-0.000 254 797 590 468 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 468(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
-0.000 254 797 590 468(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:

|-0.000 254 797 590 468| = 0.000 254 797 590 468


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: 0.000 254 797 590 468.

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) 0.000 254 797 590 468 × 2 = 0 + 0.000 509 595 180 936;
  • 2) 0.000 509 595 180 936 × 2 = 0 + 0.001 019 190 361 872;
  • 3) 0.001 019 190 361 872 × 2 = 0 + 0.002 038 380 723 744;
  • 4) 0.002 038 380 723 744 × 2 = 0 + 0.004 076 761 447 488;
  • 5) 0.004 076 761 447 488 × 2 = 0 + 0.008 153 522 894 976;
  • 6) 0.008 153 522 894 976 × 2 = 0 + 0.016 307 045 789 952;
  • 7) 0.016 307 045 789 952 × 2 = 0 + 0.032 614 091 579 904;
  • 8) 0.032 614 091 579 904 × 2 = 0 + 0.065 228 183 159 808;
  • 9) 0.065 228 183 159 808 × 2 = 0 + 0.130 456 366 319 616;
  • 10) 0.130 456 366 319 616 × 2 = 0 + 0.260 912 732 639 232;
  • 11) 0.260 912 732 639 232 × 2 = 0 + 0.521 825 465 278 464;
  • 12) 0.521 825 465 278 464 × 2 = 1 + 0.043 650 930 556 928;
  • 13) 0.043 650 930 556 928 × 2 = 0 + 0.087 301 861 113 856;
  • 14) 0.087 301 861 113 856 × 2 = 0 + 0.174 603 722 227 712;
  • 15) 0.174 603 722 227 712 × 2 = 0 + 0.349 207 444 455 424;
  • 16) 0.349 207 444 455 424 × 2 = 0 + 0.698 414 888 910 848;
  • 17) 0.698 414 888 910 848 × 2 = 1 + 0.396 829 777 821 696;
  • 18) 0.396 829 777 821 696 × 2 = 0 + 0.793 659 555 643 392;
  • 19) 0.793 659 555 643 392 × 2 = 1 + 0.587 319 111 286 784;
  • 20) 0.587 319 111 286 784 × 2 = 1 + 0.174 638 222 573 568;
  • 21) 0.174 638 222 573 568 × 2 = 0 + 0.349 276 445 147 136;
  • 22) 0.349 276 445 147 136 × 2 = 0 + 0.698 552 890 294 272;
  • 23) 0.698 552 890 294 272 × 2 = 1 + 0.397 105 780 588 544;
  • 24) 0.397 105 780 588 544 × 2 = 0 + 0.794 211 561 177 088;
  • 25) 0.794 211 561 177 088 × 2 = 1 + 0.588 423 122 354 176;
  • 26) 0.588 423 122 354 176 × 2 = 1 + 0.176 846 244 708 352;
  • 27) 0.176 846 244 708 352 × 2 = 0 + 0.353 692 489 416 704;
  • 28) 0.353 692 489 416 704 × 2 = 0 + 0.707 384 978 833 408;
  • 29) 0.707 384 978 833 408 × 2 = 1 + 0.414 769 957 666 816;
  • 30) 0.414 769 957 666 816 × 2 = 0 + 0.829 539 915 333 632;
  • 31) 0.829 539 915 333 632 × 2 = 1 + 0.659 079 830 667 264;
  • 32) 0.659 079 830 667 264 × 2 = 1 + 0.318 159 661 334 528;
  • 33) 0.318 159 661 334 528 × 2 = 0 + 0.636 319 322 669 056;
  • 34) 0.636 319 322 669 056 × 2 = 1 + 0.272 638 645 338 112;
  • 35) 0.272 638 645 338 112 × 2 = 0 + 0.545 277 290 676 224;
  • 36) 0.545 277 290 676 224 × 2 = 1 + 0.090 554 581 352 448;
  • 37) 0.090 554 581 352 448 × 2 = 0 + 0.181 109 162 704 896;
  • 38) 0.181 109 162 704 896 × 2 = 0 + 0.362 218 325 409 792;
  • 39) 0.362 218 325 409 792 × 2 = 0 + 0.724 436 650 819 584;
  • 40) 0.724 436 650 819 584 × 2 = 1 + 0.448 873 301 639 168;
  • 41) 0.448 873 301 639 168 × 2 = 0 + 0.897 746 603 278 336;
  • 42) 0.897 746 603 278 336 × 2 = 1 + 0.795 493 206 556 672;
  • 43) 0.795 493 206 556 672 × 2 = 1 + 0.590 986 413 113 344;
  • 44) 0.590 986 413 113 344 × 2 = 1 + 0.181 972 826 226 688;
  • 45) 0.181 972 826 226 688 × 2 = 0 + 0.363 945 652 453 376;
  • 46) 0.363 945 652 453 376 × 2 = 0 + 0.727 891 304 906 752;
  • 47) 0.727 891 304 906 752 × 2 = 1 + 0.455 782 609 813 504;
  • 48) 0.455 782 609 813 504 × 2 = 0 + 0.911 565 219 627 008;
  • 49) 0.911 565 219 627 008 × 2 = 1 + 0.823 130 439 254 016;
  • 50) 0.823 130 439 254 016 × 2 = 1 + 0.646 260 878 508 032;
  • 51) 0.646 260 878 508 032 × 2 = 1 + 0.292 521 757 016 064;
  • 52) 0.292 521 757 016 064 × 2 = 0 + 0.585 043 514 032 128;
  • 53) 0.585 043 514 032 128 × 2 = 1 + 0.170 087 028 064 256;
  • 54) 0.170 087 028 064 256 × 2 = 0 + 0.340 174 056 128 512;
  • 55) 0.340 174 056 128 512 × 2 = 0 + 0.680 348 112 257 024;
  • 56) 0.680 348 112 257 024 × 2 = 1 + 0.360 696 224 514 048;
  • 57) 0.360 696 224 514 048 × 2 = 0 + 0.721 392 449 028 096;
  • 58) 0.721 392 449 028 096 × 2 = 1 + 0.442 784 898 056 192;
  • 59) 0.442 784 898 056 192 × 2 = 0 + 0.885 569 796 112 384;
  • 60) 0.885 569 796 112 384 × 2 = 1 + 0.771 139 592 224 768;
  • 61) 0.771 139 592 224 768 × 2 = 1 + 0.542 279 184 449 536;
  • 62) 0.542 279 184 449 536 × 2 = 1 + 0.084 558 368 899 072;
  • 63) 0.084 558 368 899 072 × 2 = 0 + 0.169 116 737 798 144;
  • 64) 0.169 116 737 798 144 × 2 = 0 + 0.338 233 475 596 288;

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:


0.000 254 797 590 468(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100(2)

6. Positive number before normalization:

0.000 254 797 590 468(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100(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:


0.000 254 797 590 468(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100(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):
1.0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100


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. 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100 =

0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100


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) =
0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100


Decimal number -0.000 254 797 590 468 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 0111 0010 1110 1001 0101 1100

Share

» Convert decimal -0.000 254 797 590 385 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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