90.899 999 999 999 991 473 487 171 34 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 171 34(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
90.899 999 999 999 991 473 487 171 34(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: 90.
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;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 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.

90(10) =

101 1010(2)


3. Convert to binary (base 2) the fractional part: 0.899 999 999 999 991 473 487 171 34.

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) 0.899 999 999 999 991 473 487 171 34 × 2 = 1 + 0.799 999 999 999 982 946 974 342 68;
  • 2) 0.799 999 999 999 982 946 974 342 68 × 2 = 1 + 0.599 999 999 999 965 893 948 685 36;
  • 3) 0.599 999 999 999 965 893 948 685 36 × 2 = 1 + 0.199 999 999 999 931 787 897 370 72;
  • 4) 0.199 999 999 999 931 787 897 370 72 × 2 = 0 + 0.399 999 999 999 863 575 794 741 44;
  • 5) 0.399 999 999 999 863 575 794 741 44 × 2 = 0 + 0.799 999 999 999 727 151 589 482 88;
  • 6) 0.799 999 999 999 727 151 589 482 88 × 2 = 1 + 0.599 999 999 999 454 303 178 965 76;
  • 7) 0.599 999 999 999 454 303 178 965 76 × 2 = 1 + 0.199 999 999 998 908 606 357 931 52;
  • 8) 0.199 999 999 998 908 606 357 931 52 × 2 = 0 + 0.399 999 999 997 817 212 715 863 04;
  • 9) 0.399 999 999 997 817 212 715 863 04 × 2 = 0 + 0.799 999 999 995 634 425 431 726 08;
  • 10) 0.799 999 999 995 634 425 431 726 08 × 2 = 1 + 0.599 999 999 991 268 850 863 452 16;
  • 11) 0.599 999 999 991 268 850 863 452 16 × 2 = 1 + 0.199 999 999 982 537 701 726 904 32;
  • 12) 0.199 999 999 982 537 701 726 904 32 × 2 = 0 + 0.399 999 999 965 075 403 453 808 64;
  • 13) 0.399 999 999 965 075 403 453 808 64 × 2 = 0 + 0.799 999 999 930 150 806 907 617 28;
  • 14) 0.799 999 999 930 150 806 907 617 28 × 2 = 1 + 0.599 999 999 860 301 613 815 234 56;
  • 15) 0.599 999 999 860 301 613 815 234 56 × 2 = 1 + 0.199 999 999 720 603 227 630 469 12;
  • 16) 0.199 999 999 720 603 227 630 469 12 × 2 = 0 + 0.399 999 999 441 206 455 260 938 24;
  • 17) 0.399 999 999 441 206 455 260 938 24 × 2 = 0 + 0.799 999 998 882 412 910 521 876 48;
  • 18) 0.799 999 998 882 412 910 521 876 48 × 2 = 1 + 0.599 999 997 764 825 821 043 752 96;
  • 19) 0.599 999 997 764 825 821 043 752 96 × 2 = 1 + 0.199 999 995 529 651 642 087 505 92;
  • 20) 0.199 999 995 529 651 642 087 505 92 × 2 = 0 + 0.399 999 991 059 303 284 175 011 84;
  • 21) 0.399 999 991 059 303 284 175 011 84 × 2 = 0 + 0.799 999 982 118 606 568 350 023 68;
  • 22) 0.799 999 982 118 606 568 350 023 68 × 2 = 1 + 0.599 999 964 237 213 136 700 047 36;
  • 23) 0.599 999 964 237 213 136 700 047 36 × 2 = 1 + 0.199 999 928 474 426 273 400 094 72;
  • 24) 0.199 999 928 474 426 273 400 094 72 × 2 = 0 + 0.399 999 856 948 852 546 800 189 44;
  • 25) 0.399 999 856 948 852 546 800 189 44 × 2 = 0 + 0.799 999 713 897 705 093 600 378 88;
  • 26) 0.799 999 713 897 705 093 600 378 88 × 2 = 1 + 0.599 999 427 795 410 187 200 757 76;
  • 27) 0.599 999 427 795 410 187 200 757 76 × 2 = 1 + 0.199 998 855 590 820 374 401 515 52;
  • 28) 0.199 998 855 590 820 374 401 515 52 × 2 = 0 + 0.399 997 711 181 640 748 803 031 04;
  • 29) 0.399 997 711 181 640 748 803 031 04 × 2 = 0 + 0.799 995 422 363 281 497 606 062 08;
  • 30) 0.799 995 422 363 281 497 606 062 08 × 2 = 1 + 0.599 990 844 726 562 995 212 124 16;
  • 31) 0.599 990 844 726 562 995 212 124 16 × 2 = 1 + 0.199 981 689 453 125 990 424 248 32;
  • 32) 0.199 981 689 453 125 990 424 248 32 × 2 = 0 + 0.399 963 378 906 251 980 848 496 64;
  • 33) 0.399 963 378 906 251 980 848 496 64 × 2 = 0 + 0.799 926 757 812 503 961 696 993 28;
  • 34) 0.799 926 757 812 503 961 696 993 28 × 2 = 1 + 0.599 853 515 625 007 923 393 986 56;
  • 35) 0.599 853 515 625 007 923 393 986 56 × 2 = 1 + 0.199 707 031 250 015 846 787 973 12;
  • 36) 0.199 707 031 250 015 846 787 973 12 × 2 = 0 + 0.399 414 062 500 031 693 575 946 24;
  • 37) 0.399 414 062 500 031 693 575 946 24 × 2 = 0 + 0.798 828 125 000 063 387 151 892 48;
  • 38) 0.798 828 125 000 063 387 151 892 48 × 2 = 1 + 0.597 656 250 000 126 774 303 784 96;
  • 39) 0.597 656 250 000 126 774 303 784 96 × 2 = 1 + 0.195 312 500 000 253 548 607 569 92;
  • 40) 0.195 312 500 000 253 548 607 569 92 × 2 = 0 + 0.390 625 000 000 507 097 215 139 84;
  • 41) 0.390 625 000 000 507 097 215 139 84 × 2 = 0 + 0.781 250 000 001 014 194 430 279 68;
  • 42) 0.781 250 000 001 014 194 430 279 68 × 2 = 1 + 0.562 500 000 002 028 388 860 559 36;
  • 43) 0.562 500 000 002 028 388 860 559 36 × 2 = 1 + 0.125 000 000 004 056 777 721 118 72;
  • 44) 0.125 000 000 004 056 777 721 118 72 × 2 = 0 + 0.250 000 000 008 113 555 442 237 44;
  • 45) 0.250 000 000 008 113 555 442 237 44 × 2 = 0 + 0.500 000 000 016 227 110 884 474 88;
  • 46) 0.500 000 000 016 227 110 884 474 88 × 2 = 1 + 0.000 000 000 032 454 221 768 949 76;
  • 47) 0.000 000 000 032 454 221 768 949 76 × 2 = 0 + 0.000 000 000 064 908 443 537 899 52;
  • 48) 0.000 000 000 064 908 443 537 899 52 × 2 = 0 + 0.000 000 000 129 816 887 075 799 04;
  • 49) 0.000 000 000 129 816 887 075 799 04 × 2 = 0 + 0.000 000 000 259 633 774 151 598 08;
  • 50) 0.000 000 000 259 633 774 151 598 08 × 2 = 0 + 0.000 000 000 519 267 548 303 196 16;
  • 51) 0.000 000 000 519 267 548 303 196 16 × 2 = 0 + 0.000 000 001 038 535 096 606 392 32;
  • 52) 0.000 000 001 038 535 096 606 392 32 × 2 = 0 + 0.000 000 002 077 070 193 212 784 64;
  • 53) 0.000 000 002 077 070 193 212 784 64 × 2 = 0 + 0.000 000 004 154 140 386 425 569 28;

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:


0.899 999 999 999 991 473 487 171 34(10) =

0.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2)

5. Positive number before normalization:

90.899 999 999 999 991 473 487 171 34(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 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:


90.899 999 999 999 991 473 487 171 34(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 0000 0(2) × 20 =

1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 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):
1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 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. 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 000 0000 =

0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


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) =
0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001


Decimal number 90.899 999 999 999 991 473 487 171 34 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0101 - 0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001

» Convert decimal 90.899 999 999 999 991 473 487 170 7 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