654.599 999 999 999 909 050 781 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 781(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
654.599 999 999 999 909 050 781(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: 0.599 999 999 999 909 050 781.

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) 0.599 999 999 999 909 050 781 × 2 = 1 + 0.199 999 999 999 818 101 562;
  • 2) 0.199 999 999 999 818 101 562 × 2 = 0 + 0.399 999 999 999 636 203 124;
  • 3) 0.399 999 999 999 636 203 124 × 2 = 0 + 0.799 999 999 999 272 406 248;
  • 4) 0.799 999 999 999 272 406 248 × 2 = 1 + 0.599 999 999 998 544 812 496;
  • 5) 0.599 999 999 998 544 812 496 × 2 = 1 + 0.199 999 999 997 089 624 992;
  • 6) 0.199 999 999 997 089 624 992 × 2 = 0 + 0.399 999 999 994 179 249 984;
  • 7) 0.399 999 999 994 179 249 984 × 2 = 0 + 0.799 999 999 988 358 499 968;
  • 8) 0.799 999 999 988 358 499 968 × 2 = 1 + 0.599 999 999 976 716 999 936;
  • 9) 0.599 999 999 976 716 999 936 × 2 = 1 + 0.199 999 999 953 433 999 872;
  • 10) 0.199 999 999 953 433 999 872 × 2 = 0 + 0.399 999 999 906 867 999 744;
  • 11) 0.399 999 999 906 867 999 744 × 2 = 0 + 0.799 999 999 813 735 999 488;
  • 12) 0.799 999 999 813 735 999 488 × 2 = 1 + 0.599 999 999 627 471 998 976;
  • 13) 0.599 999 999 627 471 998 976 × 2 = 1 + 0.199 999 999 254 943 997 952;
  • 14) 0.199 999 999 254 943 997 952 × 2 = 0 + 0.399 999 998 509 887 995 904;
  • 15) 0.399 999 998 509 887 995 904 × 2 = 0 + 0.799 999 997 019 775 991 808;
  • 16) 0.799 999 997 019 775 991 808 × 2 = 1 + 0.599 999 994 039 551 983 616;
  • 17) 0.599 999 994 039 551 983 616 × 2 = 1 + 0.199 999 988 079 103 967 232;
  • 18) 0.199 999 988 079 103 967 232 × 2 = 0 + 0.399 999 976 158 207 934 464;
  • 19) 0.399 999 976 158 207 934 464 × 2 = 0 + 0.799 999 952 316 415 868 928;
  • 20) 0.799 999 952 316 415 868 928 × 2 = 1 + 0.599 999 904 632 831 737 856;
  • 21) 0.599 999 904 632 831 737 856 × 2 = 1 + 0.199 999 809 265 663 475 712;
  • 22) 0.199 999 809 265 663 475 712 × 2 = 0 + 0.399 999 618 531 326 951 424;
  • 23) 0.399 999 618 531 326 951 424 × 2 = 0 + 0.799 999 237 062 653 902 848;
  • 24) 0.799 999 237 062 653 902 848 × 2 = 1 + 0.599 998 474 125 307 805 696;
  • 25) 0.599 998 474 125 307 805 696 × 2 = 1 + 0.199 996 948 250 615 611 392;
  • 26) 0.199 996 948 250 615 611 392 × 2 = 0 + 0.399 993 896 501 231 222 784;
  • 27) 0.399 993 896 501 231 222 784 × 2 = 0 + 0.799 987 793 002 462 445 568;
  • 28) 0.799 987 793 002 462 445 568 × 2 = 1 + 0.599 975 586 004 924 891 136;
  • 29) 0.599 975 586 004 924 891 136 × 2 = 1 + 0.199 951 172 009 849 782 272;
  • 30) 0.199 951 172 009 849 782 272 × 2 = 0 + 0.399 902 344 019 699 564 544;
  • 31) 0.399 902 344 019 699 564 544 × 2 = 0 + 0.799 804 688 039 399 129 088;
  • 32) 0.799 804 688 039 399 129 088 × 2 = 1 + 0.599 609 376 078 798 258 176;
  • 33) 0.599 609 376 078 798 258 176 × 2 = 1 + 0.199 218 752 157 596 516 352;
  • 34) 0.199 218 752 157 596 516 352 × 2 = 0 + 0.398 437 504 315 193 032 704;
  • 35) 0.398 437 504 315 193 032 704 × 2 = 0 + 0.796 875 008 630 386 065 408;
  • 36) 0.796 875 008 630 386 065 408 × 2 = 1 + 0.593 750 017 260 772 130 816;
  • 37) 0.593 750 017 260 772 130 816 × 2 = 1 + 0.187 500 034 521 544 261 632;
  • 38) 0.187 500 034 521 544 261 632 × 2 = 0 + 0.375 000 069 043 088 523 264;
  • 39) 0.375 000 069 043 088 523 264 × 2 = 0 + 0.750 000 138 086 177 046 528;
  • 40) 0.750 000 138 086 177 046 528 × 2 = 1 + 0.500 000 276 172 354 093 056;
  • 41) 0.500 000 276 172 354 093 056 × 2 = 1 + 0.000 000 552 344 708 186 112;
  • 42) 0.000 000 552 344 708 186 112 × 2 = 0 + 0.000 001 104 689 416 372 224;
  • 43) 0.000 001 104 689 416 372 224 × 2 = 0 + 0.000 002 209 378 832 744 448;
  • 44) 0.000 002 209 378 832 744 448 × 2 = 0 + 0.000 004 418 757 665 488 896;
  • 45) 0.000 004 418 757 665 488 896 × 2 = 0 + 0.000 008 837 515 330 977 792;
  • 46) 0.000 008 837 515 330 977 792 × 2 = 0 + 0.000 017 675 030 661 955 584;
  • 47) 0.000 017 675 030 661 955 584 × 2 = 0 + 0.000 035 350 061 323 911 168;
  • 48) 0.000 035 350 061 323 911 168 × 2 = 0 + 0.000 070 700 122 647 822 336;
  • 49) 0.000 070 700 122 647 822 336 × 2 = 0 + 0.000 141 400 245 295 644 672;
  • 50) 0.000 141 400 245 295 644 672 × 2 = 0 + 0.000 282 800 490 591 289 344;
  • 51) 0.000 282 800 490 591 289 344 × 2 = 0 + 0.000 565 600 981 182 578 688;
  • 52) 0.000 565 600 981 182 578 688 × 2 = 0 + 0.001 131 201 962 365 157 376;
  • 53) 0.001 131 201 962 365 157 376 × 2 = 0 + 0.002 262 403 924 730 314 752;

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:


0.599 999 999 999 909 050 781(10) =

0.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2)

5. Positive number before normalization:

654.599 999 999 999 909 050 781(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(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:


654.599 999 999 999 909 050 781(10) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) =

10 1000 1110.1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 0000 0000 0(2) × 20 =

1.0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0000 00(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 1100 0000 0000 00


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 1100 00 0000 0000 =

0100 0111 0100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100


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 1100


Decimal number 654.599 999 999 999 909 050 781 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 1100

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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