2.356 194 490 192 344 928 844 87 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 844 87(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
2.356 194 490 192 344 928 844 87(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. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.356 194 490 192 344 928 844 87.

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) 0.356 194 490 192 344 928 844 87 × 2 = 0 + 0.712 388 980 384 689 857 689 74;
  • 2) 0.712 388 980 384 689 857 689 74 × 2 = 1 + 0.424 777 960 769 379 715 379 48;
  • 3) 0.424 777 960 769 379 715 379 48 × 2 = 0 + 0.849 555 921 538 759 430 758 96;
  • 4) 0.849 555 921 538 759 430 758 96 × 2 = 1 + 0.699 111 843 077 518 861 517 92;
  • 5) 0.699 111 843 077 518 861 517 92 × 2 = 1 + 0.398 223 686 155 037 723 035 84;
  • 6) 0.398 223 686 155 037 723 035 84 × 2 = 0 + 0.796 447 372 310 075 446 071 68;
  • 7) 0.796 447 372 310 075 446 071 68 × 2 = 1 + 0.592 894 744 620 150 892 143 36;
  • 8) 0.592 894 744 620 150 892 143 36 × 2 = 1 + 0.185 789 489 240 301 784 286 72;
  • 9) 0.185 789 489 240 301 784 286 72 × 2 = 0 + 0.371 578 978 480 603 568 573 44;
  • 10) 0.371 578 978 480 603 568 573 44 × 2 = 0 + 0.743 157 956 961 207 137 146 88;
  • 11) 0.743 157 956 961 207 137 146 88 × 2 = 1 + 0.486 315 913 922 414 274 293 76;
  • 12) 0.486 315 913 922 414 274 293 76 × 2 = 0 + 0.972 631 827 844 828 548 587 52;
  • 13) 0.972 631 827 844 828 548 587 52 × 2 = 1 + 0.945 263 655 689 657 097 175 04;
  • 14) 0.945 263 655 689 657 097 175 04 × 2 = 1 + 0.890 527 311 379 314 194 350 08;
  • 15) 0.890 527 311 379 314 194 350 08 × 2 = 1 + 0.781 054 622 758 628 388 700 16;
  • 16) 0.781 054 622 758 628 388 700 16 × 2 = 1 + 0.562 109 245 517 256 777 400 32;
  • 17) 0.562 109 245 517 256 777 400 32 × 2 = 1 + 0.124 218 491 034 513 554 800 64;
  • 18) 0.124 218 491 034 513 554 800 64 × 2 = 0 + 0.248 436 982 069 027 109 601 28;
  • 19) 0.248 436 982 069 027 109 601 28 × 2 = 0 + 0.496 873 964 138 054 219 202 56;
  • 20) 0.496 873 964 138 054 219 202 56 × 2 = 0 + 0.993 747 928 276 108 438 405 12;
  • 21) 0.993 747 928 276 108 438 405 12 × 2 = 1 + 0.987 495 856 552 216 876 810 24;
  • 22) 0.987 495 856 552 216 876 810 24 × 2 = 1 + 0.974 991 713 104 433 753 620 48;
  • 23) 0.974 991 713 104 433 753 620 48 × 2 = 1 + 0.949 983 426 208 867 507 240 96;
  • 24) 0.949 983 426 208 867 507 240 96 × 2 = 1 + 0.899 966 852 417 735 014 481 92;
  • 25) 0.899 966 852 417 735 014 481 92 × 2 = 1 + 0.799 933 704 835 470 028 963 84;
  • 26) 0.799 933 704 835 470 028 963 84 × 2 = 1 + 0.599 867 409 670 940 057 927 68;
  • 27) 0.599 867 409 670 940 057 927 68 × 2 = 1 + 0.199 734 819 341 880 115 855 36;
  • 28) 0.199 734 819 341 880 115 855 36 × 2 = 0 + 0.399 469 638 683 760 231 710 72;
  • 29) 0.399 469 638 683 760 231 710 72 × 2 = 0 + 0.798 939 277 367 520 463 421 44;
  • 30) 0.798 939 277 367 520 463 421 44 × 2 = 1 + 0.597 878 554 735 040 926 842 88;
  • 31) 0.597 878 554 735 040 926 842 88 × 2 = 1 + 0.195 757 109 470 081 853 685 76;
  • 32) 0.195 757 109 470 081 853 685 76 × 2 = 0 + 0.391 514 218 940 163 707 371 52;
  • 33) 0.391 514 218 940 163 707 371 52 × 2 = 0 + 0.783 028 437 880 327 414 743 04;
  • 34) 0.783 028 437 880 327 414 743 04 × 2 = 1 + 0.566 056 875 760 654 829 486 08;
  • 35) 0.566 056 875 760 654 829 486 08 × 2 = 1 + 0.132 113 751 521 309 658 972 16;
  • 36) 0.132 113 751 521 309 658 972 16 × 2 = 0 + 0.264 227 503 042 619 317 944 32;
  • 37) 0.264 227 503 042 619 317 944 32 × 2 = 0 + 0.528 455 006 085 238 635 888 64;
  • 38) 0.528 455 006 085 238 635 888 64 × 2 = 1 + 0.056 910 012 170 477 271 777 28;
  • 39) 0.056 910 012 170 477 271 777 28 × 2 = 0 + 0.113 820 024 340 954 543 554 56;
  • 40) 0.113 820 024 340 954 543 554 56 × 2 = 0 + 0.227 640 048 681 909 087 109 12;
  • 41) 0.227 640 048 681 909 087 109 12 × 2 = 0 + 0.455 280 097 363 818 174 218 24;
  • 42) 0.455 280 097 363 818 174 218 24 × 2 = 0 + 0.910 560 194 727 636 348 436 48;
  • 43) 0.910 560 194 727 636 348 436 48 × 2 = 1 + 0.821 120 389 455 272 696 872 96;
  • 44) 0.821 120 389 455 272 696 872 96 × 2 = 1 + 0.642 240 778 910 545 393 745 92;
  • 45) 0.642 240 778 910 545 393 745 92 × 2 = 1 + 0.284 481 557 821 090 787 491 84;
  • 46) 0.284 481 557 821 090 787 491 84 × 2 = 0 + 0.568 963 115 642 181 574 983 68;
  • 47) 0.568 963 115 642 181 574 983 68 × 2 = 1 + 0.137 926 231 284 363 149 967 36;
  • 48) 0.137 926 231 284 363 149 967 36 × 2 = 0 + 0.275 852 462 568 726 299 934 72;
  • 49) 0.275 852 462 568 726 299 934 72 × 2 = 0 + 0.551 704 925 137 452 599 869 44;
  • 50) 0.551 704 925 137 452 599 869 44 × 2 = 1 + 0.103 409 850 274 905 199 738 88;
  • 51) 0.103 409 850 274 905 199 738 88 × 2 = 0 + 0.206 819 700 549 810 399 477 76;
  • 52) 0.206 819 700 549 810 399 477 76 × 2 = 0 + 0.413 639 401 099 620 798 955 52;
  • 53) 0.413 639 401 099 620 798 955 52 × 2 = 0 + 0.827 278 802 199 241 597 911 04;

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).


4. 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:


0.356 194 490 192 344 928 844 87(10) =

0.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2)

5. Positive number before normalization:

2.356 194 490 192 344 928 844 87(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.356 194 490 192 344 928 844 87(10) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) =

10.0101 1011 0010 1111 1000 1111 1110 0110 0110 0100 0011 1010 0100 0(2) × 20 =

1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00(2) × 21


7. 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 0 (a positive number)

Exponent (unadjusted): 1

Mantissa (not normalized):
1.0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(10)


9. 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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1024(10) =

100 0000 0000(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010 00 =

0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010


Decimal number 2.356 194 490 192 344 928 844 87 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0010 1101 1001 0111 1100 0111 1111 0011 0011 0010 0001 1101 0010

» Convert decimal 2.356 194 490 192 344 928 844 67 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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