-0.000 282 005 916 65 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 916 65(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 65(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 65| = 0.000 282 005 916 65


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 65.

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 65 × 2 = 0 + 0.000 564 011 833 3;
  • 2) 0.000 564 011 833 3 × 2 = 0 + 0.001 128 023 666 6;
  • 3) 0.001 128 023 666 6 × 2 = 0 + 0.002 256 047 333 2;
  • 4) 0.002 256 047 333 2 × 2 = 0 + 0.004 512 094 666 4;
  • 5) 0.004 512 094 666 4 × 2 = 0 + 0.009 024 189 332 8;
  • 6) 0.009 024 189 332 8 × 2 = 0 + 0.018 048 378 665 6;
  • 7) 0.018 048 378 665 6 × 2 = 0 + 0.036 096 757 331 2;
  • 8) 0.036 096 757 331 2 × 2 = 0 + 0.072 193 514 662 4;
  • 9) 0.072 193 514 662 4 × 2 = 0 + 0.144 387 029 324 8;
  • 10) 0.144 387 029 324 8 × 2 = 0 + 0.288 774 058 649 6;
  • 11) 0.288 774 058 649 6 × 2 = 0 + 0.577 548 117 299 2;
  • 12) 0.577 548 117 299 2 × 2 = 1 + 0.155 096 234 598 4;
  • 13) 0.155 096 234 598 4 × 2 = 0 + 0.310 192 469 196 8;
  • 14) 0.310 192 469 196 8 × 2 = 0 + 0.620 384 938 393 6;
  • 15) 0.620 384 938 393 6 × 2 = 1 + 0.240 769 876 787 2;
  • 16) 0.240 769 876 787 2 × 2 = 0 + 0.481 539 753 574 4;
  • 17) 0.481 539 753 574 4 × 2 = 0 + 0.963 079 507 148 8;
  • 18) 0.963 079 507 148 8 × 2 = 1 + 0.926 159 014 297 6;
  • 19) 0.926 159 014 297 6 × 2 = 1 + 0.852 318 028 595 2;
  • 20) 0.852 318 028 595 2 × 2 = 1 + 0.704 636 057 190 4;
  • 21) 0.704 636 057 190 4 × 2 = 1 + 0.409 272 114 380 8;
  • 22) 0.409 272 114 380 8 × 2 = 0 + 0.818 544 228 761 6;
  • 23) 0.818 544 228 761 6 × 2 = 1 + 0.637 088 457 523 2;
  • 24) 0.637 088 457 523 2 × 2 = 1 + 0.274 176 915 046 4;
  • 25) 0.274 176 915 046 4 × 2 = 0 + 0.548 353 830 092 8;
  • 26) 0.548 353 830 092 8 × 2 = 1 + 0.096 707 660 185 6;
  • 27) 0.096 707 660 185 6 × 2 = 0 + 0.193 415 320 371 2;
  • 28) 0.193 415 320 371 2 × 2 = 0 + 0.386 830 640 742 4;
  • 29) 0.386 830 640 742 4 × 2 = 0 + 0.773 661 281 484 8;
  • 30) 0.773 661 281 484 8 × 2 = 1 + 0.547 322 562 969 6;
  • 31) 0.547 322 562 969 6 × 2 = 1 + 0.094 645 125 939 2;
  • 32) 0.094 645 125 939 2 × 2 = 0 + 0.189 290 251 878 4;
  • 33) 0.189 290 251 878 4 × 2 = 0 + 0.378 580 503 756 8;
  • 34) 0.378 580 503 756 8 × 2 = 0 + 0.757 161 007 513 6;
  • 35) 0.757 161 007 513 6 × 2 = 1 + 0.514 322 015 027 2;
  • 36) 0.514 322 015 027 2 × 2 = 1 + 0.028 644 030 054 4;
  • 37) 0.028 644 030 054 4 × 2 = 0 + 0.057 288 060 108 8;
  • 38) 0.057 288 060 108 8 × 2 = 0 + 0.114 576 120 217 6;
  • 39) 0.114 576 120 217 6 × 2 = 0 + 0.229 152 240 435 2;
  • 40) 0.229 152 240 435 2 × 2 = 0 + 0.458 304 480 870 4;
  • 41) 0.458 304 480 870 4 × 2 = 0 + 0.916 608 961 740 8;
  • 42) 0.916 608 961 740 8 × 2 = 1 + 0.833 217 923 481 6;
  • 43) 0.833 217 923 481 6 × 2 = 1 + 0.666 435 846 963 2;
  • 44) 0.666 435 846 963 2 × 2 = 1 + 0.332 871 693 926 4;
  • 45) 0.332 871 693 926 4 × 2 = 0 + 0.665 743 387 852 8;
  • 46) 0.665 743 387 852 8 × 2 = 1 + 0.331 486 775 705 6;
  • 47) 0.331 486 775 705 6 × 2 = 0 + 0.662 973 551 411 2;
  • 48) 0.662 973 551 411 2 × 2 = 1 + 0.325 947 102 822 4;
  • 49) 0.325 947 102 822 4 × 2 = 0 + 0.651 894 205 644 8;
  • 50) 0.651 894 205 644 8 × 2 = 1 + 0.303 788 411 289 6;
  • 51) 0.303 788 411 289 6 × 2 = 0 + 0.607 576 822 579 2;
  • 52) 0.607 576 822 579 2 × 2 = 1 + 0.215 153 645 158 4;
  • 53) 0.215 153 645 158 4 × 2 = 0 + 0.430 307 290 316 8;
  • 54) 0.430 307 290 316 8 × 2 = 0 + 0.860 614 580 633 6;
  • 55) 0.860 614 580 633 6 × 2 = 1 + 0.721 229 161 267 2;
  • 56) 0.721 229 161 267 2 × 2 = 1 + 0.442 458 322 534 4;
  • 57) 0.442 458 322 534 4 × 2 = 0 + 0.884 916 645 068 8;
  • 58) 0.884 916 645 068 8 × 2 = 1 + 0.769 833 290 137 6;
  • 59) 0.769 833 290 137 6 × 2 = 1 + 0.539 666 580 275 2;
  • 60) 0.539 666 580 275 2 × 2 = 1 + 0.079 333 160 550 4;
  • 61) 0.079 333 160 550 4 × 2 = 0 + 0.158 666 321 100 8;
  • 62) 0.158 666 321 100 8 × 2 = 0 + 0.317 332 642 201 6;
  • 63) 0.317 332 642 201 6 × 2 = 0 + 0.634 665 284 403 2;
  • 64) 0.634 665 284 403 2 × 2 = 1 + 0.269 330 568 806 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:


0.000 282 005 916 65(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001(2)

6. Positive number before normalization:

0.000 282 005 916 65(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001(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 65(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001(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 0111 0101 0101 0011 0111 0001


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 0111 0101 0101 0011 0111 0001 =

0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001


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 0111 0101 0101 0011 0111 0001


Decimal number -0.000 282 005 916 65 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0000 0111 0101 0101 0011 0111 0001

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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