-0.000 282 005 926 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 926(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
-0.000 282 005 926(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:

|-0.000 282 005 926| = 0.000 282 005 926


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: 0.000 282 005 926.

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) 0.000 282 005 926 × 2 = 0 + 0.000 564 011 852;
  • 2) 0.000 564 011 852 × 2 = 0 + 0.001 128 023 704;
  • 3) 0.001 128 023 704 × 2 = 0 + 0.002 256 047 408;
  • 4) 0.002 256 047 408 × 2 = 0 + 0.004 512 094 816;
  • 5) 0.004 512 094 816 × 2 = 0 + 0.009 024 189 632;
  • 6) 0.009 024 189 632 × 2 = 0 + 0.018 048 379 264;
  • 7) 0.018 048 379 264 × 2 = 0 + 0.036 096 758 528;
  • 8) 0.036 096 758 528 × 2 = 0 + 0.072 193 517 056;
  • 9) 0.072 193 517 056 × 2 = 0 + 0.144 387 034 112;
  • 10) 0.144 387 034 112 × 2 = 0 + 0.288 774 068 224;
  • 11) 0.288 774 068 224 × 2 = 0 + 0.577 548 136 448;
  • 12) 0.577 548 136 448 × 2 = 1 + 0.155 096 272 896;
  • 13) 0.155 096 272 896 × 2 = 0 + 0.310 192 545 792;
  • 14) 0.310 192 545 792 × 2 = 0 + 0.620 385 091 584;
  • 15) 0.620 385 091 584 × 2 = 1 + 0.240 770 183 168;
  • 16) 0.240 770 183 168 × 2 = 0 + 0.481 540 366 336;
  • 17) 0.481 540 366 336 × 2 = 0 + 0.963 080 732 672;
  • 18) 0.963 080 732 672 × 2 = 1 + 0.926 161 465 344;
  • 19) 0.926 161 465 344 × 2 = 1 + 0.852 322 930 688;
  • 20) 0.852 322 930 688 × 2 = 1 + 0.704 645 861 376;
  • 21) 0.704 645 861 376 × 2 = 1 + 0.409 291 722 752;
  • 22) 0.409 291 722 752 × 2 = 0 + 0.818 583 445 504;
  • 23) 0.818 583 445 504 × 2 = 1 + 0.637 166 891 008;
  • 24) 0.637 166 891 008 × 2 = 1 + 0.274 333 782 016;
  • 25) 0.274 333 782 016 × 2 = 0 + 0.548 667 564 032;
  • 26) 0.548 667 564 032 × 2 = 1 + 0.097 335 128 064;
  • 27) 0.097 335 128 064 × 2 = 0 + 0.194 670 256 128;
  • 28) 0.194 670 256 128 × 2 = 0 + 0.389 340 512 256;
  • 29) 0.389 340 512 256 × 2 = 0 + 0.778 681 024 512;
  • 30) 0.778 681 024 512 × 2 = 1 + 0.557 362 049 024;
  • 31) 0.557 362 049 024 × 2 = 1 + 0.114 724 098 048;
  • 32) 0.114 724 098 048 × 2 = 0 + 0.229 448 196 096;
  • 33) 0.229 448 196 096 × 2 = 0 + 0.458 896 392 192;
  • 34) 0.458 896 392 192 × 2 = 0 + 0.917 792 784 384;
  • 35) 0.917 792 784 384 × 2 = 1 + 0.835 585 568 768;
  • 36) 0.835 585 568 768 × 2 = 1 + 0.671 171 137 536;
  • 37) 0.671 171 137 536 × 2 = 1 + 0.342 342 275 072;
  • 38) 0.342 342 275 072 × 2 = 0 + 0.684 684 550 144;
  • 39) 0.684 684 550 144 × 2 = 1 + 0.369 369 100 288;
  • 40) 0.369 369 100 288 × 2 = 0 + 0.738 738 200 576;
  • 41) 0.738 738 200 576 × 2 = 1 + 0.477 476 401 152;
  • 42) 0.477 476 401 152 × 2 = 0 + 0.954 952 802 304;
  • 43) 0.954 952 802 304 × 2 = 1 + 0.909 905 604 608;
  • 44) 0.909 905 604 608 × 2 = 1 + 0.819 811 209 216;
  • 45) 0.819 811 209 216 × 2 = 1 + 0.639 622 418 432;
  • 46) 0.639 622 418 432 × 2 = 1 + 0.279 244 836 864;
  • 47) 0.279 244 836 864 × 2 = 0 + 0.558 489 673 728;
  • 48) 0.558 489 673 728 × 2 = 1 + 0.116 979 347 456;
  • 49) 0.116 979 347 456 × 2 = 0 + 0.233 958 694 912;
  • 50) 0.233 958 694 912 × 2 = 0 + 0.467 917 389 824;
  • 51) 0.467 917 389 824 × 2 = 0 + 0.935 834 779 648;
  • 52) 0.935 834 779 648 × 2 = 1 + 0.871 669 559 296;
  • 53) 0.871 669 559 296 × 2 = 1 + 0.743 339 118 592;
  • 54) 0.743 339 118 592 × 2 = 1 + 0.486 678 237 184;
  • 55) 0.486 678 237 184 × 2 = 0 + 0.973 356 474 368;
  • 56) 0.973 356 474 368 × 2 = 1 + 0.946 712 948 736;
  • 57) 0.946 712 948 736 × 2 = 1 + 0.893 425 897 472;
  • 58) 0.893 425 897 472 × 2 = 1 + 0.786 851 794 944;
  • 59) 0.786 851 794 944 × 2 = 1 + 0.573 703 589 888;
  • 60) 0.573 703 589 888 × 2 = 1 + 0.147 407 179 776;
  • 61) 0.147 407 179 776 × 2 = 0 + 0.294 814 359 552;
  • 62) 0.294 814 359 552 × 2 = 0 + 0.589 628 719 104;
  • 63) 0.589 628 719 104 × 2 = 1 + 0.179 257 438 208;
  • 64) 0.179 257 438 208 × 2 = 0 + 0.358 514 876 416;

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:


0.000 282 005 926(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010(2)

6. Positive number before normalization:

0.000 282 005 926(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010(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:


0.000 282 005 926(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010(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 1010 1011 1101 0001 1101 1111 0010


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 1010 1011 1101 0001 1101 1111 0010 =

0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010


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 1010 1011 1101 0001 1101 1111 0010


Decimal number -0.000 282 005 926 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 1010 1011 1101 0001 1101 1111 0010

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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