-0.000 254 797 590 385 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 254 797 590 385(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
-0.000 254 797 590 385(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:

|-0.000 254 797 590 385| = 0.000 254 797 590 385


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: 0.000 254 797 590 385.

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) 0.000 254 797 590 385 × 2 = 0 + 0.000 509 595 180 77;
  • 2) 0.000 509 595 180 77 × 2 = 0 + 0.001 019 190 361 54;
  • 3) 0.001 019 190 361 54 × 2 = 0 + 0.002 038 380 723 08;
  • 4) 0.002 038 380 723 08 × 2 = 0 + 0.004 076 761 446 16;
  • 5) 0.004 076 761 446 16 × 2 = 0 + 0.008 153 522 892 32;
  • 6) 0.008 153 522 892 32 × 2 = 0 + 0.016 307 045 784 64;
  • 7) 0.016 307 045 784 64 × 2 = 0 + 0.032 614 091 569 28;
  • 8) 0.032 614 091 569 28 × 2 = 0 + 0.065 228 183 138 56;
  • 9) 0.065 228 183 138 56 × 2 = 0 + 0.130 456 366 277 12;
  • 10) 0.130 456 366 277 12 × 2 = 0 + 0.260 912 732 554 24;
  • 11) 0.260 912 732 554 24 × 2 = 0 + 0.521 825 465 108 48;
  • 12) 0.521 825 465 108 48 × 2 = 1 + 0.043 650 930 216 96;
  • 13) 0.043 650 930 216 96 × 2 = 0 + 0.087 301 860 433 92;
  • 14) 0.087 301 860 433 92 × 2 = 0 + 0.174 603 720 867 84;
  • 15) 0.174 603 720 867 84 × 2 = 0 + 0.349 207 441 735 68;
  • 16) 0.349 207 441 735 68 × 2 = 0 + 0.698 414 883 471 36;
  • 17) 0.698 414 883 471 36 × 2 = 1 + 0.396 829 766 942 72;
  • 18) 0.396 829 766 942 72 × 2 = 0 + 0.793 659 533 885 44;
  • 19) 0.793 659 533 885 44 × 2 = 1 + 0.587 319 067 770 88;
  • 20) 0.587 319 067 770 88 × 2 = 1 + 0.174 638 135 541 76;
  • 21) 0.174 638 135 541 76 × 2 = 0 + 0.349 276 271 083 52;
  • 22) 0.349 276 271 083 52 × 2 = 0 + 0.698 552 542 167 04;
  • 23) 0.698 552 542 167 04 × 2 = 1 + 0.397 105 084 334 08;
  • 24) 0.397 105 084 334 08 × 2 = 0 + 0.794 210 168 668 16;
  • 25) 0.794 210 168 668 16 × 2 = 1 + 0.588 420 337 336 32;
  • 26) 0.588 420 337 336 32 × 2 = 1 + 0.176 840 674 672 64;
  • 27) 0.176 840 674 672 64 × 2 = 0 + 0.353 681 349 345 28;
  • 28) 0.353 681 349 345 28 × 2 = 0 + 0.707 362 698 690 56;
  • 29) 0.707 362 698 690 56 × 2 = 1 + 0.414 725 397 381 12;
  • 30) 0.414 725 397 381 12 × 2 = 0 + 0.829 450 794 762 24;
  • 31) 0.829 450 794 762 24 × 2 = 1 + 0.658 901 589 524 48;
  • 32) 0.658 901 589 524 48 × 2 = 1 + 0.317 803 179 048 96;
  • 33) 0.317 803 179 048 96 × 2 = 0 + 0.635 606 358 097 92;
  • 34) 0.635 606 358 097 92 × 2 = 1 + 0.271 212 716 195 84;
  • 35) 0.271 212 716 195 84 × 2 = 0 + 0.542 425 432 391 68;
  • 36) 0.542 425 432 391 68 × 2 = 1 + 0.084 850 864 783 36;
  • 37) 0.084 850 864 783 36 × 2 = 0 + 0.169 701 729 566 72;
  • 38) 0.169 701 729 566 72 × 2 = 0 + 0.339 403 459 133 44;
  • 39) 0.339 403 459 133 44 × 2 = 0 + 0.678 806 918 266 88;
  • 40) 0.678 806 918 266 88 × 2 = 1 + 0.357 613 836 533 76;
  • 41) 0.357 613 836 533 76 × 2 = 0 + 0.715 227 673 067 52;
  • 42) 0.715 227 673 067 52 × 2 = 1 + 0.430 455 346 135 04;
  • 43) 0.430 455 346 135 04 × 2 = 0 + 0.860 910 692 270 08;
  • 44) 0.860 910 692 270 08 × 2 = 1 + 0.721 821 384 540 16;
  • 45) 0.721 821 384 540 16 × 2 = 1 + 0.443 642 769 080 32;
  • 46) 0.443 642 769 080 32 × 2 = 0 + 0.887 285 538 160 64;
  • 47) 0.887 285 538 160 64 × 2 = 1 + 0.774 571 076 321 28;
  • 48) 0.774 571 076 321 28 × 2 = 1 + 0.549 142 152 642 56;
  • 49) 0.549 142 152 642 56 × 2 = 1 + 0.098 284 305 285 12;
  • 50) 0.098 284 305 285 12 × 2 = 0 + 0.196 568 610 570 24;
  • 51) 0.196 568 610 570 24 × 2 = 0 + 0.393 137 221 140 48;
  • 52) 0.393 137 221 140 48 × 2 = 0 + 0.786 274 442 280 96;
  • 53) 0.786 274 442 280 96 × 2 = 1 + 0.572 548 884 561 92;
  • 54) 0.572 548 884 561 92 × 2 = 1 + 0.145 097 769 123 84;
  • 55) 0.145 097 769 123 84 × 2 = 0 + 0.290 195 538 247 68;
  • 56) 0.290 195 538 247 68 × 2 = 0 + 0.580 391 076 495 36;
  • 57) 0.580 391 076 495 36 × 2 = 1 + 0.160 782 152 990 72;
  • 58) 0.160 782 152 990 72 × 2 = 0 + 0.321 564 305 981 44;
  • 59) 0.321 564 305 981 44 × 2 = 0 + 0.643 128 611 962 88;
  • 60) 0.643 128 611 962 88 × 2 = 1 + 0.286 257 223 925 76;
  • 61) 0.286 257 223 925 76 × 2 = 0 + 0.572 514 447 851 52;
  • 62) 0.572 514 447 851 52 × 2 = 1 + 0.145 028 895 703 04;
  • 63) 0.145 028 895 703 04 × 2 = 0 + 0.290 057 791 406 08;
  • 64) 0.290 057 791 406 08 × 2 = 0 + 0.580 115 582 812 16;

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:


0.000 254 797 590 385(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 1001 0100(2)

6. Positive number before normalization:

0.000 254 797 590 385(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 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:


0.000 254 797 590 385(10) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 1001 0100(2) =

0.0000 0000 0001 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 1001 0100(2) × 20 =

1.0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 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):
1.0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 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. 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 1001 0100 =

0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 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) =
0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 1001 0100


Decimal number -0.000 254 797 590 385 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0000 1011 0010 1100 1011 0101 0001 0101 1011 1000 1100 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