2.356 194 490 192 344 928 854 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.356 194 490 192 344 928 854 6(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
2.356 194 490 192 344 928 854 6(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: 0.356 194 490 192 344 928 854 6.

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) 0.356 194 490 192 344 928 854 6 × 2 = 0 + 0.712 388 980 384 689 857 709 2;
  • 2) 0.712 388 980 384 689 857 709 2 × 2 = 1 + 0.424 777 960 769 379 715 418 4;
  • 3) 0.424 777 960 769 379 715 418 4 × 2 = 0 + 0.849 555 921 538 759 430 836 8;
  • 4) 0.849 555 921 538 759 430 836 8 × 2 = 1 + 0.699 111 843 077 518 861 673 6;
  • 5) 0.699 111 843 077 518 861 673 6 × 2 = 1 + 0.398 223 686 155 037 723 347 2;
  • 6) 0.398 223 686 155 037 723 347 2 × 2 = 0 + 0.796 447 372 310 075 446 694 4;
  • 7) 0.796 447 372 310 075 446 694 4 × 2 = 1 + 0.592 894 744 620 150 893 388 8;
  • 8) 0.592 894 744 620 150 893 388 8 × 2 = 1 + 0.185 789 489 240 301 786 777 6;
  • 9) 0.185 789 489 240 301 786 777 6 × 2 = 0 + 0.371 578 978 480 603 573 555 2;
  • 10) 0.371 578 978 480 603 573 555 2 × 2 = 0 + 0.743 157 956 961 207 147 110 4;
  • 11) 0.743 157 956 961 207 147 110 4 × 2 = 1 + 0.486 315 913 922 414 294 220 8;
  • 12) 0.486 315 913 922 414 294 220 8 × 2 = 0 + 0.972 631 827 844 828 588 441 6;
  • 13) 0.972 631 827 844 828 588 441 6 × 2 = 1 + 0.945 263 655 689 657 176 883 2;
  • 14) 0.945 263 655 689 657 176 883 2 × 2 = 1 + 0.890 527 311 379 314 353 766 4;
  • 15) 0.890 527 311 379 314 353 766 4 × 2 = 1 + 0.781 054 622 758 628 707 532 8;
  • 16) 0.781 054 622 758 628 707 532 8 × 2 = 1 + 0.562 109 245 517 257 415 065 6;
  • 17) 0.562 109 245 517 257 415 065 6 × 2 = 1 + 0.124 218 491 034 514 830 131 2;
  • 18) 0.124 218 491 034 514 830 131 2 × 2 = 0 + 0.248 436 982 069 029 660 262 4;
  • 19) 0.248 436 982 069 029 660 262 4 × 2 = 0 + 0.496 873 964 138 059 320 524 8;
  • 20) 0.496 873 964 138 059 320 524 8 × 2 = 0 + 0.993 747 928 276 118 641 049 6;
  • 21) 0.993 747 928 276 118 641 049 6 × 2 = 1 + 0.987 495 856 552 237 282 099 2;
  • 22) 0.987 495 856 552 237 282 099 2 × 2 = 1 + 0.974 991 713 104 474 564 198 4;
  • 23) 0.974 991 713 104 474 564 198 4 × 2 = 1 + 0.949 983 426 208 949 128 396 8;
  • 24) 0.949 983 426 208 949 128 396 8 × 2 = 1 + 0.899 966 852 417 898 256 793 6;
  • 25) 0.899 966 852 417 898 256 793 6 × 2 = 1 + 0.799 933 704 835 796 513 587 2;
  • 26) 0.799 933 704 835 796 513 587 2 × 2 = 1 + 0.599 867 409 671 593 027 174 4;
  • 27) 0.599 867 409 671 593 027 174 4 × 2 = 1 + 0.199 734 819 343 186 054 348 8;
  • 28) 0.199 734 819 343 186 054 348 8 × 2 = 0 + 0.399 469 638 686 372 108 697 6;
  • 29) 0.399 469 638 686 372 108 697 6 × 2 = 0 + 0.798 939 277 372 744 217 395 2;
  • 30) 0.798 939 277 372 744 217 395 2 × 2 = 1 + 0.597 878 554 745 488 434 790 4;
  • 31) 0.597 878 554 745 488 434 790 4 × 2 = 1 + 0.195 757 109 490 976 869 580 8;
  • 32) 0.195 757 109 490 976 869 580 8 × 2 = 0 + 0.391 514 218 981 953 739 161 6;
  • 33) 0.391 514 218 981 953 739 161 6 × 2 = 0 + 0.783 028 437 963 907 478 323 2;
  • 34) 0.783 028 437 963 907 478 323 2 × 2 = 1 + 0.566 056 875 927 814 956 646 4;
  • 35) 0.566 056 875 927 814 956 646 4 × 2 = 1 + 0.132 113 751 855 629 913 292 8;
  • 36) 0.132 113 751 855 629 913 292 8 × 2 = 0 + 0.264 227 503 711 259 826 585 6;
  • 37) 0.264 227 503 711 259 826 585 6 × 2 = 0 + 0.528 455 007 422 519 653 171 2;
  • 38) 0.528 455 007 422 519 653 171 2 × 2 = 1 + 0.056 910 014 845 039 306 342 4;
  • 39) 0.056 910 014 845 039 306 342 4 × 2 = 0 + 0.113 820 029 690 078 612 684 8;
  • 40) 0.113 820 029 690 078 612 684 8 × 2 = 0 + 0.227 640 059 380 157 225 369 6;
  • 41) 0.227 640 059 380 157 225 369 6 × 2 = 0 + 0.455 280 118 760 314 450 739 2;
  • 42) 0.455 280 118 760 314 450 739 2 × 2 = 0 + 0.910 560 237 520 628 901 478 4;
  • 43) 0.910 560 237 520 628 901 478 4 × 2 = 1 + 0.821 120 475 041 257 802 956 8;
  • 44) 0.821 120 475 041 257 802 956 8 × 2 = 1 + 0.642 240 950 082 515 605 913 6;
  • 45) 0.642 240 950 082 515 605 913 6 × 2 = 1 + 0.284 481 900 165 031 211 827 2;
  • 46) 0.284 481 900 165 031 211 827 2 × 2 = 0 + 0.568 963 800 330 062 423 654 4;
  • 47) 0.568 963 800 330 062 423 654 4 × 2 = 1 + 0.137 927 600 660 124 847 308 8;
  • 48) 0.137 927 600 660 124 847 308 8 × 2 = 0 + 0.275 855 201 320 249 694 617 6;
  • 49) 0.275 855 201 320 249 694 617 6 × 2 = 0 + 0.551 710 402 640 499 389 235 2;
  • 50) 0.551 710 402 640 499 389 235 2 × 2 = 1 + 0.103 420 805 280 998 778 470 4;
  • 51) 0.103 420 805 280 998 778 470 4 × 2 = 0 + 0.206 841 610 561 997 556 940 8;
  • 52) 0.206 841 610 561 997 556 940 8 × 2 = 0 + 0.413 683 221 123 995 113 881 6;
  • 53) 0.413 683 221 123 995 113 881 6 × 2 = 0 + 0.827 366 442 247 990 227 763 2;

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:


0.356 194 490 192 344 928 854 6(10) =

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

5. Positive number before normalization:

2.356 194 490 192 344 928 854 6(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:


2.356 194 490 192 344 928 854 6(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 2.356 194 490 192 344 928 854 6 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