204.120 999 999 999 980 900 611 262 768 507 001 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 204.120 999 999 999 980 900 611 262 768 507 001 2(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
204.120 999 999 999 980 900 611 262 768 507 001 2(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: 204.
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;
  • 204 ÷ 2 = 102 + 0;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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.

204(10) =

1100 1100(2)


3. Convert to binary (base 2) the fractional part: 0.120 999 999 999 980 900 611 262 768 507 001 2.

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) 0.120 999 999 999 980 900 611 262 768 507 001 2 × 2 = 0 + 0.241 999 999 999 961 801 222 525 537 014 002 4;
  • 2) 0.241 999 999 999 961 801 222 525 537 014 002 4 × 2 = 0 + 0.483 999 999 999 923 602 445 051 074 028 004 8;
  • 3) 0.483 999 999 999 923 602 445 051 074 028 004 8 × 2 = 0 + 0.967 999 999 999 847 204 890 102 148 056 009 6;
  • 4) 0.967 999 999 999 847 204 890 102 148 056 009 6 × 2 = 1 + 0.935 999 999 999 694 409 780 204 296 112 019 2;
  • 5) 0.935 999 999 999 694 409 780 204 296 112 019 2 × 2 = 1 + 0.871 999 999 999 388 819 560 408 592 224 038 4;
  • 6) 0.871 999 999 999 388 819 560 408 592 224 038 4 × 2 = 1 + 0.743 999 999 998 777 639 120 817 184 448 076 8;
  • 7) 0.743 999 999 998 777 639 120 817 184 448 076 8 × 2 = 1 + 0.487 999 999 997 555 278 241 634 368 896 153 6;
  • 8) 0.487 999 999 997 555 278 241 634 368 896 153 6 × 2 = 0 + 0.975 999 999 995 110 556 483 268 737 792 307 2;
  • 9) 0.975 999 999 995 110 556 483 268 737 792 307 2 × 2 = 1 + 0.951 999 999 990 221 112 966 537 475 584 614 4;
  • 10) 0.951 999 999 990 221 112 966 537 475 584 614 4 × 2 = 1 + 0.903 999 999 980 442 225 933 074 951 169 228 8;
  • 11) 0.903 999 999 980 442 225 933 074 951 169 228 8 × 2 = 1 + 0.807 999 999 960 884 451 866 149 902 338 457 6;
  • 12) 0.807 999 999 960 884 451 866 149 902 338 457 6 × 2 = 1 + 0.615 999 999 921 768 903 732 299 804 676 915 2;
  • 13) 0.615 999 999 921 768 903 732 299 804 676 915 2 × 2 = 1 + 0.231 999 999 843 537 807 464 599 609 353 830 4;
  • 14) 0.231 999 999 843 537 807 464 599 609 353 830 4 × 2 = 0 + 0.463 999 999 687 075 614 929 199 218 707 660 8;
  • 15) 0.463 999 999 687 075 614 929 199 218 707 660 8 × 2 = 0 + 0.927 999 999 374 151 229 858 398 437 415 321 6;
  • 16) 0.927 999 999 374 151 229 858 398 437 415 321 6 × 2 = 1 + 0.855 999 998 748 302 459 716 796 874 830 643 2;
  • 17) 0.855 999 998 748 302 459 716 796 874 830 643 2 × 2 = 1 + 0.711 999 997 496 604 919 433 593 749 661 286 4;
  • 18) 0.711 999 997 496 604 919 433 593 749 661 286 4 × 2 = 1 + 0.423 999 994 993 209 838 867 187 499 322 572 8;
  • 19) 0.423 999 994 993 209 838 867 187 499 322 572 8 × 2 = 0 + 0.847 999 989 986 419 677 734 374 998 645 145 6;
  • 20) 0.847 999 989 986 419 677 734 374 998 645 145 6 × 2 = 1 + 0.695 999 979 972 839 355 468 749 997 290 291 2;
  • 21) 0.695 999 979 972 839 355 468 749 997 290 291 2 × 2 = 1 + 0.391 999 959 945 678 710 937 499 994 580 582 4;
  • 22) 0.391 999 959 945 678 710 937 499 994 580 582 4 × 2 = 0 + 0.783 999 919 891 357 421 874 999 989 161 164 8;
  • 23) 0.783 999 919 891 357 421 874 999 989 161 164 8 × 2 = 1 + 0.567 999 839 782 714 843 749 999 978 322 329 6;
  • 24) 0.567 999 839 782 714 843 749 999 978 322 329 6 × 2 = 1 + 0.135 999 679 565 429 687 499 999 956 644 659 2;
  • 25) 0.135 999 679 565 429 687 499 999 956 644 659 2 × 2 = 0 + 0.271 999 359 130 859 374 999 999 913 289 318 4;
  • 26) 0.271 999 359 130 859 374 999 999 913 289 318 4 × 2 = 0 + 0.543 998 718 261 718 749 999 999 826 578 636 8;
  • 27) 0.543 998 718 261 718 749 999 999 826 578 636 8 × 2 = 1 + 0.087 997 436 523 437 499 999 999 653 157 273 6;
  • 28) 0.087 997 436 523 437 499 999 999 653 157 273 6 × 2 = 0 + 0.175 994 873 046 874 999 999 999 306 314 547 2;
  • 29) 0.175 994 873 046 874 999 999 999 306 314 547 2 × 2 = 0 + 0.351 989 746 093 749 999 999 998 612 629 094 4;
  • 30) 0.351 989 746 093 749 999 999 998 612 629 094 4 × 2 = 0 + 0.703 979 492 187 499 999 999 997 225 258 188 8;
  • 31) 0.703 979 492 187 499 999 999 997 225 258 188 8 × 2 = 1 + 0.407 958 984 374 999 999 999 994 450 516 377 6;
  • 32) 0.407 958 984 374 999 999 999 994 450 516 377 6 × 2 = 0 + 0.815 917 968 749 999 999 999 988 901 032 755 2;
  • 33) 0.815 917 968 749 999 999 999 988 901 032 755 2 × 2 = 1 + 0.631 835 937 499 999 999 999 977 802 065 510 4;
  • 34) 0.631 835 937 499 999 999 999 977 802 065 510 4 × 2 = 1 + 0.263 671 874 999 999 999 999 955 604 131 020 8;
  • 35) 0.263 671 874 999 999 999 999 955 604 131 020 8 × 2 = 0 + 0.527 343 749 999 999 999 999 911 208 262 041 6;
  • 36) 0.527 343 749 999 999 999 999 911 208 262 041 6 × 2 = 1 + 0.054 687 499 999 999 999 999 822 416 524 083 2;
  • 37) 0.054 687 499 999 999 999 999 822 416 524 083 2 × 2 = 0 + 0.109 374 999 999 999 999 999 644 833 048 166 4;
  • 38) 0.109 374 999 999 999 999 999 644 833 048 166 4 × 2 = 0 + 0.218 749 999 999 999 999 999 289 666 096 332 8;
  • 39) 0.218 749 999 999 999 999 999 289 666 096 332 8 × 2 = 0 + 0.437 499 999 999 999 999 998 579 332 192 665 6;
  • 40) 0.437 499 999 999 999 999 998 579 332 192 665 6 × 2 = 0 + 0.874 999 999 999 999 999 997 158 664 385 331 2;
  • 41) 0.874 999 999 999 999 999 997 158 664 385 331 2 × 2 = 1 + 0.749 999 999 999 999 999 994 317 328 770 662 4;
  • 42) 0.749 999 999 999 999 999 994 317 328 770 662 4 × 2 = 1 + 0.499 999 999 999 999 999 988 634 657 541 324 8;
  • 43) 0.499 999 999 999 999 999 988 634 657 541 324 8 × 2 = 0 + 0.999 999 999 999 999 999 977 269 315 082 649 6;
  • 44) 0.999 999 999 999 999 999 977 269 315 082 649 6 × 2 = 1 + 0.999 999 999 999 999 999 954 538 630 165 299 2;
  • 45) 0.999 999 999 999 999 999 954 538 630 165 299 2 × 2 = 1 + 0.999 999 999 999 999 999 909 077 260 330 598 4;
  • 46) 0.999 999 999 999 999 999 909 077 260 330 598 4 × 2 = 1 + 0.999 999 999 999 999 999 818 154 520 661 196 8;
  • 47) 0.999 999 999 999 999 999 818 154 520 661 196 8 × 2 = 1 + 0.999 999 999 999 999 999 636 309 041 322 393 6;
  • 48) 0.999 999 999 999 999 999 636 309 041 322 393 6 × 2 = 1 + 0.999 999 999 999 999 999 272 618 082 644 787 2;
  • 49) 0.999 999 999 999 999 999 272 618 082 644 787 2 × 2 = 1 + 0.999 999 999 999 999 998 545 236 165 289 574 4;
  • 50) 0.999 999 999 999 999 998 545 236 165 289 574 4 × 2 = 1 + 0.999 999 999 999 999 997 090 472 330 579 148 8;
  • 51) 0.999 999 999 999 999 997 090 472 330 579 148 8 × 2 = 1 + 0.999 999 999 999 999 994 180 944 661 158 297 6;
  • 52) 0.999 999 999 999 999 994 180 944 661 158 297 6 × 2 = 1 + 0.999 999 999 999 999 988 361 889 322 316 595 2;
  • 53) 0.999 999 999 999 999 988 361 889 322 316 595 2 × 2 = 1 + 0.999 999 999 999 999 976 723 778 644 633 190 4;

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:


0.120 999 999 999 980 900 611 262 768 507 001 2(10) =

0.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 1111 1(2)

5. Positive number before normalization:

204.120 999 999 999 980 900 611 262 768 507 001 2(10) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


204.120 999 999 999 980 900 611 262 768 507 001 2(10) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 1111 1(2) =

1100 1100.0001 1110 1111 1001 1101 1011 0010 0010 1101 0000 1101 1111 1111 1(2) × 20 =

1.1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1111(2) × 27


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): 7

Mantissa (not normalized):
1.1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1111


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

7 + 2(11-1) - 1 =

(7 + 1 023)(10) =

1 030(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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1030(10) =

100 0000 0110(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. 1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011 1111 1111 =

1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011


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 0110

Mantissa (52 bits) =
1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011


Decimal number 204.120 999 999 999 980 900 611 262 768 507 001 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0110 - 1001 1000 0011 1101 1111 0011 1011 0110 0100 0101 1010 0001 1011

» Convert decimal 204.120 999 999 999 980 900 611 262 768 506 999 2 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