90.899 999 999 999 991 473 487 170 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 90.899 999 999 999 991 473 487 170 7(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
90.899 999 999 999 991 473 487 170 7(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: 0.899 999 999 999 991 473 487 170 7.

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) 0.899 999 999 999 991 473 487 170 7 × 2 = 1 + 0.799 999 999 999 982 946 974 341 4;
  • 2) 0.799 999 999 999 982 946 974 341 4 × 2 = 1 + 0.599 999 999 999 965 893 948 682 8;
  • 3) 0.599 999 999 999 965 893 948 682 8 × 2 = 1 + 0.199 999 999 999 931 787 897 365 6;
  • 4) 0.199 999 999 999 931 787 897 365 6 × 2 = 0 + 0.399 999 999 999 863 575 794 731 2;
  • 5) 0.399 999 999 999 863 575 794 731 2 × 2 = 0 + 0.799 999 999 999 727 151 589 462 4;
  • 6) 0.799 999 999 999 727 151 589 462 4 × 2 = 1 + 0.599 999 999 999 454 303 178 924 8;
  • 7) 0.599 999 999 999 454 303 178 924 8 × 2 = 1 + 0.199 999 999 998 908 606 357 849 6;
  • 8) 0.199 999 999 998 908 606 357 849 6 × 2 = 0 + 0.399 999 999 997 817 212 715 699 2;
  • 9) 0.399 999 999 997 817 212 715 699 2 × 2 = 0 + 0.799 999 999 995 634 425 431 398 4;
  • 10) 0.799 999 999 995 634 425 431 398 4 × 2 = 1 + 0.599 999 999 991 268 850 862 796 8;
  • 11) 0.599 999 999 991 268 850 862 796 8 × 2 = 1 + 0.199 999 999 982 537 701 725 593 6;
  • 12) 0.199 999 999 982 537 701 725 593 6 × 2 = 0 + 0.399 999 999 965 075 403 451 187 2;
  • 13) 0.399 999 999 965 075 403 451 187 2 × 2 = 0 + 0.799 999 999 930 150 806 902 374 4;
  • 14) 0.799 999 999 930 150 806 902 374 4 × 2 = 1 + 0.599 999 999 860 301 613 804 748 8;
  • 15) 0.599 999 999 860 301 613 804 748 8 × 2 = 1 + 0.199 999 999 720 603 227 609 497 6;
  • 16) 0.199 999 999 720 603 227 609 497 6 × 2 = 0 + 0.399 999 999 441 206 455 218 995 2;
  • 17) 0.399 999 999 441 206 455 218 995 2 × 2 = 0 + 0.799 999 998 882 412 910 437 990 4;
  • 18) 0.799 999 998 882 412 910 437 990 4 × 2 = 1 + 0.599 999 997 764 825 820 875 980 8;
  • 19) 0.599 999 997 764 825 820 875 980 8 × 2 = 1 + 0.199 999 995 529 651 641 751 961 6;
  • 20) 0.199 999 995 529 651 641 751 961 6 × 2 = 0 + 0.399 999 991 059 303 283 503 923 2;
  • 21) 0.399 999 991 059 303 283 503 923 2 × 2 = 0 + 0.799 999 982 118 606 567 007 846 4;
  • 22) 0.799 999 982 118 606 567 007 846 4 × 2 = 1 + 0.599 999 964 237 213 134 015 692 8;
  • 23) 0.599 999 964 237 213 134 015 692 8 × 2 = 1 + 0.199 999 928 474 426 268 031 385 6;
  • 24) 0.199 999 928 474 426 268 031 385 6 × 2 = 0 + 0.399 999 856 948 852 536 062 771 2;
  • 25) 0.399 999 856 948 852 536 062 771 2 × 2 = 0 + 0.799 999 713 897 705 072 125 542 4;
  • 26) 0.799 999 713 897 705 072 125 542 4 × 2 = 1 + 0.599 999 427 795 410 144 251 084 8;
  • 27) 0.599 999 427 795 410 144 251 084 8 × 2 = 1 + 0.199 998 855 590 820 288 502 169 6;
  • 28) 0.199 998 855 590 820 288 502 169 6 × 2 = 0 + 0.399 997 711 181 640 577 004 339 2;
  • 29) 0.399 997 711 181 640 577 004 339 2 × 2 = 0 + 0.799 995 422 363 281 154 008 678 4;
  • 30) 0.799 995 422 363 281 154 008 678 4 × 2 = 1 + 0.599 990 844 726 562 308 017 356 8;
  • 31) 0.599 990 844 726 562 308 017 356 8 × 2 = 1 + 0.199 981 689 453 124 616 034 713 6;
  • 32) 0.199 981 689 453 124 616 034 713 6 × 2 = 0 + 0.399 963 378 906 249 232 069 427 2;
  • 33) 0.399 963 378 906 249 232 069 427 2 × 2 = 0 + 0.799 926 757 812 498 464 138 854 4;
  • 34) 0.799 926 757 812 498 464 138 854 4 × 2 = 1 + 0.599 853 515 624 996 928 277 708 8;
  • 35) 0.599 853 515 624 996 928 277 708 8 × 2 = 1 + 0.199 707 031 249 993 856 555 417 6;
  • 36) 0.199 707 031 249 993 856 555 417 6 × 2 = 0 + 0.399 414 062 499 987 713 110 835 2;
  • 37) 0.399 414 062 499 987 713 110 835 2 × 2 = 0 + 0.798 828 124 999 975 426 221 670 4;
  • 38) 0.798 828 124 999 975 426 221 670 4 × 2 = 1 + 0.597 656 249 999 950 852 443 340 8;
  • 39) 0.597 656 249 999 950 852 443 340 8 × 2 = 1 + 0.195 312 499 999 901 704 886 681 6;
  • 40) 0.195 312 499 999 901 704 886 681 6 × 2 = 0 + 0.390 624 999 999 803 409 773 363 2;
  • 41) 0.390 624 999 999 803 409 773 363 2 × 2 = 0 + 0.781 249 999 999 606 819 546 726 4;
  • 42) 0.781 249 999 999 606 819 546 726 4 × 2 = 1 + 0.562 499 999 999 213 639 093 452 8;
  • 43) 0.562 499 999 999 213 639 093 452 8 × 2 = 1 + 0.124 999 999 998 427 278 186 905 6;
  • 44) 0.124 999 999 998 427 278 186 905 6 × 2 = 0 + 0.249 999 999 996 854 556 373 811 2;
  • 45) 0.249 999 999 996 854 556 373 811 2 × 2 = 0 + 0.499 999 999 993 709 112 747 622 4;
  • 46) 0.499 999 999 993 709 112 747 622 4 × 2 = 0 + 0.999 999 999 987 418 225 495 244 8;
  • 47) 0.999 999 999 987 418 225 495 244 8 × 2 = 1 + 0.999 999 999 974 836 450 990 489 6;
  • 48) 0.999 999 999 974 836 450 990 489 6 × 2 = 1 + 0.999 999 999 949 672 901 980 979 2;
  • 49) 0.999 999 999 949 672 901 980 979 2 × 2 = 1 + 0.999 999 999 899 345 803 961 958 4;
  • 50) 0.999 999 999 899 345 803 961 958 4 × 2 = 1 + 0.999 999 999 798 691 607 923 916 8;
  • 51) 0.999 999 999 798 691 607 923 916 8 × 2 = 1 + 0.999 999 999 597 383 215 847 833 6;
  • 52) 0.999 999 999 597 383 215 847 833 6 × 2 = 1 + 0.999 999 999 194 766 431 695 667 2;
  • 53) 0.999 999 999 194 766 431 695 667 2 × 2 = 1 + 0.999 999 998 389 532 863 391 334 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:


0.899 999 999 999 991 473 487 170 7(10) =

0.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2)

5. Positive number before normalization:

90.899 999 999 999 991 473 487 170 7(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(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:


90.899 999 999 999 991 473 487 170 7(10) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2) =

101 1010.1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0011 1111 1(2) × 20 =

1.0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000 1111 111(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 1000 1111 111


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 1000 111 1111 =

0110 1011 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1000


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 1000


Decimal number 90.899 999 999 999 991 473 487 170 7 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 1000

» Convert decimal 90.899 999 999 999 991 473 487 177 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