-2.447 581 244 357 921 461 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -2.447 581 244 357 921 461(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
-2.447 581 244 357 921 461(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. Start with the positive version of the number:

|-2.447 581 244 357 921 461| = 2.447 581 244 357 921 461


2. First, convert to binary (in base 2) the integer part: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.

2(10) =

10(2)


4. Convert to binary (base 2) the fractional part: 0.447 581 244 357 921 461.

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) 0.447 581 244 357 921 461 × 2 = 0 + 0.895 162 488 715 842 922;
  • 2) 0.895 162 488 715 842 922 × 2 = 1 + 0.790 324 977 431 685 844;
  • 3) 0.790 324 977 431 685 844 × 2 = 1 + 0.580 649 954 863 371 688;
  • 4) 0.580 649 954 863 371 688 × 2 = 1 + 0.161 299 909 726 743 376;
  • 5) 0.161 299 909 726 743 376 × 2 = 0 + 0.322 599 819 453 486 752;
  • 6) 0.322 599 819 453 486 752 × 2 = 0 + 0.645 199 638 906 973 504;
  • 7) 0.645 199 638 906 973 504 × 2 = 1 + 0.290 399 277 813 947 008;
  • 8) 0.290 399 277 813 947 008 × 2 = 0 + 0.580 798 555 627 894 016;
  • 9) 0.580 798 555 627 894 016 × 2 = 1 + 0.161 597 111 255 788 032;
  • 10) 0.161 597 111 255 788 032 × 2 = 0 + 0.323 194 222 511 576 064;
  • 11) 0.323 194 222 511 576 064 × 2 = 0 + 0.646 388 445 023 152 128;
  • 12) 0.646 388 445 023 152 128 × 2 = 1 + 0.292 776 890 046 304 256;
  • 13) 0.292 776 890 046 304 256 × 2 = 0 + 0.585 553 780 092 608 512;
  • 14) 0.585 553 780 092 608 512 × 2 = 1 + 0.171 107 560 185 217 024;
  • 15) 0.171 107 560 185 217 024 × 2 = 0 + 0.342 215 120 370 434 048;
  • 16) 0.342 215 120 370 434 048 × 2 = 0 + 0.684 430 240 740 868 096;
  • 17) 0.684 430 240 740 868 096 × 2 = 1 + 0.368 860 481 481 736 192;
  • 18) 0.368 860 481 481 736 192 × 2 = 0 + 0.737 720 962 963 472 384;
  • 19) 0.737 720 962 963 472 384 × 2 = 1 + 0.475 441 925 926 944 768;
  • 20) 0.475 441 925 926 944 768 × 2 = 0 + 0.950 883 851 853 889 536;
  • 21) 0.950 883 851 853 889 536 × 2 = 1 + 0.901 767 703 707 779 072;
  • 22) 0.901 767 703 707 779 072 × 2 = 1 + 0.803 535 407 415 558 144;
  • 23) 0.803 535 407 415 558 144 × 2 = 1 + 0.607 070 814 831 116 288;
  • 24) 0.607 070 814 831 116 288 × 2 = 1 + 0.214 141 629 662 232 576;
  • 25) 0.214 141 629 662 232 576 × 2 = 0 + 0.428 283 259 324 465 152;
  • 26) 0.428 283 259 324 465 152 × 2 = 0 + 0.856 566 518 648 930 304;
  • 27) 0.856 566 518 648 930 304 × 2 = 1 + 0.713 133 037 297 860 608;
  • 28) 0.713 133 037 297 860 608 × 2 = 1 + 0.426 266 074 595 721 216;
  • 29) 0.426 266 074 595 721 216 × 2 = 0 + 0.852 532 149 191 442 432;
  • 30) 0.852 532 149 191 442 432 × 2 = 1 + 0.705 064 298 382 884 864;
  • 31) 0.705 064 298 382 884 864 × 2 = 1 + 0.410 128 596 765 769 728;
  • 32) 0.410 128 596 765 769 728 × 2 = 0 + 0.820 257 193 531 539 456;
  • 33) 0.820 257 193 531 539 456 × 2 = 1 + 0.640 514 387 063 078 912;
  • 34) 0.640 514 387 063 078 912 × 2 = 1 + 0.281 028 774 126 157 824;
  • 35) 0.281 028 774 126 157 824 × 2 = 0 + 0.562 057 548 252 315 648;
  • 36) 0.562 057 548 252 315 648 × 2 = 1 + 0.124 115 096 504 631 296;
  • 37) 0.124 115 096 504 631 296 × 2 = 0 + 0.248 230 193 009 262 592;
  • 38) 0.248 230 193 009 262 592 × 2 = 0 + 0.496 460 386 018 525 184;
  • 39) 0.496 460 386 018 525 184 × 2 = 0 + 0.992 920 772 037 050 368;
  • 40) 0.992 920 772 037 050 368 × 2 = 1 + 0.985 841 544 074 100 736;
  • 41) 0.985 841 544 074 100 736 × 2 = 1 + 0.971 683 088 148 201 472;
  • 42) 0.971 683 088 148 201 472 × 2 = 1 + 0.943 366 176 296 402 944;
  • 43) 0.943 366 176 296 402 944 × 2 = 1 + 0.886 732 352 592 805 888;
  • 44) 0.886 732 352 592 805 888 × 2 = 1 + 0.773 464 705 185 611 776;
  • 45) 0.773 464 705 185 611 776 × 2 = 1 + 0.546 929 410 371 223 552;
  • 46) 0.546 929 410 371 223 552 × 2 = 1 + 0.093 858 820 742 447 104;
  • 47) 0.093 858 820 742 447 104 × 2 = 0 + 0.187 717 641 484 894 208;
  • 48) 0.187 717 641 484 894 208 × 2 = 0 + 0.375 435 282 969 788 416;
  • 49) 0.375 435 282 969 788 416 × 2 = 0 + 0.750 870 565 939 576 832;
  • 50) 0.750 870 565 939 576 832 × 2 = 1 + 0.501 741 131 879 153 664;
  • 51) 0.501 741 131 879 153 664 × 2 = 1 + 0.003 482 263 758 307 328;
  • 52) 0.003 482 263 758 307 328 × 2 = 0 + 0.006 964 527 516 614 656;
  • 53) 0.006 964 527 516 614 656 × 2 = 0 + 0.013 929 055 033 229 312;

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).


5. 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:


0.447 581 244 357 921 461(10) =

0.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2)

6. Positive number before normalization:

2.447 581 244 357 921 461(10) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2)

7. 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:


2.447 581 244 357 921 461(10) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2) =

10.0111 0010 1001 0100 1010 1111 0011 0110 1101 0001 1111 1100 0110 0(2) × 20 =

1.0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00(2) × 21


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): 1

Mantissa (not normalized):
1.0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00


9. 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)


10. 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;

11. 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)


12. 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 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011 00 =

0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
100 0000 0000

Mantissa (52 bits) =
0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011


Decimal number -2.447 581 244 357 921 461 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 0011 1001 0100 1010 0101 0111 1001 1011 0110 1000 1111 1110 0011

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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