Decimal to 64 Bit IEEE 754 Binary: Convert Number 17 119 642 101 780 175 996 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 17 119 642 101 780 175 996(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;
  • 17 119 642 101 780 175 996 ÷ 2 = 8 559 821 050 890 087 998 + 0;
  • 8 559 821 050 890 087 998 ÷ 2 = 4 279 910 525 445 043 999 + 0;
  • 4 279 910 525 445 043 999 ÷ 2 = 2 139 955 262 722 521 999 + 1;
  • 2 139 955 262 722 521 999 ÷ 2 = 1 069 977 631 361 260 999 + 1;
  • 1 069 977 631 361 260 999 ÷ 2 = 534 988 815 680 630 499 + 1;
  • 534 988 815 680 630 499 ÷ 2 = 267 494 407 840 315 249 + 1;
  • 267 494 407 840 315 249 ÷ 2 = 133 747 203 920 157 624 + 1;
  • 133 747 203 920 157 624 ÷ 2 = 66 873 601 960 078 812 + 0;
  • 66 873 601 960 078 812 ÷ 2 = 33 436 800 980 039 406 + 0;
  • 33 436 800 980 039 406 ÷ 2 = 16 718 400 490 019 703 + 0;
  • 16 718 400 490 019 703 ÷ 2 = 8 359 200 245 009 851 + 1;
  • 8 359 200 245 009 851 ÷ 2 = 4 179 600 122 504 925 + 1;
  • 4 179 600 122 504 925 ÷ 2 = 2 089 800 061 252 462 + 1;
  • 2 089 800 061 252 462 ÷ 2 = 1 044 900 030 626 231 + 0;
  • 1 044 900 030 626 231 ÷ 2 = 522 450 015 313 115 + 1;
  • 522 450 015 313 115 ÷ 2 = 261 225 007 656 557 + 1;
  • 261 225 007 656 557 ÷ 2 = 130 612 503 828 278 + 1;
  • 130 612 503 828 278 ÷ 2 = 65 306 251 914 139 + 0;
  • 65 306 251 914 139 ÷ 2 = 32 653 125 957 069 + 1;
  • 32 653 125 957 069 ÷ 2 = 16 326 562 978 534 + 1;
  • 16 326 562 978 534 ÷ 2 = 8 163 281 489 267 + 0;
  • 8 163 281 489 267 ÷ 2 = 4 081 640 744 633 + 1;
  • 4 081 640 744 633 ÷ 2 = 2 040 820 372 316 + 1;
  • 2 040 820 372 316 ÷ 2 = 1 020 410 186 158 + 0;
  • 1 020 410 186 158 ÷ 2 = 510 205 093 079 + 0;
  • 510 205 093 079 ÷ 2 = 255 102 546 539 + 1;
  • 255 102 546 539 ÷ 2 = 127 551 273 269 + 1;
  • 127 551 273 269 ÷ 2 = 63 775 636 634 + 1;
  • 63 775 636 634 ÷ 2 = 31 887 818 317 + 0;
  • 31 887 818 317 ÷ 2 = 15 943 909 158 + 1;
  • 15 943 909 158 ÷ 2 = 7 971 954 579 + 0;
  • 7 971 954 579 ÷ 2 = 3 985 977 289 + 1;
  • 3 985 977 289 ÷ 2 = 1 992 988 644 + 1;
  • 1 992 988 644 ÷ 2 = 996 494 322 + 0;
  • 996 494 322 ÷ 2 = 498 247 161 + 0;
  • 498 247 161 ÷ 2 = 249 123 580 + 1;
  • 249 123 580 ÷ 2 = 124 561 790 + 0;
  • 124 561 790 ÷ 2 = 62 280 895 + 0;
  • 62 280 895 ÷ 2 = 31 140 447 + 1;
  • 31 140 447 ÷ 2 = 15 570 223 + 1;
  • 15 570 223 ÷ 2 = 7 785 111 + 1;
  • 7 785 111 ÷ 2 = 3 892 555 + 1;
  • 3 892 555 ÷ 2 = 1 946 277 + 1;
  • 1 946 277 ÷ 2 = 973 138 + 1;
  • 973 138 ÷ 2 = 486 569 + 0;
  • 486 569 ÷ 2 = 243 284 + 1;
  • 243 284 ÷ 2 = 121 642 + 0;
  • 121 642 ÷ 2 = 60 821 + 0;
  • 60 821 ÷ 2 = 30 410 + 1;
  • 30 410 ÷ 2 = 15 205 + 0;
  • 15 205 ÷ 2 = 7 602 + 1;
  • 7 602 ÷ 2 = 3 801 + 0;
  • 3 801 ÷ 2 = 1 900 + 1;
  • 1 900 ÷ 2 = 950 + 0;
  • 950 ÷ 2 = 475 + 0;
  • 475 ÷ 2 = 237 + 1;
  • 237 ÷ 2 = 118 + 1;
  • 118 ÷ 2 = 59 + 0;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.

17 119 642 101 780 175 996(10) =


1110 1101 1001 0101 0010 1111 1100 1001 1010 1110 0110 1101 1101 1100 0111 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the left, so that only one non zero digit remains to the left of it:


17 119 642 101 780 175 996(10) =


1110 1101 1001 0101 0010 1111 1100 1001 1010 1110 0110 1101 1101 1100 0111 1100(2) =


1110 1101 1001 0101 0010 1111 1100 1001 1010 1110 0110 1101 1101 1100 0111 1100(2) × 20 =


1.1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011 1000 1111 100(2) × 263


4. 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): 63


Mantissa (not normalized):
1.1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011 1000 1111 100


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


63 + 2(11-1) - 1 =


(63 + 1 023)(10) =


1 086(10)


6. 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;
  • 1 086 ÷ 2 = 543 + 0;
  • 543 ÷ 2 = 271 + 1;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1086(10) =


100 0011 1110(2)


8. 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. 1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011 100 0111 1100 =


1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011


9. 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 0011 1110


Mantissa (52 bits) =
1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011


The base ten decimal number 17 119 642 101 780 175 996 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0011 1110 - 1101 1011 0010 1010 0101 1111 1001 0011 0101 1100 1101 1011 1011

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