64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 4.280 003 580 697 414 5 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 4.280 003 580 697 414 5(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: 4.
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;
  • 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.


4(10) =

100(2)


3. Convert to binary (base 2) the fractional part: 0.280 003 580 697 414 5.

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) 0.280 003 580 697 414 5 × 2 = 0 + 0.560 007 161 394 829;
  • 2) 0.560 007 161 394 829 × 2 = 1 + 0.120 014 322 789 658;
  • 3) 0.120 014 322 789 658 × 2 = 0 + 0.240 028 645 579 316;
  • 4) 0.240 028 645 579 316 × 2 = 0 + 0.480 057 291 158 632;
  • 5) 0.480 057 291 158 632 × 2 = 0 + 0.960 114 582 317 264;
  • 6) 0.960 114 582 317 264 × 2 = 1 + 0.920 229 164 634 528;
  • 7) 0.920 229 164 634 528 × 2 = 1 + 0.840 458 329 269 056;
  • 8) 0.840 458 329 269 056 × 2 = 1 + 0.680 916 658 538 112;
  • 9) 0.680 916 658 538 112 × 2 = 1 + 0.361 833 317 076 224;
  • 10) 0.361 833 317 076 224 × 2 = 0 + 0.723 666 634 152 448;
  • 11) 0.723 666 634 152 448 × 2 = 1 + 0.447 333 268 304 896;
  • 12) 0.447 333 268 304 896 × 2 = 0 + 0.894 666 536 609 792;
  • 13) 0.894 666 536 609 792 × 2 = 1 + 0.789 333 073 219 584;
  • 14) 0.789 333 073 219 584 × 2 = 1 + 0.578 666 146 439 168;
  • 15) 0.578 666 146 439 168 × 2 = 1 + 0.157 332 292 878 336;
  • 16) 0.157 332 292 878 336 × 2 = 0 + 0.314 664 585 756 672;
  • 17) 0.314 664 585 756 672 × 2 = 0 + 0.629 329 171 513 344;
  • 18) 0.629 329 171 513 344 × 2 = 1 + 0.258 658 343 026 688;
  • 19) 0.258 658 343 026 688 × 2 = 0 + 0.517 316 686 053 376;
  • 20) 0.517 316 686 053 376 × 2 = 1 + 0.034 633 372 106 752;
  • 21) 0.034 633 372 106 752 × 2 = 0 + 0.069 266 744 213 504;
  • 22) 0.069 266 744 213 504 × 2 = 0 + 0.138 533 488 427 008;
  • 23) 0.138 533 488 427 008 × 2 = 0 + 0.277 066 976 854 016;
  • 24) 0.277 066 976 854 016 × 2 = 0 + 0.554 133 953 708 032;
  • 25) 0.554 133 953 708 032 × 2 = 1 + 0.108 267 907 416 064;
  • 26) 0.108 267 907 416 064 × 2 = 0 + 0.216 535 814 832 128;
  • 27) 0.216 535 814 832 128 × 2 = 0 + 0.433 071 629 664 256;
  • 28) 0.433 071 629 664 256 × 2 = 0 + 0.866 143 259 328 512;
  • 29) 0.866 143 259 328 512 × 2 = 1 + 0.732 286 518 657 024;
  • 30) 0.732 286 518 657 024 × 2 = 1 + 0.464 573 037 314 048;
  • 31) 0.464 573 037 314 048 × 2 = 0 + 0.929 146 074 628 096;
  • 32) 0.929 146 074 628 096 × 2 = 1 + 0.858 292 149 256 192;
  • 33) 0.858 292 149 256 192 × 2 = 1 + 0.716 584 298 512 384;
  • 34) 0.716 584 298 512 384 × 2 = 1 + 0.433 168 597 024 768;
  • 35) 0.433 168 597 024 768 × 2 = 0 + 0.866 337 194 049 536;
  • 36) 0.866 337 194 049 536 × 2 = 1 + 0.732 674 388 099 072;
  • 37) 0.732 674 388 099 072 × 2 = 1 + 0.465 348 776 198 144;
  • 38) 0.465 348 776 198 144 × 2 = 0 + 0.930 697 552 396 288;
  • 39) 0.930 697 552 396 288 × 2 = 1 + 0.861 395 104 792 576;
  • 40) 0.861 395 104 792 576 × 2 = 1 + 0.722 790 209 585 152;
  • 41) 0.722 790 209 585 152 × 2 = 1 + 0.445 580 419 170 304;
  • 42) 0.445 580 419 170 304 × 2 = 0 + 0.891 160 838 340 608;
  • 43) 0.891 160 838 340 608 × 2 = 1 + 0.782 321 676 681 216;
  • 44) 0.782 321 676 681 216 × 2 = 1 + 0.564 643 353 362 432;
  • 45) 0.564 643 353 362 432 × 2 = 1 + 0.129 286 706 724 864;
  • 46) 0.129 286 706 724 864 × 2 = 0 + 0.258 573 413 449 728;
  • 47) 0.258 573 413 449 728 × 2 = 0 + 0.517 146 826 899 456;
  • 48) 0.517 146 826 899 456 × 2 = 1 + 0.034 293 653 798 912;
  • 49) 0.034 293 653 798 912 × 2 = 0 + 0.068 587 307 597 824;
  • 50) 0.068 587 307 597 824 × 2 = 0 + 0.137 174 615 195 648;
  • 51) 0.137 174 615 195 648 × 2 = 0 + 0.274 349 230 391 296;
  • 52) 0.274 349 230 391 296 × 2 = 0 + 0.548 698 460 782 592;
  • 53) 0.548 698 460 782 592 × 2 = 1 + 0.097 396 921 565 184;

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:


0.280 003 580 697 414 5(10) =

0.0100 0111 1010 1110 0101 0000 1000 1101 1101 1011 1011 1001 0000 1(2)


5. Positive number before normalization:

4.280 003 580 697 414 5(10) =

100.0100 0111 1010 1110 0101 0000 1000 1101 1101 1011 1011 1001 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


4.280 003 580 697 414 5(10) =

100.0100 0111 1010 1110 0101 0000 1000 1101 1101 1011 1011 1001 0000 1(2) =

100.0100 0111 1010 1110 0101 0000 1000 1101 1101 1011 1011 1001 0000 1(2) × 20 =

1.0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100 001(2) × 22


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): 2

Mantissa (not normalized):
1.0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100 001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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. 0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100 001 =

0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100


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 0001

Mantissa (52 bits) =
0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100


The base ten decimal number 4.280 003 580 697 414 5 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0001 - 0001 0001 1110 1011 1001 0100 0010 0011 0111 0110 1110 1110 0100

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

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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