-0.000 282 005 887 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 887 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
-0.000 282 005 887 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:

|-0.000 282 005 887 3| = 0.000 282 005 887 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: 0.000 282 005 887 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) 0.000 282 005 887 3 × 2 = 0 + 0.000 564 011 774 6;
  • 2) 0.000 564 011 774 6 × 2 = 0 + 0.001 128 023 549 2;
  • 3) 0.001 128 023 549 2 × 2 = 0 + 0.002 256 047 098 4;
  • 4) 0.002 256 047 098 4 × 2 = 0 + 0.004 512 094 196 8;
  • 5) 0.004 512 094 196 8 × 2 = 0 + 0.009 024 188 393 6;
  • 6) 0.009 024 188 393 6 × 2 = 0 + 0.018 048 376 787 2;
  • 7) 0.018 048 376 787 2 × 2 = 0 + 0.036 096 753 574 4;
  • 8) 0.036 096 753 574 4 × 2 = 0 + 0.072 193 507 148 8;
  • 9) 0.072 193 507 148 8 × 2 = 0 + 0.144 387 014 297 6;
  • 10) 0.144 387 014 297 6 × 2 = 0 + 0.288 774 028 595 2;
  • 11) 0.288 774 028 595 2 × 2 = 0 + 0.577 548 057 190 4;
  • 12) 0.577 548 057 190 4 × 2 = 1 + 0.155 096 114 380 8;
  • 13) 0.155 096 114 380 8 × 2 = 0 + 0.310 192 228 761 6;
  • 14) 0.310 192 228 761 6 × 2 = 0 + 0.620 384 457 523 2;
  • 15) 0.620 384 457 523 2 × 2 = 1 + 0.240 768 915 046 4;
  • 16) 0.240 768 915 046 4 × 2 = 0 + 0.481 537 830 092 8;
  • 17) 0.481 537 830 092 8 × 2 = 0 + 0.963 075 660 185 6;
  • 18) 0.963 075 660 185 6 × 2 = 1 + 0.926 151 320 371 2;
  • 19) 0.926 151 320 371 2 × 2 = 1 + 0.852 302 640 742 4;
  • 20) 0.852 302 640 742 4 × 2 = 1 + 0.704 605 281 484 8;
  • 21) 0.704 605 281 484 8 × 2 = 1 + 0.409 210 562 969 6;
  • 22) 0.409 210 562 969 6 × 2 = 0 + 0.818 421 125 939 2;
  • 23) 0.818 421 125 939 2 × 2 = 1 + 0.636 842 251 878 4;
  • 24) 0.636 842 251 878 4 × 2 = 1 + 0.273 684 503 756 8;
  • 25) 0.273 684 503 756 8 × 2 = 0 + 0.547 369 007 513 6;
  • 26) 0.547 369 007 513 6 × 2 = 1 + 0.094 738 015 027 2;
  • 27) 0.094 738 015 027 2 × 2 = 0 + 0.189 476 030 054 4;
  • 28) 0.189 476 030 054 4 × 2 = 0 + 0.378 952 060 108 8;
  • 29) 0.378 952 060 108 8 × 2 = 0 + 0.757 904 120 217 6;
  • 30) 0.757 904 120 217 6 × 2 = 1 + 0.515 808 240 435 2;
  • 31) 0.515 808 240 435 2 × 2 = 1 + 0.031 616 480 870 4;
  • 32) 0.031 616 480 870 4 × 2 = 0 + 0.063 232 961 740 8;
  • 33) 0.063 232 961 740 8 × 2 = 0 + 0.126 465 923 481 6;
  • 34) 0.126 465 923 481 6 × 2 = 0 + 0.252 931 846 963 2;
  • 35) 0.252 931 846 963 2 × 2 = 0 + 0.505 863 693 926 4;
  • 36) 0.505 863 693 926 4 × 2 = 1 + 0.011 727 387 852 8;
  • 37) 0.011 727 387 852 8 × 2 = 0 + 0.023 454 775 705 6;
  • 38) 0.023 454 775 705 6 × 2 = 0 + 0.046 909 551 411 2;
  • 39) 0.046 909 551 411 2 × 2 = 0 + 0.093 819 102 822 4;
  • 40) 0.093 819 102 822 4 × 2 = 0 + 0.187 638 205 644 8;
  • 41) 0.187 638 205 644 8 × 2 = 0 + 0.375 276 411 289 6;
  • 42) 0.375 276 411 289 6 × 2 = 0 + 0.750 552 822 579 2;
  • 43) 0.750 552 822 579 2 × 2 = 1 + 0.501 105 645 158 4;
  • 44) 0.501 105 645 158 4 × 2 = 1 + 0.002 211 290 316 8;
  • 45) 0.002 211 290 316 8 × 2 = 0 + 0.004 422 580 633 6;
  • 46) 0.004 422 580 633 6 × 2 = 0 + 0.008 845 161 267 2;
  • 47) 0.008 845 161 267 2 × 2 = 0 + 0.017 690 322 534 4;
  • 48) 0.017 690 322 534 4 × 2 = 0 + 0.035 380 645 068 8;
  • 49) 0.035 380 645 068 8 × 2 = 0 + 0.070 761 290 137 6;
  • 50) 0.070 761 290 137 6 × 2 = 0 + 0.141 522 580 275 2;
  • 51) 0.141 522 580 275 2 × 2 = 0 + 0.283 045 160 550 4;
  • 52) 0.283 045 160 550 4 × 2 = 0 + 0.566 090 321 100 8;
  • 53) 0.566 090 321 100 8 × 2 = 1 + 0.132 180 642 201 6;
  • 54) 0.132 180 642 201 6 × 2 = 0 + 0.264 361 284 403 2;
  • 55) 0.264 361 284 403 2 × 2 = 0 + 0.528 722 568 806 4;
  • 56) 0.528 722 568 806 4 × 2 = 1 + 0.057 445 137 612 8;
  • 57) 0.057 445 137 612 8 × 2 = 0 + 0.114 890 275 225 6;
  • 58) 0.114 890 275 225 6 × 2 = 0 + 0.229 780 550 451 2;
  • 59) 0.229 780 550 451 2 × 2 = 0 + 0.459 561 100 902 4;
  • 60) 0.459 561 100 902 4 × 2 = 0 + 0.919 122 201 804 8;
  • 61) 0.919 122 201 804 8 × 2 = 1 + 0.838 244 403 609 6;
  • 62) 0.838 244 403 609 6 × 2 = 1 + 0.676 488 807 219 2;
  • 63) 0.676 488 807 219 2 × 2 = 1 + 0.352 977 614 438 4;
  • 64) 0.352 977 614 438 4 × 2 = 0 + 0.705 955 228 876 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:


0.000 282 005 887 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110(2)

6. Positive number before normalization:

0.000 282 005 887 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110(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:


0.000 282 005 887 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110(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 0000 0011 0000 0000 1001 0000 1110


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 0000 0011 0000 0000 1001 0000 1110 =

0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 1110


Decimal number -0.000 282 005 887 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0000 0011 0000 0000 1001 0000 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