-0.000 282 005 943 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 943 4(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
-0.000 282 005 943 4(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:

|-0.000 282 005 943 4| = 0.000 282 005 943 4


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: 0.000 282 005 943 4.

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) 0.000 282 005 943 4 × 2 = 0 + 0.000 564 011 886 8;
  • 2) 0.000 564 011 886 8 × 2 = 0 + 0.001 128 023 773 6;
  • 3) 0.001 128 023 773 6 × 2 = 0 + 0.002 256 047 547 2;
  • 4) 0.002 256 047 547 2 × 2 = 0 + 0.004 512 095 094 4;
  • 5) 0.004 512 095 094 4 × 2 = 0 + 0.009 024 190 188 8;
  • 6) 0.009 024 190 188 8 × 2 = 0 + 0.018 048 380 377 6;
  • 7) 0.018 048 380 377 6 × 2 = 0 + 0.036 096 760 755 2;
  • 8) 0.036 096 760 755 2 × 2 = 0 + 0.072 193 521 510 4;
  • 9) 0.072 193 521 510 4 × 2 = 0 + 0.144 387 043 020 8;
  • 10) 0.144 387 043 020 8 × 2 = 0 + 0.288 774 086 041 6;
  • 11) 0.288 774 086 041 6 × 2 = 0 + 0.577 548 172 083 2;
  • 12) 0.577 548 172 083 2 × 2 = 1 + 0.155 096 344 166 4;
  • 13) 0.155 096 344 166 4 × 2 = 0 + 0.310 192 688 332 8;
  • 14) 0.310 192 688 332 8 × 2 = 0 + 0.620 385 376 665 6;
  • 15) 0.620 385 376 665 6 × 2 = 1 + 0.240 770 753 331 2;
  • 16) 0.240 770 753 331 2 × 2 = 0 + 0.481 541 506 662 4;
  • 17) 0.481 541 506 662 4 × 2 = 0 + 0.963 083 013 324 8;
  • 18) 0.963 083 013 324 8 × 2 = 1 + 0.926 166 026 649 6;
  • 19) 0.926 166 026 649 6 × 2 = 1 + 0.852 332 053 299 2;
  • 20) 0.852 332 053 299 2 × 2 = 1 + 0.704 664 106 598 4;
  • 21) 0.704 664 106 598 4 × 2 = 1 + 0.409 328 213 196 8;
  • 22) 0.409 328 213 196 8 × 2 = 0 + 0.818 656 426 393 6;
  • 23) 0.818 656 426 393 6 × 2 = 1 + 0.637 312 852 787 2;
  • 24) 0.637 312 852 787 2 × 2 = 1 + 0.274 625 705 574 4;
  • 25) 0.274 625 705 574 4 × 2 = 0 + 0.549 251 411 148 8;
  • 26) 0.549 251 411 148 8 × 2 = 1 + 0.098 502 822 297 6;
  • 27) 0.098 502 822 297 6 × 2 = 0 + 0.197 005 644 595 2;
  • 28) 0.197 005 644 595 2 × 2 = 0 + 0.394 011 289 190 4;
  • 29) 0.394 011 289 190 4 × 2 = 0 + 0.788 022 578 380 8;
  • 30) 0.788 022 578 380 8 × 2 = 1 + 0.576 045 156 761 6;
  • 31) 0.576 045 156 761 6 × 2 = 1 + 0.152 090 313 523 2;
  • 32) 0.152 090 313 523 2 × 2 = 0 + 0.304 180 627 046 4;
  • 33) 0.304 180 627 046 4 × 2 = 0 + 0.608 361 254 092 8;
  • 34) 0.608 361 254 092 8 × 2 = 1 + 0.216 722 508 185 6;
  • 35) 0.216 722 508 185 6 × 2 = 0 + 0.433 445 016 371 2;
  • 36) 0.433 445 016 371 2 × 2 = 0 + 0.866 890 032 742 4;
  • 37) 0.866 890 032 742 4 × 2 = 1 + 0.733 780 065 484 8;
  • 38) 0.733 780 065 484 8 × 2 = 1 + 0.467 560 130 969 6;
  • 39) 0.467 560 130 969 6 × 2 = 0 + 0.935 120 261 939 2;
  • 40) 0.935 120 261 939 2 × 2 = 1 + 0.870 240 523 878 4;
  • 41) 0.870 240 523 878 4 × 2 = 1 + 0.740 481 047 756 8;
  • 42) 0.740 481 047 756 8 × 2 = 1 + 0.480 962 095 513 6;
  • 43) 0.480 962 095 513 6 × 2 = 0 + 0.961 924 191 027 2;
  • 44) 0.961 924 191 027 2 × 2 = 1 + 0.923 848 382 054 4;
  • 45) 0.923 848 382 054 4 × 2 = 1 + 0.847 696 764 108 8;
  • 46) 0.847 696 764 108 8 × 2 = 1 + 0.695 393 528 217 6;
  • 47) 0.695 393 528 217 6 × 2 = 1 + 0.390 787 056 435 2;
  • 48) 0.390 787 056 435 2 × 2 = 0 + 0.781 574 112 870 4;
  • 49) 0.781 574 112 870 4 × 2 = 1 + 0.563 148 225 740 8;
  • 50) 0.563 148 225 740 8 × 2 = 1 + 0.126 296 451 481 6;
  • 51) 0.126 296 451 481 6 × 2 = 0 + 0.252 592 902 963 2;
  • 52) 0.252 592 902 963 2 × 2 = 0 + 0.505 185 805 926 4;
  • 53) 0.505 185 805 926 4 × 2 = 1 + 0.010 371 611 852 8;
  • 54) 0.010 371 611 852 8 × 2 = 0 + 0.020 743 223 705 6;
  • 55) 0.020 743 223 705 6 × 2 = 0 + 0.041 486 447 411 2;
  • 56) 0.041 486 447 411 2 × 2 = 0 + 0.082 972 894 822 4;
  • 57) 0.082 972 894 822 4 × 2 = 0 + 0.165 945 789 644 8;
  • 58) 0.165 945 789 644 8 × 2 = 0 + 0.331 891 579 289 6;
  • 59) 0.331 891 579 289 6 × 2 = 0 + 0.663 783 158 579 2;
  • 60) 0.663 783 158 579 2 × 2 = 1 + 0.327 566 317 158 4;
  • 61) 0.327 566 317 158 4 × 2 = 0 + 0.655 132 634 316 8;
  • 62) 0.655 132 634 316 8 × 2 = 1 + 0.310 265 268 633 6;
  • 63) 0.310 265 268 633 6 × 2 = 0 + 0.620 530 537 267 2;
  • 64) 0.620 530 537 267 2 × 2 = 1 + 0.241 061 074 534 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:


0.000 282 005 943 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 0101(2)

6. Positive number before normalization:

0.000 282 005 943 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 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:


0.000 282 005 943 4(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 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 0100 1101 1101 1110 1100 1000 0001 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 0100 1101 1101 1110 1100 1000 0001 0101 =

0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 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 0100 1101 1101 1110 1100 1000 0001 0101


Decimal number -0.000 282 005 943 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0100 1101 1101 1110 1100 1000 0001 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