64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 158 597.199 999 999 982 6 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 158 597.199 999 999 982 6(10) converted and written in 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: 158 597.
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;
  • 158 597 ÷ 2 = 79 298 + 1;
  • 79 298 ÷ 2 = 39 649 + 0;
  • 39 649 ÷ 2 = 19 824 + 1;
  • 19 824 ÷ 2 = 9 912 + 0;
  • 9 912 ÷ 2 = 4 956 + 0;
  • 4 956 ÷ 2 = 2 478 + 0;
  • 2 478 ÷ 2 = 1 239 + 0;
  • 1 239 ÷ 2 = 619 + 1;
  • 619 ÷ 2 = 309 + 1;
  • 309 ÷ 2 = 154 + 1;
  • 154 ÷ 2 = 77 + 0;
  • 77 ÷ 2 = 38 + 1;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 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.


158 597(10) =


10 0110 1011 1000 0101(2)


3. Convert to binary (base 2) the fractional part: 0.199 999 999 982 6.

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) 0.199 999 999 982 6 × 2 = 0 + 0.399 999 999 965 2;
  • 2) 0.399 999 999 965 2 × 2 = 0 + 0.799 999 999 930 4;
  • 3) 0.799 999 999 930 4 × 2 = 1 + 0.599 999 999 860 8;
  • 4) 0.599 999 999 860 8 × 2 = 1 + 0.199 999 999 721 6;
  • 5) 0.199 999 999 721 6 × 2 = 0 + 0.399 999 999 443 2;
  • 6) 0.399 999 999 443 2 × 2 = 0 + 0.799 999 998 886 4;
  • 7) 0.799 999 998 886 4 × 2 = 1 + 0.599 999 997 772 8;
  • 8) 0.599 999 997 772 8 × 2 = 1 + 0.199 999 995 545 6;
  • 9) 0.199 999 995 545 6 × 2 = 0 + 0.399 999 991 091 2;
  • 10) 0.399 999 991 091 2 × 2 = 0 + 0.799 999 982 182 4;
  • 11) 0.799 999 982 182 4 × 2 = 1 + 0.599 999 964 364 8;
  • 12) 0.599 999 964 364 8 × 2 = 1 + 0.199 999 928 729 6;
  • 13) 0.199 999 928 729 6 × 2 = 0 + 0.399 999 857 459 2;
  • 14) 0.399 999 857 459 2 × 2 = 0 + 0.799 999 714 918 4;
  • 15) 0.799 999 714 918 4 × 2 = 1 + 0.599 999 429 836 8;
  • 16) 0.599 999 429 836 8 × 2 = 1 + 0.199 998 859 673 6;
  • 17) 0.199 998 859 673 6 × 2 = 0 + 0.399 997 719 347 2;
  • 18) 0.399 997 719 347 2 × 2 = 0 + 0.799 995 438 694 4;
  • 19) 0.799 995 438 694 4 × 2 = 1 + 0.599 990 877 388 8;
  • 20) 0.599 990 877 388 8 × 2 = 1 + 0.199 981 754 777 6;
  • 21) 0.199 981 754 777 6 × 2 = 0 + 0.399 963 509 555 2;
  • 22) 0.399 963 509 555 2 × 2 = 0 + 0.799 927 019 110 4;
  • 23) 0.799 927 019 110 4 × 2 = 1 + 0.599 854 038 220 8;
  • 24) 0.599 854 038 220 8 × 2 = 1 + 0.199 708 076 441 6;
  • 25) 0.199 708 076 441 6 × 2 = 0 + 0.399 416 152 883 2;
  • 26) 0.399 416 152 883 2 × 2 = 0 + 0.798 832 305 766 4;
  • 27) 0.798 832 305 766 4 × 2 = 1 + 0.597 664 611 532 8;
  • 28) 0.597 664 611 532 8 × 2 = 1 + 0.195 329 223 065 6;
  • 29) 0.195 329 223 065 6 × 2 = 0 + 0.390 658 446 131 2;
  • 30) 0.390 658 446 131 2 × 2 = 0 + 0.781 316 892 262 4;
  • 31) 0.781 316 892 262 4 × 2 = 1 + 0.562 633 784 524 8;
  • 32) 0.562 633 784 524 8 × 2 = 1 + 0.125 267 569 049 6;
  • 33) 0.125 267 569 049 6 × 2 = 0 + 0.250 535 138 099 2;
  • 34) 0.250 535 138 099 2 × 2 = 0 + 0.501 070 276 198 4;
  • 35) 0.501 070 276 198 4 × 2 = 1 + 0.002 140 552 396 8;
  • 36) 0.002 140 552 396 8 × 2 = 0 + 0.004 281 104 793 6;
  • 37) 0.004 281 104 793 6 × 2 = 0 + 0.008 562 209 587 2;
  • 38) 0.008 562 209 587 2 × 2 = 0 + 0.017 124 419 174 4;
  • 39) 0.017 124 419 174 4 × 2 = 0 + 0.034 248 838 348 8;
  • 40) 0.034 248 838 348 8 × 2 = 0 + 0.068 497 676 697 6;
  • 41) 0.068 497 676 697 6 × 2 = 0 + 0.136 995 353 395 2;
  • 42) 0.136 995 353 395 2 × 2 = 0 + 0.273 990 706 790 4;
  • 43) 0.273 990 706 790 4 × 2 = 0 + 0.547 981 413 580 8;
  • 44) 0.547 981 413 580 8 × 2 = 1 + 0.095 962 827 161 6;
  • 45) 0.095 962 827 161 6 × 2 = 0 + 0.191 925 654 323 2;
  • 46) 0.191 925 654 323 2 × 2 = 0 + 0.383 851 308 646 4;
  • 47) 0.383 851 308 646 4 × 2 = 0 + 0.767 702 617 292 8;
  • 48) 0.767 702 617 292 8 × 2 = 1 + 0.535 405 234 585 6;
  • 49) 0.535 405 234 585 6 × 2 = 1 + 0.070 810 469 171 2;
  • 50) 0.070 810 469 171 2 × 2 = 0 + 0.141 620 938 342 4;
  • 51) 0.141 620 938 342 4 × 2 = 0 + 0.283 241 876 684 8;
  • 52) 0.283 241 876 684 8 × 2 = 0 + 0.566 483 753 369 6;
  • 53) 0.566 483 753 369 6 × 2 = 1 + 0.132 967 506 739 2;

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...)


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:


0.199 999 999 982 6(10) =


0.0011 0011 0011 0011 0011 0011 0011 0011 0010 0000 0001 0001 1000 1(2)


5. Positive number before normalization:

158 597.199 999 999 982 6(10) =


10 0110 1011 1000 0101.0011 0011 0011 0011 0011 0011 0011 0011 0010 0000 0001 0001 1000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the left, so that only one non zero digit remains to the left of it:


158 597.199 999 999 982 6(10) =


10 0110 1011 1000 0101.0011 0011 0011 0011 0011 0011 0011 0011 0010 0000 0001 0001 1000 1(2) =


10 0110 1011 1000 0101.0011 0011 0011 0011 0011 0011 0011 0011 0010 0000 0001 0001 1000 1(2) × 20 =


1.0011 0101 1100 0010 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 0000 1000 1100 01(2) × 217


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): 17


Mantissa (not normalized):
1.0011 0101 1100 0010 1001 1001 1001 1001 1001 1001 1001 1001 1001 0000 0000 1000 1100 01


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


17 + 2(11-1) - 1 =


(17 + 1 023)(10) =


1 040(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;
  • 1 040 ÷ 2 = 520 + 0;
  • 520 ÷ 2 = 260 + 0;
  • 260 ÷ 2 = 130 + 0;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 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) =


1040(10) =


100 0001 0000(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. 0011 0101 1100 0010 1001 1001 1001 1001 1001 1001 1001 1001 1001 00 0000 0010 0011 0001 =


0011 0101 1100 0010 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 0001 0000


Mantissa (52 bits) =
0011 0101 1100 0010 1001 1001 1001 1001 1001 1001 1001 1001 1001


The base ten decimal number 158 597.199 999 999 982 6 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 0000 - 0011 0101 1100 0010 1001 1001 1001 1001 1001 1001 1001 1001 1001

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

Number 2 968 081 121 459 873 565 900 803 878 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:19 UTC (GMT)
Number 126 895 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 495 870 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 536 980 111 220 801 564 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 82 600 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 94 226 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 8.163 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number -15 655 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 88.888 889 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
Number 104 857 168 884 986 026 393 converted from decimal system (written in base ten) to 64 bit double precision IEEE 754 binary floating point representation standard Jul 13 12:18 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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