87.549 999 999 999 997 157 898 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 87.549 999 999 999 997 157 898(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
87.549 999 999 999 997 157 898(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: 0.549 999 999 999 997 157 898.

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) 0.549 999 999 999 997 157 898 × 2 = 1 + 0.099 999 999 999 994 315 796;
  • 2) 0.099 999 999 999 994 315 796 × 2 = 0 + 0.199 999 999 999 988 631 592;
  • 3) 0.199 999 999 999 988 631 592 × 2 = 0 + 0.399 999 999 999 977 263 184;
  • 4) 0.399 999 999 999 977 263 184 × 2 = 0 + 0.799 999 999 999 954 526 368;
  • 5) 0.799 999 999 999 954 526 368 × 2 = 1 + 0.599 999 999 999 909 052 736;
  • 6) 0.599 999 999 999 909 052 736 × 2 = 1 + 0.199 999 999 999 818 105 472;
  • 7) 0.199 999 999 999 818 105 472 × 2 = 0 + 0.399 999 999 999 636 210 944;
  • 8) 0.399 999 999 999 636 210 944 × 2 = 0 + 0.799 999 999 999 272 421 888;
  • 9) 0.799 999 999 999 272 421 888 × 2 = 1 + 0.599 999 999 998 544 843 776;
  • 10) 0.599 999 999 998 544 843 776 × 2 = 1 + 0.199 999 999 997 089 687 552;
  • 11) 0.199 999 999 997 089 687 552 × 2 = 0 + 0.399 999 999 994 179 375 104;
  • 12) 0.399 999 999 994 179 375 104 × 2 = 0 + 0.799 999 999 988 358 750 208;
  • 13) 0.799 999 999 988 358 750 208 × 2 = 1 + 0.599 999 999 976 717 500 416;
  • 14) 0.599 999 999 976 717 500 416 × 2 = 1 + 0.199 999 999 953 435 000 832;
  • 15) 0.199 999 999 953 435 000 832 × 2 = 0 + 0.399 999 999 906 870 001 664;
  • 16) 0.399 999 999 906 870 001 664 × 2 = 0 + 0.799 999 999 813 740 003 328;
  • 17) 0.799 999 999 813 740 003 328 × 2 = 1 + 0.599 999 999 627 480 006 656;
  • 18) 0.599 999 999 627 480 006 656 × 2 = 1 + 0.199 999 999 254 960 013 312;
  • 19) 0.199 999 999 254 960 013 312 × 2 = 0 + 0.399 999 998 509 920 026 624;
  • 20) 0.399 999 998 509 920 026 624 × 2 = 0 + 0.799 999 997 019 840 053 248;
  • 21) 0.799 999 997 019 840 053 248 × 2 = 1 + 0.599 999 994 039 680 106 496;
  • 22) 0.599 999 994 039 680 106 496 × 2 = 1 + 0.199 999 988 079 360 212 992;
  • 23) 0.199 999 988 079 360 212 992 × 2 = 0 + 0.399 999 976 158 720 425 984;
  • 24) 0.399 999 976 158 720 425 984 × 2 = 0 + 0.799 999 952 317 440 851 968;
  • 25) 0.799 999 952 317 440 851 968 × 2 = 1 + 0.599 999 904 634 881 703 936;
  • 26) 0.599 999 904 634 881 703 936 × 2 = 1 + 0.199 999 809 269 763 407 872;
  • 27) 0.199 999 809 269 763 407 872 × 2 = 0 + 0.399 999 618 539 526 815 744;
  • 28) 0.399 999 618 539 526 815 744 × 2 = 0 + 0.799 999 237 079 053 631 488;
  • 29) 0.799 999 237 079 053 631 488 × 2 = 1 + 0.599 998 474 158 107 262 976;
  • 30) 0.599 998 474 158 107 262 976 × 2 = 1 + 0.199 996 948 316 214 525 952;
  • 31) 0.199 996 948 316 214 525 952 × 2 = 0 + 0.399 993 896 632 429 051 904;
  • 32) 0.399 993 896 632 429 051 904 × 2 = 0 + 0.799 987 793 264 858 103 808;
  • 33) 0.799 987 793 264 858 103 808 × 2 = 1 + 0.599 975 586 529 716 207 616;
  • 34) 0.599 975 586 529 716 207 616 × 2 = 1 + 0.199 951 173 059 432 415 232;
  • 35) 0.199 951 173 059 432 415 232 × 2 = 0 + 0.399 902 346 118 864 830 464;
  • 36) 0.399 902 346 118 864 830 464 × 2 = 0 + 0.799 804 692 237 729 660 928;
  • 37) 0.799 804 692 237 729 660 928 × 2 = 1 + 0.599 609 384 475 459 321 856;
  • 38) 0.599 609 384 475 459 321 856 × 2 = 1 + 0.199 218 768 950 918 643 712;
  • 39) 0.199 218 768 950 918 643 712 × 2 = 0 + 0.398 437 537 901 837 287 424;
  • 40) 0.398 437 537 901 837 287 424 × 2 = 0 + 0.796 875 075 803 674 574 848;
  • 41) 0.796 875 075 803 674 574 848 × 2 = 1 + 0.593 750 151 607 349 149 696;
  • 42) 0.593 750 151 607 349 149 696 × 2 = 1 + 0.187 500 303 214 698 299 392;
  • 43) 0.187 500 303 214 698 299 392 × 2 = 0 + 0.375 000 606 429 396 598 784;
  • 44) 0.375 000 606 429 396 598 784 × 2 = 0 + 0.750 001 212 858 793 197 568;
  • 45) 0.750 001 212 858 793 197 568 × 2 = 1 + 0.500 002 425 717 586 395 136;
  • 46) 0.500 002 425 717 586 395 136 × 2 = 1 + 0.000 004 851 435 172 790 272;
  • 47) 0.000 004 851 435 172 790 272 × 2 = 0 + 0.000 009 702 870 345 580 544;
  • 48) 0.000 009 702 870 345 580 544 × 2 = 0 + 0.000 019 405 740 691 161 088;
  • 49) 0.000 019 405 740 691 161 088 × 2 = 0 + 0.000 038 811 481 382 322 176;
  • 50) 0.000 038 811 481 382 322 176 × 2 = 0 + 0.000 077 622 962 764 644 352;
  • 51) 0.000 077 622 962 764 644 352 × 2 = 0 + 0.000 155 245 925 529 288 704;
  • 52) 0.000 155 245 925 529 288 704 × 2 = 0 + 0.000 310 491 851 058 577 408;
  • 53) 0.000 310 491 851 058 577 408 × 2 = 0 + 0.000 620 983 702 117 154 816;

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:


0.549 999 999 999 997 157 898(10) =

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

5. Positive number before normalization:

87.549 999 999 999 997 157 898(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:


87.549 999 999 999 997 157 898(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 87.549 999 999 999 997 157 898 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 87.549 999 999 999 997 157 852 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