3.123 145 321 678 979 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.123 145 321 678 979 9(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
3.123 145 321 678 979 9(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.123 145 321 678 979 9.

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) 0.123 145 321 678 979 9 × 2 = 0 + 0.246 290 643 357 959 8;
  • 2) 0.246 290 643 357 959 8 × 2 = 0 + 0.492 581 286 715 919 6;
  • 3) 0.492 581 286 715 919 6 × 2 = 0 + 0.985 162 573 431 839 2;
  • 4) 0.985 162 573 431 839 2 × 2 = 1 + 0.970 325 146 863 678 4;
  • 5) 0.970 325 146 863 678 4 × 2 = 1 + 0.940 650 293 727 356 8;
  • 6) 0.940 650 293 727 356 8 × 2 = 1 + 0.881 300 587 454 713 6;
  • 7) 0.881 300 587 454 713 6 × 2 = 1 + 0.762 601 174 909 427 2;
  • 8) 0.762 601 174 909 427 2 × 2 = 1 + 0.525 202 349 818 854 4;
  • 9) 0.525 202 349 818 854 4 × 2 = 1 + 0.050 404 699 637 708 8;
  • 10) 0.050 404 699 637 708 8 × 2 = 0 + 0.100 809 399 275 417 6;
  • 11) 0.100 809 399 275 417 6 × 2 = 0 + 0.201 618 798 550 835 2;
  • 12) 0.201 618 798 550 835 2 × 2 = 0 + 0.403 237 597 101 670 4;
  • 13) 0.403 237 597 101 670 4 × 2 = 0 + 0.806 475 194 203 340 8;
  • 14) 0.806 475 194 203 340 8 × 2 = 1 + 0.612 950 388 406 681 6;
  • 15) 0.612 950 388 406 681 6 × 2 = 1 + 0.225 900 776 813 363 2;
  • 16) 0.225 900 776 813 363 2 × 2 = 0 + 0.451 801 553 626 726 4;
  • 17) 0.451 801 553 626 726 4 × 2 = 0 + 0.903 603 107 253 452 8;
  • 18) 0.903 603 107 253 452 8 × 2 = 1 + 0.807 206 214 506 905 6;
  • 19) 0.807 206 214 506 905 6 × 2 = 1 + 0.614 412 429 013 811 2;
  • 20) 0.614 412 429 013 811 2 × 2 = 1 + 0.228 824 858 027 622 4;
  • 21) 0.228 824 858 027 622 4 × 2 = 0 + 0.457 649 716 055 244 8;
  • 22) 0.457 649 716 055 244 8 × 2 = 0 + 0.915 299 432 110 489 6;
  • 23) 0.915 299 432 110 489 6 × 2 = 1 + 0.830 598 864 220 979 2;
  • 24) 0.830 598 864 220 979 2 × 2 = 1 + 0.661 197 728 441 958 4;
  • 25) 0.661 197 728 441 958 4 × 2 = 1 + 0.322 395 456 883 916 8;
  • 26) 0.322 395 456 883 916 8 × 2 = 0 + 0.644 790 913 767 833 6;
  • 27) 0.644 790 913 767 833 6 × 2 = 1 + 0.289 581 827 535 667 2;
  • 28) 0.289 581 827 535 667 2 × 2 = 0 + 0.579 163 655 071 334 4;
  • 29) 0.579 163 655 071 334 4 × 2 = 1 + 0.158 327 310 142 668 8;
  • 30) 0.158 327 310 142 668 8 × 2 = 0 + 0.316 654 620 285 337 6;
  • 31) 0.316 654 620 285 337 6 × 2 = 0 + 0.633 309 240 570 675 2;
  • 32) 0.633 309 240 570 675 2 × 2 = 1 + 0.266 618 481 141 350 4;
  • 33) 0.266 618 481 141 350 4 × 2 = 0 + 0.533 236 962 282 700 8;
  • 34) 0.533 236 962 282 700 8 × 2 = 1 + 0.066 473 924 565 401 6;
  • 35) 0.066 473 924 565 401 6 × 2 = 0 + 0.132 947 849 130 803 2;
  • 36) 0.132 947 849 130 803 2 × 2 = 0 + 0.265 895 698 261 606 4;
  • 37) 0.265 895 698 261 606 4 × 2 = 0 + 0.531 791 396 523 212 8;
  • 38) 0.531 791 396 523 212 8 × 2 = 1 + 0.063 582 793 046 425 6;
  • 39) 0.063 582 793 046 425 6 × 2 = 0 + 0.127 165 586 092 851 2;
  • 40) 0.127 165 586 092 851 2 × 2 = 0 + 0.254 331 172 185 702 4;
  • 41) 0.254 331 172 185 702 4 × 2 = 0 + 0.508 662 344 371 404 8;
  • 42) 0.508 662 344 371 404 8 × 2 = 1 + 0.017 324 688 742 809 6;
  • 43) 0.017 324 688 742 809 6 × 2 = 0 + 0.034 649 377 485 619 2;
  • 44) 0.034 649 377 485 619 2 × 2 = 0 + 0.069 298 754 971 238 4;
  • 45) 0.069 298 754 971 238 4 × 2 = 0 + 0.138 597 509 942 476 8;
  • 46) 0.138 597 509 942 476 8 × 2 = 0 + 0.277 195 019 884 953 6;
  • 47) 0.277 195 019 884 953 6 × 2 = 0 + 0.554 390 039 769 907 2;
  • 48) 0.554 390 039 769 907 2 × 2 = 1 + 0.108 780 079 539 814 4;
  • 49) 0.108 780 079 539 814 4 × 2 = 0 + 0.217 560 159 079 628 8;
  • 50) 0.217 560 159 079 628 8 × 2 = 0 + 0.435 120 318 159 257 6;
  • 51) 0.435 120 318 159 257 6 × 2 = 0 + 0.870 240 636 318 515 2;
  • 52) 0.870 240 636 318 515 2 × 2 = 1 + 0.740 481 272 637 030 4;
  • 53) 0.740 481 272 637 030 4 × 2 = 1 + 0.480 962 545 274 060 8;

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:


0.123 145 321 678 979 9(10) =

0.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0001 0001 1(2)

5. Positive number before normalization:

3.123 145 321 678 979 9(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0001 0001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.123 145 321 678 979 9(10) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0001 0001 1(2) =

11.0001 1111 1000 0110 0111 0011 1010 1001 0100 0100 0100 0001 0001 1(2) × 20 =

1.1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 1000 11(2) × 21


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): 1

Mantissa (not normalized):
1.1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 1000 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 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) =

1024(10) =

100 0000 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. 1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 1000 11 =

1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 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 0000

Mantissa (52 bits) =
1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 1000


Decimal number 3.123 145 321 678 979 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1000 1111 1100 0011 0011 1001 1101 0100 1010 0010 0010 0000 1000

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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