654.599 999 999 999 909 050 362 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 362(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
654.599 999 999 999 909 050 362(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: 654.
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;
  • 654 ÷ 2 = 327 + 0;
  • 327 ÷ 2 = 163 + 1;
  • 163 ÷ 2 = 81 + 1;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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.

654(10) =

10 1000 1110(2)


3. Convert to binary (base 2) the fractional part: 0.599 999 999 999 909 050 362.

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) 0.599 999 999 999 909 050 362 × 2 = 1 + 0.199 999 999 999 818 100 724;
  • 2) 0.199 999 999 999 818 100 724 × 2 = 0 + 0.399 999 999 999 636 201 448;
  • 3) 0.399 999 999 999 636 201 448 × 2 = 0 + 0.799 999 999 999 272 402 896;
  • 4) 0.799 999 999 999 272 402 896 × 2 = 1 + 0.599 999 999 998 544 805 792;
  • 5) 0.599 999 999 998 544 805 792 × 2 = 1 + 0.199 999 999 997 089 611 584;
  • 6) 0.199 999 999 997 089 611 584 × 2 = 0 + 0.399 999 999 994 179 223 168;
  • 7) 0.399 999 999 994 179 223 168 × 2 = 0 + 0.799 999 999 988 358 446 336;
  • 8) 0.799 999 999 988 358 446 336 × 2 = 1 + 0.599 999 999 976 716 892 672;
  • 9) 0.599 999 999 976 716 892 672 × 2 = 1 + 0.199 999 999 953 433 785 344;
  • 10) 0.199 999 999 953 433 785 344 × 2 = 0 + 0.399 999 999 906 867 570 688;
  • 11) 0.399 999 999 906 867 570 688 × 2 = 0 + 0.799 999 999 813 735 141 376;
  • 12) 0.799 999 999 813 735 141 376 × 2 = 1 + 0.599 999 999 627 470 282 752;
  • 13) 0.599 999 999 627 470 282 752 × 2 = 1 + 0.199 999 999 254 940 565 504;
  • 14) 0.199 999 999 254 940 565 504 × 2 = 0 + 0.399 999 998 509 881 131 008;
  • 15) 0.399 999 998 509 881 131 008 × 2 = 0 + 0.799 999 997 019 762 262 016;
  • 16) 0.799 999 997 019 762 262 016 × 2 = 1 + 0.599 999 994 039 524 524 032;
  • 17) 0.599 999 994 039 524 524 032 × 2 = 1 + 0.199 999 988 079 049 048 064;
  • 18) 0.199 999 988 079 049 048 064 × 2 = 0 + 0.399 999 976 158 098 096 128;
  • 19) 0.399 999 976 158 098 096 128 × 2 = 0 + 0.799 999 952 316 196 192 256;
  • 20) 0.799 999 952 316 196 192 256 × 2 = 1 + 0.599 999 904 632 392 384 512;
  • 21) 0.599 999 904 632 392 384 512 × 2 = 1 + 0.199 999 809 264 784 769 024;
  • 22) 0.199 999 809 264 784 769 024 × 2 = 0 + 0.399 999 618 529 569 538 048;
  • 23) 0.399 999 618 529 569 538 048 × 2 = 0 + 0.799 999 237 059 139 076 096;
  • 24) 0.799 999 237 059 139 076 096 × 2 = 1 + 0.599 998 474 118 278 152 192;
  • 25) 0.599 998 474 118 278 152 192 × 2 = 1 + 0.199 996 948 236 556 304 384;
  • 26) 0.199 996 948 236 556 304 384 × 2 = 0 + 0.399 993 896 473 112 608 768;
  • 27) 0.399 993 896 473 112 608 768 × 2 = 0 + 0.799 987 792 946 225 217 536;
  • 28) 0.799 987 792 946 225 217 536 × 2 = 1 + 0.599 975 585 892 450 435 072;
  • 29) 0.599 975 585 892 450 435 072 × 2 = 1 + 0.199 951 171 784 900 870 144;
  • 30) 0.199 951 171 784 900 870 144 × 2 = 0 + 0.399 902 343 569 801 740 288;
  • 31) 0.399 902 343 569 801 740 288 × 2 = 0 + 0.799 804 687 139 603 480 576;
  • 32) 0.799 804 687 139 603 480 576 × 2 = 1 + 0.599 609 374 279 206 961 152;
  • 33) 0.599 609 374 279 206 961 152 × 2 = 1 + 0.199 218 748 558 413 922 304;
  • 34) 0.199 218 748 558 413 922 304 × 2 = 0 + 0.398 437 497 116 827 844 608;
  • 35) 0.398 437 497 116 827 844 608 × 2 = 0 + 0.796 874 994 233 655 689 216;
  • 36) 0.796 874 994 233 655 689 216 × 2 = 1 + 0.593 749 988 467 311 378 432;
  • 37) 0.593 749 988 467 311 378 432 × 2 = 1 + 0.187 499 976 934 622 756 864;
  • 38) 0.187 499 976 934 622 756 864 × 2 = 0 + 0.374 999 953 869 245 513 728;
  • 39) 0.374 999 953 869 245 513 728 × 2 = 0 + 0.749 999 907 738 491 027 456;
  • 40) 0.749 999 907 738 491 027 456 × 2 = 1 + 0.499 999 815 476 982 054 912;
  • 41) 0.499 999 815 476 982 054 912 × 2 = 0 + 0.999 999 630 953 964 109 824;
  • 42) 0.999 999 630 953 964 109 824 × 2 = 1 + 0.999 999 261 907 928 219 648;
  • 43) 0.999 999 261 907 928 219 648 × 2 = 1 + 0.999 998 523 815 856 439 296;
  • 44) 0.999 998 523 815 856 439 296 × 2 = 1 + 0.999 997 047 631 712 878 592;
  • 45) 0.999 997 047 631 712 878 592 × 2 = 1 + 0.999 994 095 263 425 757 184;
  • 46) 0.999 994 095 263 425 757 184 × 2 = 1 + 0.999 988 190 526 851 514 368;
  • 47) 0.999 988 190 526 851 514 368 × 2 = 1 + 0.999 976 381 053 703 028 736;
  • 48) 0.999 976 381 053 703 028 736 × 2 = 1 + 0.999 952 762 107 406 057 472;
  • 49) 0.999 952 762 107 406 057 472 × 2 = 1 + 0.999 905 524 214 812 114 944;
  • 50) 0.999 905 524 214 812 114 944 × 2 = 1 + 0.999 811 048 429 624 229 888;
  • 51) 0.999 811 048 429 624 229 888 × 2 = 1 + 0.999 622 096 859 248 459 776;
  • 52) 0.999 622 096 859 248 459 776 × 2 = 1 + 0.999 244 193 718 496 919 552;
  • 53) 0.999 244 193 718 496 919 552 × 2 = 1 + 0.998 488 387 436 993 839 104;

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:


0.599 999 999 999 909 050 362(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2)

5. Positive number before normalization:

654.599 999 999 999 909 050 362(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the left, so that only one non zero digit remains to the left of it:


654.599 999 999 999 909 050 362(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 1111 1(2) × 20 =

1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 11(2) × 29


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): 9

Mantissa (not normalized):
1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 1111 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

9 + 2(11-1) - 1 =

(9 + 1 023)(10) =

1 032(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;
  • 1 032 ÷ 2 = 516 + 0;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 129 ÷ 2 = 64 + 1;
  • 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) =

1032(10) =

100 0000 1000(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. 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011 11 1111 1111 =

0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 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 1000

Mantissa (52 bits) =
0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011


Decimal number 654.599 999 999 999 909 050 362 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 1000 - 0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1011

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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