654.599 999 999 999 909 050 63 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 63(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 63(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 63.

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 63 × 2 = 1 + 0.199 999 999 999 818 101 26;
  • 2) 0.199 999 999 999 818 101 26 × 2 = 0 + 0.399 999 999 999 636 202 52;
  • 3) 0.399 999 999 999 636 202 52 × 2 = 0 + 0.799 999 999 999 272 405 04;
  • 4) 0.799 999 999 999 272 405 04 × 2 = 1 + 0.599 999 999 998 544 810 08;
  • 5) 0.599 999 999 998 544 810 08 × 2 = 1 + 0.199 999 999 997 089 620 16;
  • 6) 0.199 999 999 997 089 620 16 × 2 = 0 + 0.399 999 999 994 179 240 32;
  • 7) 0.399 999 999 994 179 240 32 × 2 = 0 + 0.799 999 999 988 358 480 64;
  • 8) 0.799 999 999 988 358 480 64 × 2 = 1 + 0.599 999 999 976 716 961 28;
  • 9) 0.599 999 999 976 716 961 28 × 2 = 1 + 0.199 999 999 953 433 922 56;
  • 10) 0.199 999 999 953 433 922 56 × 2 = 0 + 0.399 999 999 906 867 845 12;
  • 11) 0.399 999 999 906 867 845 12 × 2 = 0 + 0.799 999 999 813 735 690 24;
  • 12) 0.799 999 999 813 735 690 24 × 2 = 1 + 0.599 999 999 627 471 380 48;
  • 13) 0.599 999 999 627 471 380 48 × 2 = 1 + 0.199 999 999 254 942 760 96;
  • 14) 0.199 999 999 254 942 760 96 × 2 = 0 + 0.399 999 998 509 885 521 92;
  • 15) 0.399 999 998 509 885 521 92 × 2 = 0 + 0.799 999 997 019 771 043 84;
  • 16) 0.799 999 997 019 771 043 84 × 2 = 1 + 0.599 999 994 039 542 087 68;
  • 17) 0.599 999 994 039 542 087 68 × 2 = 1 + 0.199 999 988 079 084 175 36;
  • 18) 0.199 999 988 079 084 175 36 × 2 = 0 + 0.399 999 976 158 168 350 72;
  • 19) 0.399 999 976 158 168 350 72 × 2 = 0 + 0.799 999 952 316 336 701 44;
  • 20) 0.799 999 952 316 336 701 44 × 2 = 1 + 0.599 999 904 632 673 402 88;
  • 21) 0.599 999 904 632 673 402 88 × 2 = 1 + 0.199 999 809 265 346 805 76;
  • 22) 0.199 999 809 265 346 805 76 × 2 = 0 + 0.399 999 618 530 693 611 52;
  • 23) 0.399 999 618 530 693 611 52 × 2 = 0 + 0.799 999 237 061 387 223 04;
  • 24) 0.799 999 237 061 387 223 04 × 2 = 1 + 0.599 998 474 122 774 446 08;
  • 25) 0.599 998 474 122 774 446 08 × 2 = 1 + 0.199 996 948 245 548 892 16;
  • 26) 0.199 996 948 245 548 892 16 × 2 = 0 + 0.399 993 896 491 097 784 32;
  • 27) 0.399 993 896 491 097 784 32 × 2 = 0 + 0.799 987 792 982 195 568 64;
  • 28) 0.799 987 792 982 195 568 64 × 2 = 1 + 0.599 975 585 964 391 137 28;
  • 29) 0.599 975 585 964 391 137 28 × 2 = 1 + 0.199 951 171 928 782 274 56;
  • 30) 0.199 951 171 928 782 274 56 × 2 = 0 + 0.399 902 343 857 564 549 12;
  • 31) 0.399 902 343 857 564 549 12 × 2 = 0 + 0.799 804 687 715 129 098 24;
  • 32) 0.799 804 687 715 129 098 24 × 2 = 1 + 0.599 609 375 430 258 196 48;
  • 33) 0.599 609 375 430 258 196 48 × 2 = 1 + 0.199 218 750 860 516 392 96;
  • 34) 0.199 218 750 860 516 392 96 × 2 = 0 + 0.398 437 501 721 032 785 92;
  • 35) 0.398 437 501 721 032 785 92 × 2 = 0 + 0.796 875 003 442 065 571 84;
  • 36) 0.796 875 003 442 065 571 84 × 2 = 1 + 0.593 750 006 884 131 143 68;
  • 37) 0.593 750 006 884 131 143 68 × 2 = 1 + 0.187 500 013 768 262 287 36;
  • 38) 0.187 500 013 768 262 287 36 × 2 = 0 + 0.375 000 027 536 524 574 72;
  • 39) 0.375 000 027 536 524 574 72 × 2 = 0 + 0.750 000 055 073 049 149 44;
  • 40) 0.750 000 055 073 049 149 44 × 2 = 1 + 0.500 000 110 146 098 298 88;
  • 41) 0.500 000 110 146 098 298 88 × 2 = 1 + 0.000 000 220 292 196 597 76;
  • 42) 0.000 000 220 292 196 597 76 × 2 = 0 + 0.000 000 440 584 393 195 52;
  • 43) 0.000 000 440 584 393 195 52 × 2 = 0 + 0.000 000 881 168 786 391 04;
  • 44) 0.000 000 881 168 786 391 04 × 2 = 0 + 0.000 001 762 337 572 782 08;
  • 45) 0.000 001 762 337 572 782 08 × 2 = 0 + 0.000 003 524 675 145 564 16;
  • 46) 0.000 003 524 675 145 564 16 × 2 = 0 + 0.000 007 049 350 291 128 32;
  • 47) 0.000 007 049 350 291 128 32 × 2 = 0 + 0.000 014 098 700 582 256 64;
  • 48) 0.000 014 098 700 582 256 64 × 2 = 0 + 0.000 028 197 401 164 513 28;
  • 49) 0.000 028 197 401 164 513 28 × 2 = 0 + 0.000 056 394 802 329 026 56;
  • 50) 0.000 056 394 802 329 026 56 × 2 = 0 + 0.000 112 789 604 658 053 12;
  • 51) 0.000 112 789 604 658 053 12 × 2 = 0 + 0.000 225 579 209 316 106 24;
  • 52) 0.000 225 579 209 316 106 24 × 2 = 0 + 0.000 451 158 418 632 212 48;
  • 53) 0.000 451 158 418 632 212 48 × 2 = 0 + 0.000 902 316 837 264 424 96;

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 63(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 63(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 63(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 63 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

Share

» Convert decimal 654.599 999 999 999 909 051 2 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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