-0.013 661 441 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.013 661 441(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
-0.013 661 441(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:

|-0.013 661 441| = 0.013 661 441


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: 0.013 661 441.

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) 0.013 661 441 × 2 = 0 + 0.027 322 882;
  • 2) 0.027 322 882 × 2 = 0 + 0.054 645 764;
  • 3) 0.054 645 764 × 2 = 0 + 0.109 291 528;
  • 4) 0.109 291 528 × 2 = 0 + 0.218 583 056;
  • 5) 0.218 583 056 × 2 = 0 + 0.437 166 112;
  • 6) 0.437 166 112 × 2 = 0 + 0.874 332 224;
  • 7) 0.874 332 224 × 2 = 1 + 0.748 664 448;
  • 8) 0.748 664 448 × 2 = 1 + 0.497 328 896;
  • 9) 0.497 328 896 × 2 = 0 + 0.994 657 792;
  • 10) 0.994 657 792 × 2 = 1 + 0.989 315 584;
  • 11) 0.989 315 584 × 2 = 1 + 0.978 631 168;
  • 12) 0.978 631 168 × 2 = 1 + 0.957 262 336;
  • 13) 0.957 262 336 × 2 = 1 + 0.914 524 672;
  • 14) 0.914 524 672 × 2 = 1 + 0.829 049 344;
  • 15) 0.829 049 344 × 2 = 1 + 0.658 098 688;
  • 16) 0.658 098 688 × 2 = 1 + 0.316 197 376;
  • 17) 0.316 197 376 × 2 = 0 + 0.632 394 752;
  • 18) 0.632 394 752 × 2 = 1 + 0.264 789 504;
  • 19) 0.264 789 504 × 2 = 0 + 0.529 579 008;
  • 20) 0.529 579 008 × 2 = 1 + 0.059 158 016;
  • 21) 0.059 158 016 × 2 = 0 + 0.118 316 032;
  • 22) 0.118 316 032 × 2 = 0 + 0.236 632 064;
  • 23) 0.236 632 064 × 2 = 0 + 0.473 264 128;
  • 24) 0.473 264 128 × 2 = 0 + 0.946 528 256;
  • 25) 0.946 528 256 × 2 = 1 + 0.893 056 512;
  • 26) 0.893 056 512 × 2 = 1 + 0.786 113 024;
  • 27) 0.786 113 024 × 2 = 1 + 0.572 226 048;
  • 28) 0.572 226 048 × 2 = 1 + 0.144 452 096;
  • 29) 0.144 452 096 × 2 = 0 + 0.288 904 192;
  • 30) 0.288 904 192 × 2 = 0 + 0.577 808 384;
  • 31) 0.577 808 384 × 2 = 1 + 0.155 616 768;
  • 32) 0.155 616 768 × 2 = 0 + 0.311 233 536;
  • 33) 0.311 233 536 × 2 = 0 + 0.622 467 072;
  • 34) 0.622 467 072 × 2 = 1 + 0.244 934 144;
  • 35) 0.244 934 144 × 2 = 0 + 0.489 868 288;
  • 36) 0.489 868 288 × 2 = 0 + 0.979 736 576;
  • 37) 0.979 736 576 × 2 = 1 + 0.959 473 152;
  • 38) 0.959 473 152 × 2 = 1 + 0.918 946 304;
  • 39) 0.918 946 304 × 2 = 1 + 0.837 892 608;
  • 40) 0.837 892 608 × 2 = 1 + 0.675 785 216;
  • 41) 0.675 785 216 × 2 = 1 + 0.351 570 432;
  • 42) 0.351 570 432 × 2 = 0 + 0.703 140 864;
  • 43) 0.703 140 864 × 2 = 1 + 0.406 281 728;
  • 44) 0.406 281 728 × 2 = 0 + 0.812 563 456;
  • 45) 0.812 563 456 × 2 = 1 + 0.625 126 912;
  • 46) 0.625 126 912 × 2 = 1 + 0.250 253 824;
  • 47) 0.250 253 824 × 2 = 0 + 0.500 507 648;
  • 48) 0.500 507 648 × 2 = 1 + 0.001 015 296;
  • 49) 0.001 015 296 × 2 = 0 + 0.002 030 592;
  • 50) 0.002 030 592 × 2 = 0 + 0.004 061 184;
  • 51) 0.004 061 184 × 2 = 0 + 0.008 122 368;
  • 52) 0.008 122 368 × 2 = 0 + 0.016 244 736;
  • 53) 0.016 244 736 × 2 = 0 + 0.032 489 472;
  • 54) 0.032 489 472 × 2 = 0 + 0.064 978 944;
  • 55) 0.064 978 944 × 2 = 0 + 0.129 957 888;
  • 56) 0.129 957 888 × 2 = 0 + 0.259 915 776;
  • 57) 0.259 915 776 × 2 = 0 + 0.519 831 552;
  • 58) 0.519 831 552 × 2 = 1 + 0.039 663 104;
  • 59) 0.039 663 104 × 2 = 0 + 0.079 326 208;

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:


0.013 661 441(10) =

0.0000 0011 0111 1111 0101 0000 1111 0010 0100 1111 1010 1101 0000 0000 010(2)

6. Positive number before normalization:

0.013 661 441(10) =

0.0000 0011 0111 1111 0101 0000 1111 0010 0100 1111 1010 1101 0000 0000 010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.013 661 441(10) =

0.0000 0011 0111 1111 0101 0000 1111 0010 0100 1111 1010 1101 0000 0000 010(2) =

0.0000 0011 0111 1111 0101 0000 1111 0010 0100 1111 1010 1101 0000 0000 010(2) × 20 =

1.1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010(2) × 2-7


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): -7

Mantissa (not normalized):
1.1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010 =

1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010


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 1000

Mantissa (52 bits) =
1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010


Decimal number -0.013 661 441 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 1011 1111 1010 1000 0111 1001 0010 0111 1101 0110 1000 0000 0010

» Convert decimal -0.013 661 342 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