2.356 194 490 192 344 928 847 071 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 847 071(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
2.356 194 490 192 344 928 847 071(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: 0.356 194 490 192 344 928 847 071.

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) 0.356 194 490 192 344 928 847 071 × 2 = 0 + 0.712 388 980 384 689 857 694 142;
  • 2) 0.712 388 980 384 689 857 694 142 × 2 = 1 + 0.424 777 960 769 379 715 388 284;
  • 3) 0.424 777 960 769 379 715 388 284 × 2 = 0 + 0.849 555 921 538 759 430 776 568;
  • 4) 0.849 555 921 538 759 430 776 568 × 2 = 1 + 0.699 111 843 077 518 861 553 136;
  • 5) 0.699 111 843 077 518 861 553 136 × 2 = 1 + 0.398 223 686 155 037 723 106 272;
  • 6) 0.398 223 686 155 037 723 106 272 × 2 = 0 + 0.796 447 372 310 075 446 212 544;
  • 7) 0.796 447 372 310 075 446 212 544 × 2 = 1 + 0.592 894 744 620 150 892 425 088;
  • 8) 0.592 894 744 620 150 892 425 088 × 2 = 1 + 0.185 789 489 240 301 784 850 176;
  • 9) 0.185 789 489 240 301 784 850 176 × 2 = 0 + 0.371 578 978 480 603 569 700 352;
  • 10) 0.371 578 978 480 603 569 700 352 × 2 = 0 + 0.743 157 956 961 207 139 400 704;
  • 11) 0.743 157 956 961 207 139 400 704 × 2 = 1 + 0.486 315 913 922 414 278 801 408;
  • 12) 0.486 315 913 922 414 278 801 408 × 2 = 0 + 0.972 631 827 844 828 557 602 816;
  • 13) 0.972 631 827 844 828 557 602 816 × 2 = 1 + 0.945 263 655 689 657 115 205 632;
  • 14) 0.945 263 655 689 657 115 205 632 × 2 = 1 + 0.890 527 311 379 314 230 411 264;
  • 15) 0.890 527 311 379 314 230 411 264 × 2 = 1 + 0.781 054 622 758 628 460 822 528;
  • 16) 0.781 054 622 758 628 460 822 528 × 2 = 1 + 0.562 109 245 517 256 921 645 056;
  • 17) 0.562 109 245 517 256 921 645 056 × 2 = 1 + 0.124 218 491 034 513 843 290 112;
  • 18) 0.124 218 491 034 513 843 290 112 × 2 = 0 + 0.248 436 982 069 027 686 580 224;
  • 19) 0.248 436 982 069 027 686 580 224 × 2 = 0 + 0.496 873 964 138 055 373 160 448;
  • 20) 0.496 873 964 138 055 373 160 448 × 2 = 0 + 0.993 747 928 276 110 746 320 896;
  • 21) 0.993 747 928 276 110 746 320 896 × 2 = 1 + 0.987 495 856 552 221 492 641 792;
  • 22) 0.987 495 856 552 221 492 641 792 × 2 = 1 + 0.974 991 713 104 442 985 283 584;
  • 23) 0.974 991 713 104 442 985 283 584 × 2 = 1 + 0.949 983 426 208 885 970 567 168;
  • 24) 0.949 983 426 208 885 970 567 168 × 2 = 1 + 0.899 966 852 417 771 941 134 336;
  • 25) 0.899 966 852 417 771 941 134 336 × 2 = 1 + 0.799 933 704 835 543 882 268 672;
  • 26) 0.799 933 704 835 543 882 268 672 × 2 = 1 + 0.599 867 409 671 087 764 537 344;
  • 27) 0.599 867 409 671 087 764 537 344 × 2 = 1 + 0.199 734 819 342 175 529 074 688;
  • 28) 0.199 734 819 342 175 529 074 688 × 2 = 0 + 0.399 469 638 684 351 058 149 376;
  • 29) 0.399 469 638 684 351 058 149 376 × 2 = 0 + 0.798 939 277 368 702 116 298 752;
  • 30) 0.798 939 277 368 702 116 298 752 × 2 = 1 + 0.597 878 554 737 404 232 597 504;
  • 31) 0.597 878 554 737 404 232 597 504 × 2 = 1 + 0.195 757 109 474 808 465 195 008;
  • 32) 0.195 757 109 474 808 465 195 008 × 2 = 0 + 0.391 514 218 949 616 930 390 016;
  • 33) 0.391 514 218 949 616 930 390 016 × 2 = 0 + 0.783 028 437 899 233 860 780 032;
  • 34) 0.783 028 437 899 233 860 780 032 × 2 = 1 + 0.566 056 875 798 467 721 560 064;
  • 35) 0.566 056 875 798 467 721 560 064 × 2 = 1 + 0.132 113 751 596 935 443 120 128;
  • 36) 0.132 113 751 596 935 443 120 128 × 2 = 0 + 0.264 227 503 193 870 886 240 256;
  • 37) 0.264 227 503 193 870 886 240 256 × 2 = 0 + 0.528 455 006 387 741 772 480 512;
  • 38) 0.528 455 006 387 741 772 480 512 × 2 = 1 + 0.056 910 012 775 483 544 961 024;
  • 39) 0.056 910 012 775 483 544 961 024 × 2 = 0 + 0.113 820 025 550 967 089 922 048;
  • 40) 0.113 820 025 550 967 089 922 048 × 2 = 0 + 0.227 640 051 101 934 179 844 096;
  • 41) 0.227 640 051 101 934 179 844 096 × 2 = 0 + 0.455 280 102 203 868 359 688 192;
  • 42) 0.455 280 102 203 868 359 688 192 × 2 = 0 + 0.910 560 204 407 736 719 376 384;
  • 43) 0.910 560 204 407 736 719 376 384 × 2 = 1 + 0.821 120 408 815 473 438 752 768;
  • 44) 0.821 120 408 815 473 438 752 768 × 2 = 1 + 0.642 240 817 630 946 877 505 536;
  • 45) 0.642 240 817 630 946 877 505 536 × 2 = 1 + 0.284 481 635 261 893 755 011 072;
  • 46) 0.284 481 635 261 893 755 011 072 × 2 = 0 + 0.568 963 270 523 787 510 022 144;
  • 47) 0.568 963 270 523 787 510 022 144 × 2 = 1 + 0.137 926 541 047 575 020 044 288;
  • 48) 0.137 926 541 047 575 020 044 288 × 2 = 0 + 0.275 853 082 095 150 040 088 576;
  • 49) 0.275 853 082 095 150 040 088 576 × 2 = 0 + 0.551 706 164 190 300 080 177 152;
  • 50) 0.551 706 164 190 300 080 177 152 × 2 = 1 + 0.103 412 328 380 600 160 354 304;
  • 51) 0.103 412 328 380 600 160 354 304 × 2 = 0 + 0.206 824 656 761 200 320 708 608;
  • 52) 0.206 824 656 761 200 320 708 608 × 2 = 0 + 0.413 649 313 522 400 641 417 216;
  • 53) 0.413 649 313 522 400 641 417 216 × 2 = 0 + 0.827 298 627 044 801 282 834 432;

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:


0.356 194 490 192 344 928 847 071(10) =

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

5. Positive number before normalization:

2.356 194 490 192 344 928 847 071(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:


2.356 194 490 192 344 928 847 071(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 2.356 194 490 192 344 928 847 071 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