87.549 999 999 999 997 157 832 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 87.549 999 999 999 997 157 832 2(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
87.549 999 999 999 997 157 832 2(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: 87.
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;
  • 87 ÷ 2 = 43 + 1;
  • 43 ÷ 2 = 21 + 1;
  • 21 ÷ 2 = 10 + 1;
  • 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.

87(10) =

101 0111(2)


3. Convert to binary (base 2) the fractional part: 0.549 999 999 999 997 157 832 2.

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) 0.549 999 999 999 997 157 832 2 × 2 = 1 + 0.099 999 999 999 994 315 664 4;
  • 2) 0.099 999 999 999 994 315 664 4 × 2 = 0 + 0.199 999 999 999 988 631 328 8;
  • 3) 0.199 999 999 999 988 631 328 8 × 2 = 0 + 0.399 999 999 999 977 262 657 6;
  • 4) 0.399 999 999 999 977 262 657 6 × 2 = 0 + 0.799 999 999 999 954 525 315 2;
  • 5) 0.799 999 999 999 954 525 315 2 × 2 = 1 + 0.599 999 999 999 909 050 630 4;
  • 6) 0.599 999 999 999 909 050 630 4 × 2 = 1 + 0.199 999 999 999 818 101 260 8;
  • 7) 0.199 999 999 999 818 101 260 8 × 2 = 0 + 0.399 999 999 999 636 202 521 6;
  • 8) 0.399 999 999 999 636 202 521 6 × 2 = 0 + 0.799 999 999 999 272 405 043 2;
  • 9) 0.799 999 999 999 272 405 043 2 × 2 = 1 + 0.599 999 999 998 544 810 086 4;
  • 10) 0.599 999 999 998 544 810 086 4 × 2 = 1 + 0.199 999 999 997 089 620 172 8;
  • 11) 0.199 999 999 997 089 620 172 8 × 2 = 0 + 0.399 999 999 994 179 240 345 6;
  • 12) 0.399 999 999 994 179 240 345 6 × 2 = 0 + 0.799 999 999 988 358 480 691 2;
  • 13) 0.799 999 999 988 358 480 691 2 × 2 = 1 + 0.599 999 999 976 716 961 382 4;
  • 14) 0.599 999 999 976 716 961 382 4 × 2 = 1 + 0.199 999 999 953 433 922 764 8;
  • 15) 0.199 999 999 953 433 922 764 8 × 2 = 0 + 0.399 999 999 906 867 845 529 6;
  • 16) 0.399 999 999 906 867 845 529 6 × 2 = 0 + 0.799 999 999 813 735 691 059 2;
  • 17) 0.799 999 999 813 735 691 059 2 × 2 = 1 + 0.599 999 999 627 471 382 118 4;
  • 18) 0.599 999 999 627 471 382 118 4 × 2 = 1 + 0.199 999 999 254 942 764 236 8;
  • 19) 0.199 999 999 254 942 764 236 8 × 2 = 0 + 0.399 999 998 509 885 528 473 6;
  • 20) 0.399 999 998 509 885 528 473 6 × 2 = 0 + 0.799 999 997 019 771 056 947 2;
  • 21) 0.799 999 997 019 771 056 947 2 × 2 = 1 + 0.599 999 994 039 542 113 894 4;
  • 22) 0.599 999 994 039 542 113 894 4 × 2 = 1 + 0.199 999 988 079 084 227 788 8;
  • 23) 0.199 999 988 079 084 227 788 8 × 2 = 0 + 0.399 999 976 158 168 455 577 6;
  • 24) 0.399 999 976 158 168 455 577 6 × 2 = 0 + 0.799 999 952 316 336 911 155 2;
  • 25) 0.799 999 952 316 336 911 155 2 × 2 = 1 + 0.599 999 904 632 673 822 310 4;
  • 26) 0.599 999 904 632 673 822 310 4 × 2 = 1 + 0.199 999 809 265 347 644 620 8;
  • 27) 0.199 999 809 265 347 644 620 8 × 2 = 0 + 0.399 999 618 530 695 289 241 6;
  • 28) 0.399 999 618 530 695 289 241 6 × 2 = 0 + 0.799 999 237 061 390 578 483 2;
  • 29) 0.799 999 237 061 390 578 483 2 × 2 = 1 + 0.599 998 474 122 781 156 966 4;
  • 30) 0.599 998 474 122 781 156 966 4 × 2 = 1 + 0.199 996 948 245 562 313 932 8;
  • 31) 0.199 996 948 245 562 313 932 8 × 2 = 0 + 0.399 993 896 491 124 627 865 6;
  • 32) 0.399 993 896 491 124 627 865 6 × 2 = 0 + 0.799 987 792 982 249 255 731 2;
  • 33) 0.799 987 792 982 249 255 731 2 × 2 = 1 + 0.599 975 585 964 498 511 462 4;
  • 34) 0.599 975 585 964 498 511 462 4 × 2 = 1 + 0.199 951 171 928 997 022 924 8;
  • 35) 0.199 951 171 928 997 022 924 8 × 2 = 0 + 0.399 902 343 857 994 045 849 6;
  • 36) 0.399 902 343 857 994 045 849 6 × 2 = 0 + 0.799 804 687 715 988 091 699 2;
  • 37) 0.799 804 687 715 988 091 699 2 × 2 = 1 + 0.599 609 375 431 976 183 398 4;
  • 38) 0.599 609 375 431 976 183 398 4 × 2 = 1 + 0.199 218 750 863 952 366 796 8;
  • 39) 0.199 218 750 863 952 366 796 8 × 2 = 0 + 0.398 437 501 727 904 733 593 6;
  • 40) 0.398 437 501 727 904 733 593 6 × 2 = 0 + 0.796 875 003 455 809 467 187 2;
  • 41) 0.796 875 003 455 809 467 187 2 × 2 = 1 + 0.593 750 006 911 618 934 374 4;
  • 42) 0.593 750 006 911 618 934 374 4 × 2 = 1 + 0.187 500 013 823 237 868 748 8;
  • 43) 0.187 500 013 823 237 868 748 8 × 2 = 0 + 0.375 000 027 646 475 737 497 6;
  • 44) 0.375 000 027 646 475 737 497 6 × 2 = 0 + 0.750 000 055 292 951 474 995 2;
  • 45) 0.750 000 055 292 951 474 995 2 × 2 = 1 + 0.500 000 110 585 902 949 990 4;
  • 46) 0.500 000 110 585 902 949 990 4 × 2 = 1 + 0.000 000 221 171 805 899 980 8;
  • 47) 0.000 000 221 171 805 899 980 8 × 2 = 0 + 0.000 000 442 343 611 799 961 6;
  • 48) 0.000 000 442 343 611 799 961 6 × 2 = 0 + 0.000 000 884 687 223 599 923 2;
  • 49) 0.000 000 884 687 223 599 923 2 × 2 = 0 + 0.000 001 769 374 447 199 846 4;
  • 50) 0.000 001 769 374 447 199 846 4 × 2 = 0 + 0.000 003 538 748 894 399 692 8;
  • 51) 0.000 003 538 748 894 399 692 8 × 2 = 0 + 0.000 007 077 497 788 799 385 6;
  • 52) 0.000 007 077 497 788 799 385 6 × 2 = 0 + 0.000 014 154 995 577 598 771 2;
  • 53) 0.000 014 154 995 577 598 771 2 × 2 = 0 + 0.000 028 309 991 155 197 542 4;

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:


0.549 999 999 999 997 157 832 2(10) =

0.1000 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2)

5. Positive number before normalization:

87.549 999 999 999 997 157 832 2(10) =

101 0111.1000 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


87.549 999 999 999 997 157 832 2(10) =

101 0111.1000 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2) =

101 0111.1000 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 1100 0000 0(2) × 20 =

1.0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 000(2) × 26


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): 6

Mantissa (not normalized):
1.0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0000 000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(11-1) - 1 =

(6 + 1 023)(10) =

1 029(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;
  • 1 029 ÷ 2 = 514 + 1;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1029(10) =

100 0000 0101(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. 0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 000 0000 =

0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011


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 0101

Mantissa (52 bits) =
0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011


Decimal number 87.549 999 999 999 997 157 832 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0101 1110 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011

» Convert decimal 87.549 999 999 999 997 157 841 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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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