-0.365 481 236 494 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.365 481 236 494 5(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
-0.365 481 236 494 5(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:

|-0.365 481 236 494 5| = 0.365 481 236 494 5


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

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.365 481 236 494 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) 0.365 481 236 494 5 × 2 = 0 + 0.730 962 472 989;
  • 2) 0.730 962 472 989 × 2 = 1 + 0.461 924 945 978;
  • 3) 0.461 924 945 978 × 2 = 0 + 0.923 849 891 956;
  • 4) 0.923 849 891 956 × 2 = 1 + 0.847 699 783 912;
  • 5) 0.847 699 783 912 × 2 = 1 + 0.695 399 567 824;
  • 6) 0.695 399 567 824 × 2 = 1 + 0.390 799 135 648;
  • 7) 0.390 799 135 648 × 2 = 0 + 0.781 598 271 296;
  • 8) 0.781 598 271 296 × 2 = 1 + 0.563 196 542 592;
  • 9) 0.563 196 542 592 × 2 = 1 + 0.126 393 085 184;
  • 10) 0.126 393 085 184 × 2 = 0 + 0.252 786 170 368;
  • 11) 0.252 786 170 368 × 2 = 0 + 0.505 572 340 736;
  • 12) 0.505 572 340 736 × 2 = 1 + 0.011 144 681 472;
  • 13) 0.011 144 681 472 × 2 = 0 + 0.022 289 362 944;
  • 14) 0.022 289 362 944 × 2 = 0 + 0.044 578 725 888;
  • 15) 0.044 578 725 888 × 2 = 0 + 0.089 157 451 776;
  • 16) 0.089 157 451 776 × 2 = 0 + 0.178 314 903 552;
  • 17) 0.178 314 903 552 × 2 = 0 + 0.356 629 807 104;
  • 18) 0.356 629 807 104 × 2 = 0 + 0.713 259 614 208;
  • 19) 0.713 259 614 208 × 2 = 1 + 0.426 519 228 416;
  • 20) 0.426 519 228 416 × 2 = 0 + 0.853 038 456 832;
  • 21) 0.853 038 456 832 × 2 = 1 + 0.706 076 913 664;
  • 22) 0.706 076 913 664 × 2 = 1 + 0.412 153 827 328;
  • 23) 0.412 153 827 328 × 2 = 0 + 0.824 307 654 656;
  • 24) 0.824 307 654 656 × 2 = 1 + 0.648 615 309 312;
  • 25) 0.648 615 309 312 × 2 = 1 + 0.297 230 618 624;
  • 26) 0.297 230 618 624 × 2 = 0 + 0.594 461 237 248;
  • 27) 0.594 461 237 248 × 2 = 1 + 0.188 922 474 496;
  • 28) 0.188 922 474 496 × 2 = 0 + 0.377 844 948 992;
  • 29) 0.377 844 948 992 × 2 = 0 + 0.755 689 897 984;
  • 30) 0.755 689 897 984 × 2 = 1 + 0.511 379 795 968;
  • 31) 0.511 379 795 968 × 2 = 1 + 0.022 759 591 936;
  • 32) 0.022 759 591 936 × 2 = 0 + 0.045 519 183 872;
  • 33) 0.045 519 183 872 × 2 = 0 + 0.091 038 367 744;
  • 34) 0.091 038 367 744 × 2 = 0 + 0.182 076 735 488;
  • 35) 0.182 076 735 488 × 2 = 0 + 0.364 153 470 976;
  • 36) 0.364 153 470 976 × 2 = 0 + 0.728 306 941 952;
  • 37) 0.728 306 941 952 × 2 = 1 + 0.456 613 883 904;
  • 38) 0.456 613 883 904 × 2 = 0 + 0.913 227 767 808;
  • 39) 0.913 227 767 808 × 2 = 1 + 0.826 455 535 616;
  • 40) 0.826 455 535 616 × 2 = 1 + 0.652 911 071 232;
  • 41) 0.652 911 071 232 × 2 = 1 + 0.305 822 142 464;
  • 42) 0.305 822 142 464 × 2 = 0 + 0.611 644 284 928;
  • 43) 0.611 644 284 928 × 2 = 1 + 0.223 288 569 856;
  • 44) 0.223 288 569 856 × 2 = 0 + 0.446 577 139 712;
  • 45) 0.446 577 139 712 × 2 = 0 + 0.893 154 279 424;
  • 46) 0.893 154 279 424 × 2 = 1 + 0.786 308 558 848;
  • 47) 0.786 308 558 848 × 2 = 1 + 0.572 617 117 696;
  • 48) 0.572 617 117 696 × 2 = 1 + 0.145 234 235 392;
  • 49) 0.145 234 235 392 × 2 = 0 + 0.290 468 470 784;
  • 50) 0.290 468 470 784 × 2 = 0 + 0.580 936 941 568;
  • 51) 0.580 936 941 568 × 2 = 1 + 0.161 873 883 136;
  • 52) 0.161 873 883 136 × 2 = 0 + 0.323 747 766 272;
  • 53) 0.323 747 766 272 × 2 = 0 + 0.647 495 532 544;
  • 54) 0.647 495 532 544 × 2 = 1 + 0.294 991 065 088;

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:


0.365 481 236 494 5(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 1011 1010 0111 0010 01(2)

6. Positive number before normalization:

0.365 481 236 494 5(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 1011 1010 0111 0010 01(2)

7. Normalize the binary representation of the number.

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


0.365 481 236 494 5(10) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 1011 1010 0111 0010 01(2) =

0.0101 1101 1001 0000 0010 1101 1010 0110 0000 1011 1010 0111 0010 01(2) × 20 =

1.0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001(2) × 2-2


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): -2

Mantissa (not normalized):
1.0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001


9. 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) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1021(10) =

011 1111 1101(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001 =

0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001


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) =
011 1111 1101

Mantissa (52 bits) =
0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001


Decimal number -0.365 481 236 494 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 0111 0110 0100 0000 1011 0110 1001 1000 0010 1110 1001 1100 1001

» Convert decimal -0.365 481 236 49 to 64 bit double precision IEEE 754 binary floating point representation standard

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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