2.356 194 490 192 344 928 846 982 536 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 846 982 536 59(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
2.356 194 490 192 344 928 846 982 536 59(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: 0.356 194 490 192 344 928 846 982 536 59.

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) 0.356 194 490 192 344 928 846 982 536 59 × 2 = 0 + 0.712 388 980 384 689 857 693 965 073 18;
  • 2) 0.712 388 980 384 689 857 693 965 073 18 × 2 = 1 + 0.424 777 960 769 379 715 387 930 146 36;
  • 3) 0.424 777 960 769 379 715 387 930 146 36 × 2 = 0 + 0.849 555 921 538 759 430 775 860 292 72;
  • 4) 0.849 555 921 538 759 430 775 860 292 72 × 2 = 1 + 0.699 111 843 077 518 861 551 720 585 44;
  • 5) 0.699 111 843 077 518 861 551 720 585 44 × 2 = 1 + 0.398 223 686 155 037 723 103 441 170 88;
  • 6) 0.398 223 686 155 037 723 103 441 170 88 × 2 = 0 + 0.796 447 372 310 075 446 206 882 341 76;
  • 7) 0.796 447 372 310 075 446 206 882 341 76 × 2 = 1 + 0.592 894 744 620 150 892 413 764 683 52;
  • 8) 0.592 894 744 620 150 892 413 764 683 52 × 2 = 1 + 0.185 789 489 240 301 784 827 529 367 04;
  • 9) 0.185 789 489 240 301 784 827 529 367 04 × 2 = 0 + 0.371 578 978 480 603 569 655 058 734 08;
  • 10) 0.371 578 978 480 603 569 655 058 734 08 × 2 = 0 + 0.743 157 956 961 207 139 310 117 468 16;
  • 11) 0.743 157 956 961 207 139 310 117 468 16 × 2 = 1 + 0.486 315 913 922 414 278 620 234 936 32;
  • 12) 0.486 315 913 922 414 278 620 234 936 32 × 2 = 0 + 0.972 631 827 844 828 557 240 469 872 64;
  • 13) 0.972 631 827 844 828 557 240 469 872 64 × 2 = 1 + 0.945 263 655 689 657 114 480 939 745 28;
  • 14) 0.945 263 655 689 657 114 480 939 745 28 × 2 = 1 + 0.890 527 311 379 314 228 961 879 490 56;
  • 15) 0.890 527 311 379 314 228 961 879 490 56 × 2 = 1 + 0.781 054 622 758 628 457 923 758 981 12;
  • 16) 0.781 054 622 758 628 457 923 758 981 12 × 2 = 1 + 0.562 109 245 517 256 915 847 517 962 24;
  • 17) 0.562 109 245 517 256 915 847 517 962 24 × 2 = 1 + 0.124 218 491 034 513 831 695 035 924 48;
  • 18) 0.124 218 491 034 513 831 695 035 924 48 × 2 = 0 + 0.248 436 982 069 027 663 390 071 848 96;
  • 19) 0.248 436 982 069 027 663 390 071 848 96 × 2 = 0 + 0.496 873 964 138 055 326 780 143 697 92;
  • 20) 0.496 873 964 138 055 326 780 143 697 92 × 2 = 0 + 0.993 747 928 276 110 653 560 287 395 84;
  • 21) 0.993 747 928 276 110 653 560 287 395 84 × 2 = 1 + 0.987 495 856 552 221 307 120 574 791 68;
  • 22) 0.987 495 856 552 221 307 120 574 791 68 × 2 = 1 + 0.974 991 713 104 442 614 241 149 583 36;
  • 23) 0.974 991 713 104 442 614 241 149 583 36 × 2 = 1 + 0.949 983 426 208 885 228 482 299 166 72;
  • 24) 0.949 983 426 208 885 228 482 299 166 72 × 2 = 1 + 0.899 966 852 417 770 456 964 598 333 44;
  • 25) 0.899 966 852 417 770 456 964 598 333 44 × 2 = 1 + 0.799 933 704 835 540 913 929 196 666 88;
  • 26) 0.799 933 704 835 540 913 929 196 666 88 × 2 = 1 + 0.599 867 409 671 081 827 858 393 333 76;
  • 27) 0.599 867 409 671 081 827 858 393 333 76 × 2 = 1 + 0.199 734 819 342 163 655 716 786 667 52;
  • 28) 0.199 734 819 342 163 655 716 786 667 52 × 2 = 0 + 0.399 469 638 684 327 311 433 573 335 04;
  • 29) 0.399 469 638 684 327 311 433 573 335 04 × 2 = 0 + 0.798 939 277 368 654 622 867 146 670 08;
  • 30) 0.798 939 277 368 654 622 867 146 670 08 × 2 = 1 + 0.597 878 554 737 309 245 734 293 340 16;
  • 31) 0.597 878 554 737 309 245 734 293 340 16 × 2 = 1 + 0.195 757 109 474 618 491 468 586 680 32;
  • 32) 0.195 757 109 474 618 491 468 586 680 32 × 2 = 0 + 0.391 514 218 949 236 982 937 173 360 64;
  • 33) 0.391 514 218 949 236 982 937 173 360 64 × 2 = 0 + 0.783 028 437 898 473 965 874 346 721 28;
  • 34) 0.783 028 437 898 473 965 874 346 721 28 × 2 = 1 + 0.566 056 875 796 947 931 748 693 442 56;
  • 35) 0.566 056 875 796 947 931 748 693 442 56 × 2 = 1 + 0.132 113 751 593 895 863 497 386 885 12;
  • 36) 0.132 113 751 593 895 863 497 386 885 12 × 2 = 0 + 0.264 227 503 187 791 726 994 773 770 24;
  • 37) 0.264 227 503 187 791 726 994 773 770 24 × 2 = 0 + 0.528 455 006 375 583 453 989 547 540 48;
  • 38) 0.528 455 006 375 583 453 989 547 540 48 × 2 = 1 + 0.056 910 012 751 166 907 979 095 080 96;
  • 39) 0.056 910 012 751 166 907 979 095 080 96 × 2 = 0 + 0.113 820 025 502 333 815 958 190 161 92;
  • 40) 0.113 820 025 502 333 815 958 190 161 92 × 2 = 0 + 0.227 640 051 004 667 631 916 380 323 84;
  • 41) 0.227 640 051 004 667 631 916 380 323 84 × 2 = 0 + 0.455 280 102 009 335 263 832 760 647 68;
  • 42) 0.455 280 102 009 335 263 832 760 647 68 × 2 = 0 + 0.910 560 204 018 670 527 665 521 295 36;
  • 43) 0.910 560 204 018 670 527 665 521 295 36 × 2 = 1 + 0.821 120 408 037 341 055 331 042 590 72;
  • 44) 0.821 120 408 037 341 055 331 042 590 72 × 2 = 1 + 0.642 240 816 074 682 110 662 085 181 44;
  • 45) 0.642 240 816 074 682 110 662 085 181 44 × 2 = 1 + 0.284 481 632 149 364 221 324 170 362 88;
  • 46) 0.284 481 632 149 364 221 324 170 362 88 × 2 = 0 + 0.568 963 264 298 728 442 648 340 725 76;
  • 47) 0.568 963 264 298 728 442 648 340 725 76 × 2 = 1 + 0.137 926 528 597 456 885 296 681 451 52;
  • 48) 0.137 926 528 597 456 885 296 681 451 52 × 2 = 0 + 0.275 853 057 194 913 770 593 362 903 04;
  • 49) 0.275 853 057 194 913 770 593 362 903 04 × 2 = 0 + 0.551 706 114 389 827 541 186 725 806 08;
  • 50) 0.551 706 114 389 827 541 186 725 806 08 × 2 = 1 + 0.103 412 228 779 655 082 373 451 612 16;
  • 51) 0.103 412 228 779 655 082 373 451 612 16 × 2 = 0 + 0.206 824 457 559 310 164 746 903 224 32;
  • 52) 0.206 824 457 559 310 164 746 903 224 32 × 2 = 0 + 0.413 648 915 118 620 329 493 806 448 64;
  • 53) 0.413 648 915 118 620 329 493 806 448 64 × 2 = 0 + 0.827 297 830 237 240 658 987 612 897 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).


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:


0.356 194 490 192 344 928 846 982 536 59(10) =

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

5. Positive number before normalization:

2.356 194 490 192 344 928 846 982 536 59(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:


2.356 194 490 192 344 928 846 982 536 59(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 2.356 194 490 192 344 928 846 982 536 59 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

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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