-3.402 725 076 143 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -3.402 725 076 143(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
-3.402 725 076 143(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:

|-3.402 725 076 143| = 3.402 725 076 143


2. 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;

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.

3(10) =

11(2)


4. Convert to binary (base 2) the fractional part: 0.402 725 076 143.

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) 0.402 725 076 143 × 2 = 0 + 0.805 450 152 286;
  • 2) 0.805 450 152 286 × 2 = 1 + 0.610 900 304 572;
  • 3) 0.610 900 304 572 × 2 = 1 + 0.221 800 609 144;
  • 4) 0.221 800 609 144 × 2 = 0 + 0.443 601 218 288;
  • 5) 0.443 601 218 288 × 2 = 0 + 0.887 202 436 576;
  • 6) 0.887 202 436 576 × 2 = 1 + 0.774 404 873 152;
  • 7) 0.774 404 873 152 × 2 = 1 + 0.548 809 746 304;
  • 8) 0.548 809 746 304 × 2 = 1 + 0.097 619 492 608;
  • 9) 0.097 619 492 608 × 2 = 0 + 0.195 238 985 216;
  • 10) 0.195 238 985 216 × 2 = 0 + 0.390 477 970 432;
  • 11) 0.390 477 970 432 × 2 = 0 + 0.780 955 940 864;
  • 12) 0.780 955 940 864 × 2 = 1 + 0.561 911 881 728;
  • 13) 0.561 911 881 728 × 2 = 1 + 0.123 823 763 456;
  • 14) 0.123 823 763 456 × 2 = 0 + 0.247 647 526 912;
  • 15) 0.247 647 526 912 × 2 = 0 + 0.495 295 053 824;
  • 16) 0.495 295 053 824 × 2 = 0 + 0.990 590 107 648;
  • 17) 0.990 590 107 648 × 2 = 1 + 0.981 180 215 296;
  • 18) 0.981 180 215 296 × 2 = 1 + 0.962 360 430 592;
  • 19) 0.962 360 430 592 × 2 = 1 + 0.924 720 861 184;
  • 20) 0.924 720 861 184 × 2 = 1 + 0.849 441 722 368;
  • 21) 0.849 441 722 368 × 2 = 1 + 0.698 883 444 736;
  • 22) 0.698 883 444 736 × 2 = 1 + 0.397 766 889 472;
  • 23) 0.397 766 889 472 × 2 = 0 + 0.795 533 778 944;
  • 24) 0.795 533 778 944 × 2 = 1 + 0.591 067 557 888;
  • 25) 0.591 067 557 888 × 2 = 1 + 0.182 135 115 776;
  • 26) 0.182 135 115 776 × 2 = 0 + 0.364 270 231 552;
  • 27) 0.364 270 231 552 × 2 = 0 + 0.728 540 463 104;
  • 28) 0.728 540 463 104 × 2 = 1 + 0.457 080 926 208;
  • 29) 0.457 080 926 208 × 2 = 0 + 0.914 161 852 416;
  • 30) 0.914 161 852 416 × 2 = 1 + 0.828 323 704 832;
  • 31) 0.828 323 704 832 × 2 = 1 + 0.656 647 409 664;
  • 32) 0.656 647 409 664 × 2 = 1 + 0.313 294 819 328;
  • 33) 0.313 294 819 328 × 2 = 0 + 0.626 589 638 656;
  • 34) 0.626 589 638 656 × 2 = 1 + 0.253 179 277 312;
  • 35) 0.253 179 277 312 × 2 = 0 + 0.506 358 554 624;
  • 36) 0.506 358 554 624 × 2 = 1 + 0.012 717 109 248;
  • 37) 0.012 717 109 248 × 2 = 0 + 0.025 434 218 496;
  • 38) 0.025 434 218 496 × 2 = 0 + 0.050 868 436 992;
  • 39) 0.050 868 436 992 × 2 = 0 + 0.101 736 873 984;
  • 40) 0.101 736 873 984 × 2 = 0 + 0.203 473 747 968;
  • 41) 0.203 473 747 968 × 2 = 0 + 0.406 947 495 936;
  • 42) 0.406 947 495 936 × 2 = 0 + 0.813 894 991 872;
  • 43) 0.813 894 991 872 × 2 = 1 + 0.627 789 983 744;
  • 44) 0.627 789 983 744 × 2 = 1 + 0.255 579 967 488;
  • 45) 0.255 579 967 488 × 2 = 0 + 0.511 159 934 976;
  • 46) 0.511 159 934 976 × 2 = 1 + 0.022 319 869 952;
  • 47) 0.022 319 869 952 × 2 = 0 + 0.044 639 739 904;
  • 48) 0.044 639 739 904 × 2 = 0 + 0.089 279 479 808;
  • 49) 0.089 279 479 808 × 2 = 0 + 0.178 558 959 616;
  • 50) 0.178 558 959 616 × 2 = 0 + 0.357 117 919 232;
  • 51) 0.357 117 919 232 × 2 = 0 + 0.714 235 838 464;
  • 52) 0.714 235 838 464 × 2 = 1 + 0.428 471 676 928;
  • 53) 0.428 471 676 928 × 2 = 0 + 0.856 943 353 856;

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:


0.402 725 076 143(10) =

0.0110 0111 0001 1000 1111 1101 1001 0111 0101 0000 0011 0100 0001 0(2)

6. Positive number before normalization:

3.402 725 076 143(10) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0101 0000 0011 0100 0001 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:


3.402 725 076 143(10) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0101 0000 0011 0100 0001 0(2) =

11.0110 0111 0001 1000 1111 1101 1001 0111 0101 0000 0011 0100 0001 0(2) × 20 =

1.1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000 10(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):
1.1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000 10


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. 1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000 10 =

1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000


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) =
1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000


Decimal number -3.402 725 076 143 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0000 - 1011 0011 1000 1100 0111 1110 1100 1011 1010 1000 0001 1010 0000

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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