0.025 208 056 217 6 × 2 = 0 + 0.050 416 112 435 2;
  • 38) 0.050 416 112 435 2 × 2 = 0 + 0.100 832 224 870 4;
  • 39) 0.100 832 224 870 4 × 2 = 0 + 0.201 664 449 740 8;
  • 40) 0.201 664 449 740 8 × 2 = 0 + 0.403 328 899 481 6;
  • 41) 0.403 328 899 481 6 × 2 = 0 + 0.806 657 798 963 2;
  • 42) 0.806 657 798 963 2 × 2 = 1 + 0.613 315 597 926 4;
  • 43) 0.613 315 597 926 4 × 2 = 1 + 0.226 631 195 852 8;
  • 44) 0.226 631 195 852 8 × 2 = 0 + 0.453 262 391 705 6;
  • 45) 0.453 262 391 705 6 × 2 = 0 + 0.906 524 783 411 2;
  • 46) 0.906 524 783 411 2 × 2 = 1 + 0.813 049 566 822 4;
  • 47) 0.813 049 566 822 4 × 2 = 1 + 0.626 099 133 644 8;
  • 48) 0.626 099 133 644 8 × 2 = 1 + 0.252 198 267 289 6;
  • 49) 0.252 198 267 289 6 × 2 = 0 + 0.504 396 534 579 2;
  • 50) 0.504 396 534 579 2 × 2 = 1 + 0.008 793 069 158 4;
  • 51) 0.008 793 069 158 4 × 2 = 0 + 0.017 586 138 316 8;
  • 52) 0.017 586 138 316 8 × 2 = 0 + 0.035 172 276 633 6;
  • 53) 0.035 172 276 633 6 × 2 = 0 + 0.070 344 553 267 2;
  • 54) 0.070 344 553 267 2 × 2 = 0 + 0.140 689 106 534 4;
  • 55) 0.140 689 106 534 4 × 2 = 0 + 0.281 378 213 068 8;
  • 56) 0.281 378 213 068 8 × 2 = 0 + 0.562 756 426 137 6;
  • 57) 0.562 756 426 137 6 × 2 = 1 + 0.125 512 852 275 2;
  • 58) 0.125 512 852 275 2 × 2 = 0 + 0.251 025 704 550 4;
  • 59) 0.251 025 704 550 4 × 2 = 0 + 0.502 051 409 100 8;
  • 60) 0.502 051 409 100 8 × 2 = 1 + 0.004 102 818 201 6;
  • 61) 0.004 102 818 201 6 × 2 = 0 + 0.008 205 636 403 2;
  • 62) 0.008 205 636 403 2 × 2 = 0 + 0.016 411 272 806 4;
  • 63) 0.016 411 272 806 4 × 2 = 0 + 0.032 822 545 612 8;
  • 64) 0.032 822 545 612 8 × 2 = 0 + 0.065 645 091 225 6;

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:


0.000 282 005 916 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000(2)

6. Positive number before normalization:

0.000 282 005 916 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000(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:


0.000 282 005 916 6(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000(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 0011 0000 0110 0111 0100 0000 1001 0000


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 0011 0000 0110 0111 0100 0000 1001 0000 =

0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000


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 0011 0000 0110 0111 0100 0000 1001 0000


Decimal number -0.000 282 005 916 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0000 0110 0111 0100 0000 1001 0000

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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