-89.099 999 999 999 994 174 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -89.099 999 999 999 994 174(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
-89.099 999 999 999 994 174(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. Start with the positive version of the number:

|-89.099 999 999 999 994 174| = 89.099 999 999 999 994 174


2. First, convert to binary (in base 2) the integer part: 89.
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;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

89(10) =

101 1001(2)


4. Convert to binary (base 2) the fractional part: 0.099 999 999 999 994 174.

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) 0.099 999 999 999 994 174 × 2 = 0 + 0.199 999 999 999 988 348;
  • 2) 0.199 999 999 999 988 348 × 2 = 0 + 0.399 999 999 999 976 696;
  • 3) 0.399 999 999 999 976 696 × 2 = 0 + 0.799 999 999 999 953 392;
  • 4) 0.799 999 999 999 953 392 × 2 = 1 + 0.599 999 999 999 906 784;
  • 5) 0.599 999 999 999 906 784 × 2 = 1 + 0.199 999 999 999 813 568;
  • 6) 0.199 999 999 999 813 568 × 2 = 0 + 0.399 999 999 999 627 136;
  • 7) 0.399 999 999 999 627 136 × 2 = 0 + 0.799 999 999 999 254 272;
  • 8) 0.799 999 999 999 254 272 × 2 = 1 + 0.599 999 999 998 508 544;
  • 9) 0.599 999 999 998 508 544 × 2 = 1 + 0.199 999 999 997 017 088;
  • 10) 0.199 999 999 997 017 088 × 2 = 0 + 0.399 999 999 994 034 176;
  • 11) 0.399 999 999 994 034 176 × 2 = 0 + 0.799 999 999 988 068 352;
  • 12) 0.799 999 999 988 068 352 × 2 = 1 + 0.599 999 999 976 136 704;
  • 13) 0.599 999 999 976 136 704 × 2 = 1 + 0.199 999 999 952 273 408;
  • 14) 0.199 999 999 952 273 408 × 2 = 0 + 0.399 999 999 904 546 816;
  • 15) 0.399 999 999 904 546 816 × 2 = 0 + 0.799 999 999 809 093 632;
  • 16) 0.799 999 999 809 093 632 × 2 = 1 + 0.599 999 999 618 187 264;
  • 17) 0.599 999 999 618 187 264 × 2 = 1 + 0.199 999 999 236 374 528;
  • 18) 0.199 999 999 236 374 528 × 2 = 0 + 0.399 999 998 472 749 056;
  • 19) 0.399 999 998 472 749 056 × 2 = 0 + 0.799 999 996 945 498 112;
  • 20) 0.799 999 996 945 498 112 × 2 = 1 + 0.599 999 993 890 996 224;
  • 21) 0.599 999 993 890 996 224 × 2 = 1 + 0.199 999 987 781 992 448;
  • 22) 0.199 999 987 781 992 448 × 2 = 0 + 0.399 999 975 563 984 896;
  • 23) 0.399 999 975 563 984 896 × 2 = 0 + 0.799 999 951 127 969 792;
  • 24) 0.799 999 951 127 969 792 × 2 = 1 + 0.599 999 902 255 939 584;
  • 25) 0.599 999 902 255 939 584 × 2 = 1 + 0.199 999 804 511 879 168;
  • 26) 0.199 999 804 511 879 168 × 2 = 0 + 0.399 999 609 023 758 336;
  • 27) 0.399 999 609 023 758 336 × 2 = 0 + 0.799 999 218 047 516 672;
  • 28) 0.799 999 218 047 516 672 × 2 = 1 + 0.599 998 436 095 033 344;
  • 29) 0.599 998 436 095 033 344 × 2 = 1 + 0.199 996 872 190 066 688;
  • 30) 0.199 996 872 190 066 688 × 2 = 0 + 0.399 993 744 380 133 376;
  • 31) 0.399 993 744 380 133 376 × 2 = 0 + 0.799 987 488 760 266 752;
  • 32) 0.799 987 488 760 266 752 × 2 = 1 + 0.599 974 977 520 533 504;
  • 33) 0.599 974 977 520 533 504 × 2 = 1 + 0.199 949 955 041 067 008;
  • 34) 0.199 949 955 041 067 008 × 2 = 0 + 0.399 899 910 082 134 016;
  • 35) 0.399 899 910 082 134 016 × 2 = 0 + 0.799 799 820 164 268 032;
  • 36) 0.799 799 820 164 268 032 × 2 = 1 + 0.599 599 640 328 536 064;
  • 37) 0.599 599 640 328 536 064 × 2 = 1 + 0.199 199 280 657 072 128;
  • 38) 0.199 199 280 657 072 128 × 2 = 0 + 0.398 398 561 314 144 256;
  • 39) 0.398 398 561 314 144 256 × 2 = 0 + 0.796 797 122 628 288 512;
  • 40) 0.796 797 122 628 288 512 × 2 = 1 + 0.593 594 245 256 577 024;
  • 41) 0.593 594 245 256 577 024 × 2 = 1 + 0.187 188 490 513 154 048;
  • 42) 0.187 188 490 513 154 048 × 2 = 0 + 0.374 376 981 026 308 096;
  • 43) 0.374 376 981 026 308 096 × 2 = 0 + 0.748 753 962 052 616 192;
  • 44) 0.748 753 962 052 616 192 × 2 = 1 + 0.497 507 924 105 232 384;
  • 45) 0.497 507 924 105 232 384 × 2 = 0 + 0.995 015 848 210 464 768;
  • 46) 0.995 015 848 210 464 768 × 2 = 1 + 0.990 031 696 420 929 536;
  • 47) 0.990 031 696 420 929 536 × 2 = 1 + 0.980 063 392 841 859 072;
  • 48) 0.980 063 392 841 859 072 × 2 = 1 + 0.960 126 785 683 718 144;
  • 49) 0.960 126 785 683 718 144 × 2 = 1 + 0.920 253 571 367 436 288;
  • 50) 0.920 253 571 367 436 288 × 2 = 1 + 0.840 507 142 734 872 576;
  • 51) 0.840 507 142 734 872 576 × 2 = 1 + 0.681 014 285 469 745 152;
  • 52) 0.681 014 285 469 745 152 × 2 = 1 + 0.362 028 570 939 490 304;
  • 53) 0.362 028 570 939 490 304 × 2 = 0 + 0.724 057 141 878 980 608;

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).


5. 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:


0.099 999 999 999 994 174(10) =

0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 0(2)

6. Positive number before normalization:

89.099 999 999 999 994 174(10) =

101 1001.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 0(2)

7. 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:


89.099 999 999 999 994 174(10) =

101 1001.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 0(2) =

101 1001.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 0111 1111 0(2) × 20 =

1.0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101 1111 110(2) × 26


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): 6

Mantissa (not normalized):
1.0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101 1111 110


9. 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)


10. 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;

11. 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)


12. 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. 0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101 111 1110 =

0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0101

Mantissa (52 bits) =
0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101


Decimal number -89.099 999 999 999 994 174 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0101 - 0110 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101

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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