-0.000 282 005 871 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 871 3(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
-0.000 282 005 871 3(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:

|-0.000 282 005 871 3| = 0.000 282 005 871 3


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: 0.000 282 005 871 3.

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) 0.000 282 005 871 3 × 2 = 0 + 0.000 564 011 742 6;
  • 2) 0.000 564 011 742 6 × 2 = 0 + 0.001 128 023 485 2;
  • 3) 0.001 128 023 485 2 × 2 = 0 + 0.002 256 046 970 4;
  • 4) 0.002 256 046 970 4 × 2 = 0 + 0.004 512 093 940 8;
  • 5) 0.004 512 093 940 8 × 2 = 0 + 0.009 024 187 881 6;
  • 6) 0.009 024 187 881 6 × 2 = 0 + 0.018 048 375 763 2;
  • 7) 0.018 048 375 763 2 × 2 = 0 + 0.036 096 751 526 4;
  • 8) 0.036 096 751 526 4 × 2 = 0 + 0.072 193 503 052 8;
  • 9) 0.072 193 503 052 8 × 2 = 0 + 0.144 387 006 105 6;
  • 10) 0.144 387 006 105 6 × 2 = 0 + 0.288 774 012 211 2;
  • 11) 0.288 774 012 211 2 × 2 = 0 + 0.577 548 024 422 4;
  • 12) 0.577 548 024 422 4 × 2 = 1 + 0.155 096 048 844 8;
  • 13) 0.155 096 048 844 8 × 2 = 0 + 0.310 192 097 689 6;
  • 14) 0.310 192 097 689 6 × 2 = 0 + 0.620 384 195 379 2;
  • 15) 0.620 384 195 379 2 × 2 = 1 + 0.240 768 390 758 4;
  • 16) 0.240 768 390 758 4 × 2 = 0 + 0.481 536 781 516 8;
  • 17) 0.481 536 781 516 8 × 2 = 0 + 0.963 073 563 033 6;
  • 18) 0.963 073 563 033 6 × 2 = 1 + 0.926 147 126 067 2;
  • 19) 0.926 147 126 067 2 × 2 = 1 + 0.852 294 252 134 4;
  • 20) 0.852 294 252 134 4 × 2 = 1 + 0.704 588 504 268 8;
  • 21) 0.704 588 504 268 8 × 2 = 1 + 0.409 177 008 537 6;
  • 22) 0.409 177 008 537 6 × 2 = 0 + 0.818 354 017 075 2;
  • 23) 0.818 354 017 075 2 × 2 = 1 + 0.636 708 034 150 4;
  • 24) 0.636 708 034 150 4 × 2 = 1 + 0.273 416 068 300 8;
  • 25) 0.273 416 068 300 8 × 2 = 0 + 0.546 832 136 601 6;
  • 26) 0.546 832 136 601 6 × 2 = 1 + 0.093 664 273 203 2;
  • 27) 0.093 664 273 203 2 × 2 = 0 + 0.187 328 546 406 4;
  • 28) 0.187 328 546 406 4 × 2 = 0 + 0.374 657 092 812 8;
  • 29) 0.374 657 092 812 8 × 2 = 0 + 0.749 314 185 625 6;
  • 30) 0.749 314 185 625 6 × 2 = 1 + 0.498 628 371 251 2;
  • 31) 0.498 628 371 251 2 × 2 = 0 + 0.997 256 742 502 4;
  • 32) 0.997 256 742 502 4 × 2 = 1 + 0.994 513 485 004 8;
  • 33) 0.994 513 485 004 8 × 2 = 1 + 0.989 026 970 009 6;
  • 34) 0.989 026 970 009 6 × 2 = 1 + 0.978 053 940 019 2;
  • 35) 0.978 053 940 019 2 × 2 = 1 + 0.956 107 880 038 4;
  • 36) 0.956 107 880 038 4 × 2 = 1 + 0.912 215 760 076 8;
  • 37) 0.912 215 760 076 8 × 2 = 1 + 0.824 431 520 153 6;
  • 38) 0.824 431 520 153 6 × 2 = 1 + 0.648 863 040 307 2;
  • 39) 0.648 863 040 307 2 × 2 = 1 + 0.297 726 080 614 4;
  • 40) 0.297 726 080 614 4 × 2 = 0 + 0.595 452 161 228 8;
  • 41) 0.595 452 161 228 8 × 2 = 1 + 0.190 904 322 457 6;
  • 42) 0.190 904 322 457 6 × 2 = 0 + 0.381 808 644 915 2;
  • 43) 0.381 808 644 915 2 × 2 = 0 + 0.763 617 289 830 4;
  • 44) 0.763 617 289 830 4 × 2 = 1 + 0.527 234 579 660 8;
  • 45) 0.527 234 579 660 8 × 2 = 1 + 0.054 469 159 321 6;
  • 46) 0.054 469 159 321 6 × 2 = 0 + 0.108 938 318 643 2;
  • 47) 0.108 938 318 643 2 × 2 = 0 + 0.217 876 637 286 4;
  • 48) 0.217 876 637 286 4 × 2 = 0 + 0.435 753 274 572 8;
  • 49) 0.435 753 274 572 8 × 2 = 0 + 0.871 506 549 145 6;
  • 50) 0.871 506 549 145 6 × 2 = 1 + 0.743 013 098 291 2;
  • 51) 0.743 013 098 291 2 × 2 = 1 + 0.486 026 196 582 4;
  • 52) 0.486 026 196 582 4 × 2 = 0 + 0.972 052 393 164 8;
  • 53) 0.972 052 393 164 8 × 2 = 1 + 0.944 104 786 329 6;
  • 54) 0.944 104 786 329 6 × 2 = 1 + 0.888 209 572 659 2;
  • 55) 0.888 209 572 659 2 × 2 = 1 + 0.776 419 145 318 4;
  • 56) 0.776 419 145 318 4 × 2 = 1 + 0.552 838 290 636 8;
  • 57) 0.552 838 290 636 8 × 2 = 1 + 0.105 676 581 273 6;
  • 58) 0.105 676 581 273 6 × 2 = 0 + 0.211 353 162 547 2;
  • 59) 0.211 353 162 547 2 × 2 = 0 + 0.422 706 325 094 4;
  • 60) 0.422 706 325 094 4 × 2 = 0 + 0.845 412 650 188 8;
  • 61) 0.845 412 650 188 8 × 2 = 1 + 0.690 825 300 377 6;
  • 62) 0.690 825 300 377 6 × 2 = 1 + 0.381 650 600 755 2;
  • 63) 0.381 650 600 755 2 × 2 = 0 + 0.763 301 201 510 4;
  • 64) 0.763 301 201 510 4 × 2 = 1 + 0.526 602 403 020 8;

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:


0.000 282 005 871 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101(2)

6. Positive number before normalization:

0.000 282 005 871 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101(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:


0.000 282 005 871 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101(2) × 20 =

1.0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101(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 0101 1111 1110 1001 1000 0110 1111 1000 1101


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 0101 1111 1110 1001 1000 0110 1111 1000 1101 =

0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101


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 0101 1111 1110 1001 1000 0110 1111 1000 1101


Decimal number -0.000 282 005 871 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 1111 1110 1001 1000 0110 1111 1000 1101

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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