654.599 999 999 999 909 050 646 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 654.599 999 999 999 909 050 646(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
654.599 999 999 999 909 050 646(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: 0.599 999 999 999 909 050 646.

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) 0.599 999 999 999 909 050 646 × 2 = 1 + 0.199 999 999 999 818 101 292;
  • 2) 0.199 999 999 999 818 101 292 × 2 = 0 + 0.399 999 999 999 636 202 584;
  • 3) 0.399 999 999 999 636 202 584 × 2 = 0 + 0.799 999 999 999 272 405 168;
  • 4) 0.799 999 999 999 272 405 168 × 2 = 1 + 0.599 999 999 998 544 810 336;
  • 5) 0.599 999 999 998 544 810 336 × 2 = 1 + 0.199 999 999 997 089 620 672;
  • 6) 0.199 999 999 997 089 620 672 × 2 = 0 + 0.399 999 999 994 179 241 344;
  • 7) 0.399 999 999 994 179 241 344 × 2 = 0 + 0.799 999 999 988 358 482 688;
  • 8) 0.799 999 999 988 358 482 688 × 2 = 1 + 0.599 999 999 976 716 965 376;
  • 9) 0.599 999 999 976 716 965 376 × 2 = 1 + 0.199 999 999 953 433 930 752;
  • 10) 0.199 999 999 953 433 930 752 × 2 = 0 + 0.399 999 999 906 867 861 504;
  • 11) 0.399 999 999 906 867 861 504 × 2 = 0 + 0.799 999 999 813 735 723 008;
  • 12) 0.799 999 999 813 735 723 008 × 2 = 1 + 0.599 999 999 627 471 446 016;
  • 13) 0.599 999 999 627 471 446 016 × 2 = 1 + 0.199 999 999 254 942 892 032;
  • 14) 0.199 999 999 254 942 892 032 × 2 = 0 + 0.399 999 998 509 885 784 064;
  • 15) 0.399 999 998 509 885 784 064 × 2 = 0 + 0.799 999 997 019 771 568 128;
  • 16) 0.799 999 997 019 771 568 128 × 2 = 1 + 0.599 999 994 039 543 136 256;
  • 17) 0.599 999 994 039 543 136 256 × 2 = 1 + 0.199 999 988 079 086 272 512;
  • 18) 0.199 999 988 079 086 272 512 × 2 = 0 + 0.399 999 976 158 172 545 024;
  • 19) 0.399 999 976 158 172 545 024 × 2 = 0 + 0.799 999 952 316 345 090 048;
  • 20) 0.799 999 952 316 345 090 048 × 2 = 1 + 0.599 999 904 632 690 180 096;
  • 21) 0.599 999 904 632 690 180 096 × 2 = 1 + 0.199 999 809 265 380 360 192;
  • 22) 0.199 999 809 265 380 360 192 × 2 = 0 + 0.399 999 618 530 760 720 384;
  • 23) 0.399 999 618 530 760 720 384 × 2 = 0 + 0.799 999 237 061 521 440 768;
  • 24) 0.799 999 237 061 521 440 768 × 2 = 1 + 0.599 998 474 123 042 881 536;
  • 25) 0.599 998 474 123 042 881 536 × 2 = 1 + 0.199 996 948 246 085 763 072;
  • 26) 0.199 996 948 246 085 763 072 × 2 = 0 + 0.399 993 896 492 171 526 144;
  • 27) 0.399 993 896 492 171 526 144 × 2 = 0 + 0.799 987 792 984 343 052 288;
  • 28) 0.799 987 792 984 343 052 288 × 2 = 1 + 0.599 975 585 968 686 104 576;
  • 29) 0.599 975 585 968 686 104 576 × 2 = 1 + 0.199 951 171 937 372 209 152;
  • 30) 0.199 951 171 937 372 209 152 × 2 = 0 + 0.399 902 343 874 744 418 304;
  • 31) 0.399 902 343 874 744 418 304 × 2 = 0 + 0.799 804 687 749 488 836 608;
  • 32) 0.799 804 687 749 488 836 608 × 2 = 1 + 0.599 609 375 498 977 673 216;
  • 33) 0.599 609 375 498 977 673 216 × 2 = 1 + 0.199 218 750 997 955 346 432;
  • 34) 0.199 218 750 997 955 346 432 × 2 = 0 + 0.398 437 501 995 910 692 864;
  • 35) 0.398 437 501 995 910 692 864 × 2 = 0 + 0.796 875 003 991 821 385 728;
  • 36) 0.796 875 003 991 821 385 728 × 2 = 1 + 0.593 750 007 983 642 771 456;
  • 37) 0.593 750 007 983 642 771 456 × 2 = 1 + 0.187 500 015 967 285 542 912;
  • 38) 0.187 500 015 967 285 542 912 × 2 = 0 + 0.375 000 031 934 571 085 824;
  • 39) 0.375 000 031 934 571 085 824 × 2 = 0 + 0.750 000 063 869 142 171 648;
  • 40) 0.750 000 063 869 142 171 648 × 2 = 1 + 0.500 000 127 738 284 343 296;
  • 41) 0.500 000 127 738 284 343 296 × 2 = 1 + 0.000 000 255 476 568 686 592;
  • 42) 0.000 000 255 476 568 686 592 × 2 = 0 + 0.000 000 510 953 137 373 184;
  • 43) 0.000 000 510 953 137 373 184 × 2 = 0 + 0.000 001 021 906 274 746 368;
  • 44) 0.000 001 021 906 274 746 368 × 2 = 0 + 0.000 002 043 812 549 492 736;
  • 45) 0.000 002 043 812 549 492 736 × 2 = 0 + 0.000 004 087 625 098 985 472;
  • 46) 0.000 004 087 625 098 985 472 × 2 = 0 + 0.000 008 175 250 197 970 944;
  • 47) 0.000 008 175 250 197 970 944 × 2 = 0 + 0.000 016 350 500 395 941 888;
  • 48) 0.000 016 350 500 395 941 888 × 2 = 0 + 0.000 032 701 000 791 883 776;
  • 49) 0.000 032 701 000 791 883 776 × 2 = 0 + 0.000 065 402 001 583 767 552;
  • 50) 0.000 065 402 001 583 767 552 × 2 = 0 + 0.000 130 804 003 167 535 104;
  • 51) 0.000 130 804 003 167 535 104 × 2 = 0 + 0.000 261 608 006 335 070 208;
  • 52) 0.000 261 608 006 335 070 208 × 2 = 0 + 0.000 523 216 012 670 140 416;
  • 53) 0.000 523 216 012 670 140 416 × 2 = 0 + 0.001 046 432 025 340 280 832;

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:


0.599 999 999 999 909 050 646(10) =

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

5. Positive number before normalization:

654.599 999 999 999 909 050 646(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:


654.599 999 999 999 909 050 646(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 654.599 999 999 999 909 050 646 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