-0.000 164 778 450 863 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 778 450 863 64(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
-0.000 164 778 450 863 64(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:

|-0.000 164 778 450 863 64| = 0.000 164 778 450 863 64


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: 0.000 164 778 450 863 64.

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) 0.000 164 778 450 863 64 × 2 = 0 + 0.000 329 556 901 727 28;
  • 2) 0.000 329 556 901 727 28 × 2 = 0 + 0.000 659 113 803 454 56;
  • 3) 0.000 659 113 803 454 56 × 2 = 0 + 0.001 318 227 606 909 12;
  • 4) 0.001 318 227 606 909 12 × 2 = 0 + 0.002 636 455 213 818 24;
  • 5) 0.002 636 455 213 818 24 × 2 = 0 + 0.005 272 910 427 636 48;
  • 6) 0.005 272 910 427 636 48 × 2 = 0 + 0.010 545 820 855 272 96;
  • 7) 0.010 545 820 855 272 96 × 2 = 0 + 0.021 091 641 710 545 92;
  • 8) 0.021 091 641 710 545 92 × 2 = 0 + 0.042 183 283 421 091 84;
  • 9) 0.042 183 283 421 091 84 × 2 = 0 + 0.084 366 566 842 183 68;
  • 10) 0.084 366 566 842 183 68 × 2 = 0 + 0.168 733 133 684 367 36;
  • 11) 0.168 733 133 684 367 36 × 2 = 0 + 0.337 466 267 368 734 72;
  • 12) 0.337 466 267 368 734 72 × 2 = 0 + 0.674 932 534 737 469 44;
  • 13) 0.674 932 534 737 469 44 × 2 = 1 + 0.349 865 069 474 938 88;
  • 14) 0.349 865 069 474 938 88 × 2 = 0 + 0.699 730 138 949 877 76;
  • 15) 0.699 730 138 949 877 76 × 2 = 1 + 0.399 460 277 899 755 52;
  • 16) 0.399 460 277 899 755 52 × 2 = 0 + 0.798 920 555 799 511 04;
  • 17) 0.798 920 555 799 511 04 × 2 = 1 + 0.597 841 111 599 022 08;
  • 18) 0.597 841 111 599 022 08 × 2 = 1 + 0.195 682 223 198 044 16;
  • 19) 0.195 682 223 198 044 16 × 2 = 0 + 0.391 364 446 396 088 32;
  • 20) 0.391 364 446 396 088 32 × 2 = 0 + 0.782 728 892 792 176 64;
  • 21) 0.782 728 892 792 176 64 × 2 = 1 + 0.565 457 785 584 353 28;
  • 22) 0.565 457 785 584 353 28 × 2 = 1 + 0.130 915 571 168 706 56;
  • 23) 0.130 915 571 168 706 56 × 2 = 0 + 0.261 831 142 337 413 12;
  • 24) 0.261 831 142 337 413 12 × 2 = 0 + 0.523 662 284 674 826 24;
  • 25) 0.523 662 284 674 826 24 × 2 = 1 + 0.047 324 569 349 652 48;
  • 26) 0.047 324 569 349 652 48 × 2 = 0 + 0.094 649 138 699 304 96;
  • 27) 0.094 649 138 699 304 96 × 2 = 0 + 0.189 298 277 398 609 92;
  • 28) 0.189 298 277 398 609 92 × 2 = 0 + 0.378 596 554 797 219 84;
  • 29) 0.378 596 554 797 219 84 × 2 = 0 + 0.757 193 109 594 439 68;
  • 30) 0.757 193 109 594 439 68 × 2 = 1 + 0.514 386 219 188 879 36;
  • 31) 0.514 386 219 188 879 36 × 2 = 1 + 0.028 772 438 377 758 72;
  • 32) 0.028 772 438 377 758 72 × 2 = 0 + 0.057 544 876 755 517 44;
  • 33) 0.057 544 876 755 517 44 × 2 = 0 + 0.115 089 753 511 034 88;
  • 34) 0.115 089 753 511 034 88 × 2 = 0 + 0.230 179 507 022 069 76;
  • 35) 0.230 179 507 022 069 76 × 2 = 0 + 0.460 359 014 044 139 52;
  • 36) 0.460 359 014 044 139 52 × 2 = 0 + 0.920 718 028 088 279 04;
  • 37) 0.920 718 028 088 279 04 × 2 = 1 + 0.841 436 056 176 558 08;
  • 38) 0.841 436 056 176 558 08 × 2 = 1 + 0.682 872 112 353 116 16;
  • 39) 0.682 872 112 353 116 16 × 2 = 1 + 0.365 744 224 706 232 32;
  • 40) 0.365 744 224 706 232 32 × 2 = 0 + 0.731 488 449 412 464 64;
  • 41) 0.731 488 449 412 464 64 × 2 = 1 + 0.462 976 898 824 929 28;
  • 42) 0.462 976 898 824 929 28 × 2 = 0 + 0.925 953 797 649 858 56;
  • 43) 0.925 953 797 649 858 56 × 2 = 1 + 0.851 907 595 299 717 12;
  • 44) 0.851 907 595 299 717 12 × 2 = 1 + 0.703 815 190 599 434 24;
  • 45) 0.703 815 190 599 434 24 × 2 = 1 + 0.407 630 381 198 868 48;
  • 46) 0.407 630 381 198 868 48 × 2 = 0 + 0.815 260 762 397 736 96;
  • 47) 0.815 260 762 397 736 96 × 2 = 1 + 0.630 521 524 795 473 92;
  • 48) 0.630 521 524 795 473 92 × 2 = 1 + 0.261 043 049 590 947 84;
  • 49) 0.261 043 049 590 947 84 × 2 = 0 + 0.522 086 099 181 895 68;
  • 50) 0.522 086 099 181 895 68 × 2 = 1 + 0.044 172 198 363 791 36;
  • 51) 0.044 172 198 363 791 36 × 2 = 0 + 0.088 344 396 727 582 72;
  • 52) 0.088 344 396 727 582 72 × 2 = 0 + 0.176 688 793 455 165 44;
  • 53) 0.176 688 793 455 165 44 × 2 = 0 + 0.353 377 586 910 330 88;
  • 54) 0.353 377 586 910 330 88 × 2 = 0 + 0.706 755 173 820 661 76;
  • 55) 0.706 755 173 820 661 76 × 2 = 1 + 0.413 510 347 641 323 52;
  • 56) 0.413 510 347 641 323 52 × 2 = 0 + 0.827 020 695 282 647 04;
  • 57) 0.827 020 695 282 647 04 × 2 = 1 + 0.654 041 390 565 294 08;
  • 58) 0.654 041 390 565 294 08 × 2 = 1 + 0.308 082 781 130 588 16;
  • 59) 0.308 082 781 130 588 16 × 2 = 0 + 0.616 165 562 261 176 32;
  • 60) 0.616 165 562 261 176 32 × 2 = 1 + 0.232 331 124 522 352 64;
  • 61) 0.232 331 124 522 352 64 × 2 = 0 + 0.464 662 249 044 705 28;
  • 62) 0.464 662 249 044 705 28 × 2 = 0 + 0.929 324 498 089 410 56;
  • 63) 0.929 324 498 089 410 56 × 2 = 1 + 0.858 648 996 178 821 12;
  • 64) 0.858 648 996 178 821 12 × 2 = 1 + 0.717 297 992 357 642 24;
  • 65) 0.717 297 992 357 642 24 × 2 = 1 + 0.434 595 984 715 284 48;

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:


0.000 164 778 450 863 64(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0100 0010 1101 0011 1(2)

6. Positive number before normalization:

0.000 164 778 450 863 64(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0100 0010 1101 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 164 778 450 863 64(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0100 0010 1101 0011 1(2) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0100 0010 1101 0011 1(2) × 20 =

1.0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111(2) × 2-13


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): -13

Mantissa (not normalized):
1.0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111 =

0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111


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 0010

Mantissa (52 bits) =
0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111


Decimal number -0.000 164 778 450 863 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0101 1001 1001 0000 1100 0001 1101 0111 0110 1000 0101 1010 0111

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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