-0.000 282 005 964 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 964(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
-0.000 282 005 964(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:

|-0.000 282 005 964| = 0.000 282 005 964


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: 0.000 282 005 964.

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) 0.000 282 005 964 × 2 = 0 + 0.000 564 011 928;
  • 2) 0.000 564 011 928 × 2 = 0 + 0.001 128 023 856;
  • 3) 0.001 128 023 856 × 2 = 0 + 0.002 256 047 712;
  • 4) 0.002 256 047 712 × 2 = 0 + 0.004 512 095 424;
  • 5) 0.004 512 095 424 × 2 = 0 + 0.009 024 190 848;
  • 6) 0.009 024 190 848 × 2 = 0 + 0.018 048 381 696;
  • 7) 0.018 048 381 696 × 2 = 0 + 0.036 096 763 392;
  • 8) 0.036 096 763 392 × 2 = 0 + 0.072 193 526 784;
  • 9) 0.072 193 526 784 × 2 = 0 + 0.144 387 053 568;
  • 10) 0.144 387 053 568 × 2 = 0 + 0.288 774 107 136;
  • 11) 0.288 774 107 136 × 2 = 0 + 0.577 548 214 272;
  • 12) 0.577 548 214 272 × 2 = 1 + 0.155 096 428 544;
  • 13) 0.155 096 428 544 × 2 = 0 + 0.310 192 857 088;
  • 14) 0.310 192 857 088 × 2 = 0 + 0.620 385 714 176;
  • 15) 0.620 385 714 176 × 2 = 1 + 0.240 771 428 352;
  • 16) 0.240 771 428 352 × 2 = 0 + 0.481 542 856 704;
  • 17) 0.481 542 856 704 × 2 = 0 + 0.963 085 713 408;
  • 18) 0.963 085 713 408 × 2 = 1 + 0.926 171 426 816;
  • 19) 0.926 171 426 816 × 2 = 1 + 0.852 342 853 632;
  • 20) 0.852 342 853 632 × 2 = 1 + 0.704 685 707 264;
  • 21) 0.704 685 707 264 × 2 = 1 + 0.409 371 414 528;
  • 22) 0.409 371 414 528 × 2 = 0 + 0.818 742 829 056;
  • 23) 0.818 742 829 056 × 2 = 1 + 0.637 485 658 112;
  • 24) 0.637 485 658 112 × 2 = 1 + 0.274 971 316 224;
  • 25) 0.274 971 316 224 × 2 = 0 + 0.549 942 632 448;
  • 26) 0.549 942 632 448 × 2 = 1 + 0.099 885 264 896;
  • 27) 0.099 885 264 896 × 2 = 0 + 0.199 770 529 792;
  • 28) 0.199 770 529 792 × 2 = 0 + 0.399 541 059 584;
  • 29) 0.399 541 059 584 × 2 = 0 + 0.799 082 119 168;
  • 30) 0.799 082 119 168 × 2 = 1 + 0.598 164 238 336;
  • 31) 0.598 164 238 336 × 2 = 1 + 0.196 328 476 672;
  • 32) 0.196 328 476 672 × 2 = 0 + 0.392 656 953 344;
  • 33) 0.392 656 953 344 × 2 = 0 + 0.785 313 906 688;
  • 34) 0.785 313 906 688 × 2 = 1 + 0.570 627 813 376;
  • 35) 0.570 627 813 376 × 2 = 1 + 0.141 255 626 752;
  • 36) 0.141 255 626 752 × 2 = 0 + 0.282 511 253 504;
  • 37) 0.282 511 253 504 × 2 = 0 + 0.565 022 507 008;
  • 38) 0.565 022 507 008 × 2 = 1 + 0.130 045 014 016;
  • 39) 0.130 045 014 016 × 2 = 0 + 0.260 090 028 032;
  • 40) 0.260 090 028 032 × 2 = 0 + 0.520 180 056 064;
  • 41) 0.520 180 056 064 × 2 = 1 + 0.040 360 112 128;
  • 42) 0.040 360 112 128 × 2 = 0 + 0.080 720 224 256;
  • 43) 0.080 720 224 256 × 2 = 0 + 0.161 440 448 512;
  • 44) 0.161 440 448 512 × 2 = 0 + 0.322 880 897 024;
  • 45) 0.322 880 897 024 × 2 = 0 + 0.645 761 794 048;
  • 46) 0.645 761 794 048 × 2 = 1 + 0.291 523 588 096;
  • 47) 0.291 523 588 096 × 2 = 0 + 0.583 047 176 192;
  • 48) 0.583 047 176 192 × 2 = 1 + 0.166 094 352 384;
  • 49) 0.166 094 352 384 × 2 = 0 + 0.332 188 704 768;
  • 50) 0.332 188 704 768 × 2 = 0 + 0.664 377 409 536;
  • 51) 0.664 377 409 536 × 2 = 1 + 0.328 754 819 072;
  • 52) 0.328 754 819 072 × 2 = 0 + 0.657 509 638 144;
  • 53) 0.657 509 638 144 × 2 = 1 + 0.315 019 276 288;
  • 54) 0.315 019 276 288 × 2 = 0 + 0.630 038 552 576;
  • 55) 0.630 038 552 576 × 2 = 1 + 0.260 077 105 152;
  • 56) 0.260 077 105 152 × 2 = 0 + 0.520 154 210 304;
  • 57) 0.520 154 210 304 × 2 = 1 + 0.040 308 420 608;
  • 58) 0.040 308 420 608 × 2 = 0 + 0.080 616 841 216;
  • 59) 0.080 616 841 216 × 2 = 0 + 0.161 233 682 432;
  • 60) 0.161 233 682 432 × 2 = 0 + 0.322 467 364 864;
  • 61) 0.322 467 364 864 × 2 = 0 + 0.644 934 729 728;
  • 62) 0.644 934 729 728 × 2 = 1 + 0.289 869 459 456;
  • 63) 0.289 869 459 456 × 2 = 0 + 0.579 738 918 912;
  • 64) 0.579 738 918 912 × 2 = 1 + 0.159 477 837 824;

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:


0.000 282 005 964(10) =

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

6. Positive number before normalization:

0.000 282 005 964(10) =

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


0.000 282 005 964(10) =

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

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

1.0010 0111 1011 0100 0110 0110 0100 1000 0101 0010 1010 1000 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 0110 0100 1000 0101 0010 1010 1000 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 0110 0100 1000 0101 0010 1010 1000 0101 =

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


Decimal number -0.000 282 005 964 converted to 64 bit double precision IEEE 754 binary floating point representation:

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