87.549 999 999 999 997 157 841 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 87.549 999 999 999 997 157 841 5(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
87.549 999 999 999 997 157 841 5(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: 0.549 999 999 999 997 157 841 5.

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) 0.549 999 999 999 997 157 841 5 × 2 = 1 + 0.099 999 999 999 994 315 683;
  • 2) 0.099 999 999 999 994 315 683 × 2 = 0 + 0.199 999 999 999 988 631 366;
  • 3) 0.199 999 999 999 988 631 366 × 2 = 0 + 0.399 999 999 999 977 262 732;
  • 4) 0.399 999 999 999 977 262 732 × 2 = 0 + 0.799 999 999 999 954 525 464;
  • 5) 0.799 999 999 999 954 525 464 × 2 = 1 + 0.599 999 999 999 909 050 928;
  • 6) 0.599 999 999 999 909 050 928 × 2 = 1 + 0.199 999 999 999 818 101 856;
  • 7) 0.199 999 999 999 818 101 856 × 2 = 0 + 0.399 999 999 999 636 203 712;
  • 8) 0.399 999 999 999 636 203 712 × 2 = 0 + 0.799 999 999 999 272 407 424;
  • 9) 0.799 999 999 999 272 407 424 × 2 = 1 + 0.599 999 999 998 544 814 848;
  • 10) 0.599 999 999 998 544 814 848 × 2 = 1 + 0.199 999 999 997 089 629 696;
  • 11) 0.199 999 999 997 089 629 696 × 2 = 0 + 0.399 999 999 994 179 259 392;
  • 12) 0.399 999 999 994 179 259 392 × 2 = 0 + 0.799 999 999 988 358 518 784;
  • 13) 0.799 999 999 988 358 518 784 × 2 = 1 + 0.599 999 999 976 717 037 568;
  • 14) 0.599 999 999 976 717 037 568 × 2 = 1 + 0.199 999 999 953 434 075 136;
  • 15) 0.199 999 999 953 434 075 136 × 2 = 0 + 0.399 999 999 906 868 150 272;
  • 16) 0.399 999 999 906 868 150 272 × 2 = 0 + 0.799 999 999 813 736 300 544;
  • 17) 0.799 999 999 813 736 300 544 × 2 = 1 + 0.599 999 999 627 472 601 088;
  • 18) 0.599 999 999 627 472 601 088 × 2 = 1 + 0.199 999 999 254 945 202 176;
  • 19) 0.199 999 999 254 945 202 176 × 2 = 0 + 0.399 999 998 509 890 404 352;
  • 20) 0.399 999 998 509 890 404 352 × 2 = 0 + 0.799 999 997 019 780 808 704;
  • 21) 0.799 999 997 019 780 808 704 × 2 = 1 + 0.599 999 994 039 561 617 408;
  • 22) 0.599 999 994 039 561 617 408 × 2 = 1 + 0.199 999 988 079 123 234 816;
  • 23) 0.199 999 988 079 123 234 816 × 2 = 0 + 0.399 999 976 158 246 469 632;
  • 24) 0.399 999 976 158 246 469 632 × 2 = 0 + 0.799 999 952 316 492 939 264;
  • 25) 0.799 999 952 316 492 939 264 × 2 = 1 + 0.599 999 904 632 985 878 528;
  • 26) 0.599 999 904 632 985 878 528 × 2 = 1 + 0.199 999 809 265 971 757 056;
  • 27) 0.199 999 809 265 971 757 056 × 2 = 0 + 0.399 999 618 531 943 514 112;
  • 28) 0.399 999 618 531 943 514 112 × 2 = 0 + 0.799 999 237 063 887 028 224;
  • 29) 0.799 999 237 063 887 028 224 × 2 = 1 + 0.599 998 474 127 774 056 448;
  • 30) 0.599 998 474 127 774 056 448 × 2 = 1 + 0.199 996 948 255 548 112 896;
  • 31) 0.199 996 948 255 548 112 896 × 2 = 0 + 0.399 993 896 511 096 225 792;
  • 32) 0.399 993 896 511 096 225 792 × 2 = 0 + 0.799 987 793 022 192 451 584;
  • 33) 0.799 987 793 022 192 451 584 × 2 = 1 + 0.599 975 586 044 384 903 168;
  • 34) 0.599 975 586 044 384 903 168 × 2 = 1 + 0.199 951 172 088 769 806 336;
  • 35) 0.199 951 172 088 769 806 336 × 2 = 0 + 0.399 902 344 177 539 612 672;
  • 36) 0.399 902 344 177 539 612 672 × 2 = 0 + 0.799 804 688 355 079 225 344;
  • 37) 0.799 804 688 355 079 225 344 × 2 = 1 + 0.599 609 376 710 158 450 688;
  • 38) 0.599 609 376 710 158 450 688 × 2 = 1 + 0.199 218 753 420 316 901 376;
  • 39) 0.199 218 753 420 316 901 376 × 2 = 0 + 0.398 437 506 840 633 802 752;
  • 40) 0.398 437 506 840 633 802 752 × 2 = 0 + 0.796 875 013 681 267 605 504;
  • 41) 0.796 875 013 681 267 605 504 × 2 = 1 + 0.593 750 027 362 535 211 008;
  • 42) 0.593 750 027 362 535 211 008 × 2 = 1 + 0.187 500 054 725 070 422 016;
  • 43) 0.187 500 054 725 070 422 016 × 2 = 0 + 0.375 000 109 450 140 844 032;
  • 44) 0.375 000 109 450 140 844 032 × 2 = 0 + 0.750 000 218 900 281 688 064;
  • 45) 0.750 000 218 900 281 688 064 × 2 = 1 + 0.500 000 437 800 563 376 128;
  • 46) 0.500 000 437 800 563 376 128 × 2 = 1 + 0.000 000 875 601 126 752 256;
  • 47) 0.000 000 875 601 126 752 256 × 2 = 0 + 0.000 001 751 202 253 504 512;
  • 48) 0.000 001 751 202 253 504 512 × 2 = 0 + 0.000 003 502 404 507 009 024;
  • 49) 0.000 003 502 404 507 009 024 × 2 = 0 + 0.000 007 004 809 014 018 048;
  • 50) 0.000 007 004 809 014 018 048 × 2 = 0 + 0.000 014 009 618 028 036 096;
  • 51) 0.000 014 009 618 028 036 096 × 2 = 0 + 0.000 028 019 236 056 072 192;
  • 52) 0.000 028 019 236 056 072 192 × 2 = 0 + 0.000 056 038 472 112 144 384;
  • 53) 0.000 056 038 472 112 144 384 × 2 = 0 + 0.000 112 076 944 224 288 768;

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:


0.549 999 999 999 997 157 841 5(10) =

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

5. Positive number before normalization:

87.549 999 999 999 997 157 841 5(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:


87.549 999 999 999 997 157 841 5(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 87.549 999 999 999 997 157 841 5 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 87.549 999 999 999 997 157 849 3 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