-0.000 282 005 954 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 954(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
-0.000 282 005 954(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:

|-0.000 282 005 954| = 0.000 282 005 954


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: 0.000 282 005 954.

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) 0.000 282 005 954 × 2 = 0 + 0.000 564 011 908;
  • 2) 0.000 564 011 908 × 2 = 0 + 0.001 128 023 816;
  • 3) 0.001 128 023 816 × 2 = 0 + 0.002 256 047 632;
  • 4) 0.002 256 047 632 × 2 = 0 + 0.004 512 095 264;
  • 5) 0.004 512 095 264 × 2 = 0 + 0.009 024 190 528;
  • 6) 0.009 024 190 528 × 2 = 0 + 0.018 048 381 056;
  • 7) 0.018 048 381 056 × 2 = 0 + 0.036 096 762 112;
  • 8) 0.036 096 762 112 × 2 = 0 + 0.072 193 524 224;
  • 9) 0.072 193 524 224 × 2 = 0 + 0.144 387 048 448;
  • 10) 0.144 387 048 448 × 2 = 0 + 0.288 774 096 896;
  • 11) 0.288 774 096 896 × 2 = 0 + 0.577 548 193 792;
  • 12) 0.577 548 193 792 × 2 = 1 + 0.155 096 387 584;
  • 13) 0.155 096 387 584 × 2 = 0 + 0.310 192 775 168;
  • 14) 0.310 192 775 168 × 2 = 0 + 0.620 385 550 336;
  • 15) 0.620 385 550 336 × 2 = 1 + 0.240 771 100 672;
  • 16) 0.240 771 100 672 × 2 = 0 + 0.481 542 201 344;
  • 17) 0.481 542 201 344 × 2 = 0 + 0.963 084 402 688;
  • 18) 0.963 084 402 688 × 2 = 1 + 0.926 168 805 376;
  • 19) 0.926 168 805 376 × 2 = 1 + 0.852 337 610 752;
  • 20) 0.852 337 610 752 × 2 = 1 + 0.704 675 221 504;
  • 21) 0.704 675 221 504 × 2 = 1 + 0.409 350 443 008;
  • 22) 0.409 350 443 008 × 2 = 0 + 0.818 700 886 016;
  • 23) 0.818 700 886 016 × 2 = 1 + 0.637 401 772 032;
  • 24) 0.637 401 772 032 × 2 = 1 + 0.274 803 544 064;
  • 25) 0.274 803 544 064 × 2 = 0 + 0.549 607 088 128;
  • 26) 0.549 607 088 128 × 2 = 1 + 0.099 214 176 256;
  • 27) 0.099 214 176 256 × 2 = 0 + 0.198 428 352 512;
  • 28) 0.198 428 352 512 × 2 = 0 + 0.396 856 705 024;
  • 29) 0.396 856 705 024 × 2 = 0 + 0.793 713 410 048;
  • 30) 0.793 713 410 048 × 2 = 1 + 0.587 426 820 096;
  • 31) 0.587 426 820 096 × 2 = 1 + 0.174 853 640 192;
  • 32) 0.174 853 640 192 × 2 = 0 + 0.349 707 280 384;
  • 33) 0.349 707 280 384 × 2 = 0 + 0.699 414 560 768;
  • 34) 0.699 414 560 768 × 2 = 1 + 0.398 829 121 536;
  • 35) 0.398 829 121 536 × 2 = 0 + 0.797 658 243 072;
  • 36) 0.797 658 243 072 × 2 = 1 + 0.595 316 486 144;
  • 37) 0.595 316 486 144 × 2 = 1 + 0.190 632 972 288;
  • 38) 0.190 632 972 288 × 2 = 0 + 0.381 265 944 576;
  • 39) 0.381 265 944 576 × 2 = 0 + 0.762 531 889 152;
  • 40) 0.762 531 889 152 × 2 = 1 + 0.525 063 778 304;
  • 41) 0.525 063 778 304 × 2 = 1 + 0.050 127 556 608;
  • 42) 0.050 127 556 608 × 2 = 0 + 0.100 255 113 216;
  • 43) 0.100 255 113 216 × 2 = 0 + 0.200 510 226 432;
  • 44) 0.200 510 226 432 × 2 = 0 + 0.401 020 452 864;
  • 45) 0.401 020 452 864 × 2 = 0 + 0.802 040 905 728;
  • 46) 0.802 040 905 728 × 2 = 1 + 0.604 081 811 456;
  • 47) 0.604 081 811 456 × 2 = 1 + 0.208 163 622 912;
  • 48) 0.208 163 622 912 × 2 = 0 + 0.416 327 245 824;
  • 49) 0.416 327 245 824 × 2 = 0 + 0.832 654 491 648;
  • 50) 0.832 654 491 648 × 2 = 1 + 0.665 308 983 296;
  • 51) 0.665 308 983 296 × 2 = 1 + 0.330 617 966 592;
  • 52) 0.330 617 966 592 × 2 = 0 + 0.661 235 933 184;
  • 53) 0.661 235 933 184 × 2 = 1 + 0.322 471 866 368;
  • 54) 0.322 471 866 368 × 2 = 0 + 0.644 943 732 736;
  • 55) 0.644 943 732 736 × 2 = 1 + 0.289 887 465 472;
  • 56) 0.289 887 465 472 × 2 = 0 + 0.579 774 930 944;
  • 57) 0.579 774 930 944 × 2 = 1 + 0.159 549 861 888;
  • 58) 0.159 549 861 888 × 2 = 0 + 0.319 099 723 776;
  • 59) 0.319 099 723 776 × 2 = 0 + 0.638 199 447 552;
  • 60) 0.638 199 447 552 × 2 = 1 + 0.276 398 895 104;
  • 61) 0.276 398 895 104 × 2 = 0 + 0.552 797 790 208;
  • 62) 0.552 797 790 208 × 2 = 1 + 0.105 595 580 416;
  • 63) 0.105 595 580 416 × 2 = 0 + 0.211 191 160 832;
  • 64) 0.211 191 160 832 × 2 = 0 + 0.422 382 321 664;

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:


0.000 282 005 954(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100(2)

6. Positive number before normalization:

0.000 282 005 954(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100(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:


0.000 282 005 954(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100(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 0101 1001 1000 0110 0110 1010 1001 0100


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 0101 1001 1000 0110 0110 1010 1001 0100 =

0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100


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 0101 1001 1000 0110 0110 1010 1001 0100


Decimal number -0.000 282 005 954 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1001 1000 0110 0110 1010 1001 0100

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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