-0.000 164 778 450 862 79 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 778 450 862 79(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
-0.000 164 778 450 862 79(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:

|-0.000 164 778 450 862 79| = 0.000 164 778 450 862 79


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: 0.000 164 778 450 862 79.

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) 0.000 164 778 450 862 79 × 2 = 0 + 0.000 329 556 901 725 58;
  • 2) 0.000 329 556 901 725 58 × 2 = 0 + 0.000 659 113 803 451 16;
  • 3) 0.000 659 113 803 451 16 × 2 = 0 + 0.001 318 227 606 902 32;
  • 4) 0.001 318 227 606 902 32 × 2 = 0 + 0.002 636 455 213 804 64;
  • 5) 0.002 636 455 213 804 64 × 2 = 0 + 0.005 272 910 427 609 28;
  • 6) 0.005 272 910 427 609 28 × 2 = 0 + 0.010 545 820 855 218 56;
  • 7) 0.010 545 820 855 218 56 × 2 = 0 + 0.021 091 641 710 437 12;
  • 8) 0.021 091 641 710 437 12 × 2 = 0 + 0.042 183 283 420 874 24;
  • 9) 0.042 183 283 420 874 24 × 2 = 0 + 0.084 366 566 841 748 48;
  • 10) 0.084 366 566 841 748 48 × 2 = 0 + 0.168 733 133 683 496 96;
  • 11) 0.168 733 133 683 496 96 × 2 = 0 + 0.337 466 267 366 993 92;
  • 12) 0.337 466 267 366 993 92 × 2 = 0 + 0.674 932 534 733 987 84;
  • 13) 0.674 932 534 733 987 84 × 2 = 1 + 0.349 865 069 467 975 68;
  • 14) 0.349 865 069 467 975 68 × 2 = 0 + 0.699 730 138 935 951 36;
  • 15) 0.699 730 138 935 951 36 × 2 = 1 + 0.399 460 277 871 902 72;
  • 16) 0.399 460 277 871 902 72 × 2 = 0 + 0.798 920 555 743 805 44;
  • 17) 0.798 920 555 743 805 44 × 2 = 1 + 0.597 841 111 487 610 88;
  • 18) 0.597 841 111 487 610 88 × 2 = 1 + 0.195 682 222 975 221 76;
  • 19) 0.195 682 222 975 221 76 × 2 = 0 + 0.391 364 445 950 443 52;
  • 20) 0.391 364 445 950 443 52 × 2 = 0 + 0.782 728 891 900 887 04;
  • 21) 0.782 728 891 900 887 04 × 2 = 1 + 0.565 457 783 801 774 08;
  • 22) 0.565 457 783 801 774 08 × 2 = 1 + 0.130 915 567 603 548 16;
  • 23) 0.130 915 567 603 548 16 × 2 = 0 + 0.261 831 135 207 096 32;
  • 24) 0.261 831 135 207 096 32 × 2 = 0 + 0.523 662 270 414 192 64;
  • 25) 0.523 662 270 414 192 64 × 2 = 1 + 0.047 324 540 828 385 28;
  • 26) 0.047 324 540 828 385 28 × 2 = 0 + 0.094 649 081 656 770 56;
  • 27) 0.094 649 081 656 770 56 × 2 = 0 + 0.189 298 163 313 541 12;
  • 28) 0.189 298 163 313 541 12 × 2 = 0 + 0.378 596 326 627 082 24;
  • 29) 0.378 596 326 627 082 24 × 2 = 0 + 0.757 192 653 254 164 48;
  • 30) 0.757 192 653 254 164 48 × 2 = 1 + 0.514 385 306 508 328 96;
  • 31) 0.514 385 306 508 328 96 × 2 = 1 + 0.028 770 613 016 657 92;
  • 32) 0.028 770 613 016 657 92 × 2 = 0 + 0.057 541 226 033 315 84;
  • 33) 0.057 541 226 033 315 84 × 2 = 0 + 0.115 082 452 066 631 68;
  • 34) 0.115 082 452 066 631 68 × 2 = 0 + 0.230 164 904 133 263 36;
  • 35) 0.230 164 904 133 263 36 × 2 = 0 + 0.460 329 808 266 526 72;
  • 36) 0.460 329 808 266 526 72 × 2 = 0 + 0.920 659 616 533 053 44;
  • 37) 0.920 659 616 533 053 44 × 2 = 1 + 0.841 319 233 066 106 88;
  • 38) 0.841 319 233 066 106 88 × 2 = 1 + 0.682 638 466 132 213 76;
  • 39) 0.682 638 466 132 213 76 × 2 = 1 + 0.365 276 932 264 427 52;
  • 40) 0.365 276 932 264 427 52 × 2 = 0 + 0.730 553 864 528 855 04;
  • 41) 0.730 553 864 528 855 04 × 2 = 1 + 0.461 107 729 057 710 08;
  • 42) 0.461 107 729 057 710 08 × 2 = 0 + 0.922 215 458 115 420 16;
  • 43) 0.922 215 458 115 420 16 × 2 = 1 + 0.844 430 916 230 840 32;
  • 44) 0.844 430 916 230 840 32 × 2 = 1 + 0.688 861 832 461 680 64;
  • 45) 0.688 861 832 461 680 64 × 2 = 1 + 0.377 723 664 923 361 28;
  • 46) 0.377 723 664 923 361 28 × 2 = 0 + 0.755 447 329 846 722 56;
  • 47) 0.755 447 329 846 722 56 × 2 = 1 + 0.510 894 659 693 445 12;
  • 48) 0.510 894 659 693 445 12 × 2 = 1 + 0.021 789 319 386 890 24;
  • 49) 0.021 789 319 386 890 24 × 2 = 0 + 0.043 578 638 773 780 48;
  • 50) 0.043 578 638 773 780 48 × 2 = 0 + 0.087 157 277 547 560 96;
  • 51) 0.087 157 277 547 560 96 × 2 = 0 + 0.174 314 555 095 121 92;
  • 52) 0.174 314 555 095 121 92 × 2 = 0 + 0.348 629 110 190 243 84;
  • 53) 0.348 629 110 190 243 84 × 2 = 0 + 0.697 258 220 380 487 68;
  • 54) 0.697 258 220 380 487 68 × 2 = 1 + 0.394 516 440 760 975 36;
  • 55) 0.394 516 440 760 975 36 × 2 = 0 + 0.789 032 881 521 950 72;
  • 56) 0.789 032 881 521 950 72 × 2 = 1 + 0.578 065 763 043 901 44;
  • 57) 0.578 065 763 043 901 44 × 2 = 1 + 0.156 131 526 087 802 88;
  • 58) 0.156 131 526 087 802 88 × 2 = 0 + 0.312 263 052 175 605 76;
  • 59) 0.312 263 052 175 605 76 × 2 = 0 + 0.624 526 104 351 211 52;
  • 60) 0.624 526 104 351 211 52 × 2 = 1 + 0.249 052 208 702 423 04;
  • 61) 0.249 052 208 702 423 04 × 2 = 0 + 0.498 104 417 404 846 08;
  • 62) 0.498 104 417 404 846 08 × 2 = 0 + 0.996 208 834 809 692 16;
  • 63) 0.996 208 834 809 692 16 × 2 = 1 + 0.992 417 669 619 384 32;
  • 64) 0.992 417 669 619 384 32 × 2 = 1 + 0.984 835 339 238 768 64;
  • 65) 0.984 835 339 238 768 64 × 2 = 1 + 0.969 670 678 477 537 28;

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:


0.000 164 778 450 862 79(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0000 0101 1001 0011 1(2)

6. Positive number before normalization:

0.000 164 778 450 862 79(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0000 0101 1001 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:


0.000 164 778 450 862 79(10) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0000 0101 1001 0011 1(2) =

0.0000 0000 0000 1010 1100 1100 1000 0110 0000 1110 1011 1011 0000 0101 1001 0011 1(2) × 20 =

1.0101 1001 1001 0000 1100 0001 1101 0111 0110 0000 1011 0010 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 0000 1011 0010 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 0000 1011 0010 0111 =

0101 1001 1001 0000 1100 0001 1101 0111 0110 0000 1011 0010 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 0000 1011 0010 0111


Decimal number -0.000 164 778 450 862 79 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 0000 1011 0010 0111

Share

» Convert decimal -0.000 164 778 450 863 35 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 04, 2026 [April]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: April - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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