-0.000 282 006 419 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 419(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
-0.000 282 006 419(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:

|-0.000 282 006 419| = 0.000 282 006 419


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: 0.000 282 006 419.

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) 0.000 282 006 419 × 2 = 0 + 0.000 564 012 838;
  • 2) 0.000 564 012 838 × 2 = 0 + 0.001 128 025 676;
  • 3) 0.001 128 025 676 × 2 = 0 + 0.002 256 051 352;
  • 4) 0.002 256 051 352 × 2 = 0 + 0.004 512 102 704;
  • 5) 0.004 512 102 704 × 2 = 0 + 0.009 024 205 408;
  • 6) 0.009 024 205 408 × 2 = 0 + 0.018 048 410 816;
  • 7) 0.018 048 410 816 × 2 = 0 + 0.036 096 821 632;
  • 8) 0.036 096 821 632 × 2 = 0 + 0.072 193 643 264;
  • 9) 0.072 193 643 264 × 2 = 0 + 0.144 387 286 528;
  • 10) 0.144 387 286 528 × 2 = 0 + 0.288 774 573 056;
  • 11) 0.288 774 573 056 × 2 = 0 + 0.577 549 146 112;
  • 12) 0.577 549 146 112 × 2 = 1 + 0.155 098 292 224;
  • 13) 0.155 098 292 224 × 2 = 0 + 0.310 196 584 448;
  • 14) 0.310 196 584 448 × 2 = 0 + 0.620 393 168 896;
  • 15) 0.620 393 168 896 × 2 = 1 + 0.240 786 337 792;
  • 16) 0.240 786 337 792 × 2 = 0 + 0.481 572 675 584;
  • 17) 0.481 572 675 584 × 2 = 0 + 0.963 145 351 168;
  • 18) 0.963 145 351 168 × 2 = 1 + 0.926 290 702 336;
  • 19) 0.926 290 702 336 × 2 = 1 + 0.852 581 404 672;
  • 20) 0.852 581 404 672 × 2 = 1 + 0.705 162 809 344;
  • 21) 0.705 162 809 344 × 2 = 1 + 0.410 325 618 688;
  • 22) 0.410 325 618 688 × 2 = 0 + 0.820 651 237 376;
  • 23) 0.820 651 237 376 × 2 = 1 + 0.641 302 474 752;
  • 24) 0.641 302 474 752 × 2 = 1 + 0.282 604 949 504;
  • 25) 0.282 604 949 504 × 2 = 0 + 0.565 209 899 008;
  • 26) 0.565 209 899 008 × 2 = 1 + 0.130 419 798 016;
  • 27) 0.130 419 798 016 × 2 = 0 + 0.260 839 596 032;
  • 28) 0.260 839 596 032 × 2 = 0 + 0.521 679 192 064;
  • 29) 0.521 679 192 064 × 2 = 1 + 0.043 358 384 128;
  • 30) 0.043 358 384 128 × 2 = 0 + 0.086 716 768 256;
  • 31) 0.086 716 768 256 × 2 = 0 + 0.173 433 536 512;
  • 32) 0.173 433 536 512 × 2 = 0 + 0.346 867 073 024;
  • 33) 0.346 867 073 024 × 2 = 0 + 0.693 734 146 048;
  • 34) 0.693 734 146 048 × 2 = 1 + 0.387 468 292 096;
  • 35) 0.387 468 292 096 × 2 = 0 + 0.774 936 584 192;
  • 36) 0.774 936 584 192 × 2 = 1 + 0.549 873 168 384;
  • 37) 0.549 873 168 384 × 2 = 1 + 0.099 746 336 768;
  • 38) 0.099 746 336 768 × 2 = 0 + 0.199 492 673 536;
  • 39) 0.199 492 673 536 × 2 = 0 + 0.398 985 347 072;
  • 40) 0.398 985 347 072 × 2 = 0 + 0.797 970 694 144;
  • 41) 0.797 970 694 144 × 2 = 1 + 0.595 941 388 288;
  • 42) 0.595 941 388 288 × 2 = 1 + 0.191 882 776 576;
  • 43) 0.191 882 776 576 × 2 = 0 + 0.383 765 553 152;
  • 44) 0.383 765 553 152 × 2 = 0 + 0.767 531 106 304;
  • 45) 0.767 531 106 304 × 2 = 1 + 0.535 062 212 608;
  • 46) 0.535 062 212 608 × 2 = 1 + 0.070 124 425 216;
  • 47) 0.070 124 425 216 × 2 = 0 + 0.140 248 850 432;
  • 48) 0.140 248 850 432 × 2 = 0 + 0.280 497 700 864;
  • 49) 0.280 497 700 864 × 2 = 0 + 0.560 995 401 728;
  • 50) 0.560 995 401 728 × 2 = 1 + 0.121 990 803 456;
  • 51) 0.121 990 803 456 × 2 = 0 + 0.243 981 606 912;
  • 52) 0.243 981 606 912 × 2 = 0 + 0.487 963 213 824;
  • 53) 0.487 963 213 824 × 2 = 0 + 0.975 926 427 648;
  • 54) 0.975 926 427 648 × 2 = 1 + 0.951 852 855 296;
  • 55) 0.951 852 855 296 × 2 = 1 + 0.903 705 710 592;
  • 56) 0.903 705 710 592 × 2 = 1 + 0.807 411 421 184;
  • 57) 0.807 411 421 184 × 2 = 1 + 0.614 822 842 368;
  • 58) 0.614 822 842 368 × 2 = 1 + 0.229 645 684 736;
  • 59) 0.229 645 684 736 × 2 = 0 + 0.459 291 369 472;
  • 60) 0.459 291 369 472 × 2 = 0 + 0.918 582 738 944;
  • 61) 0.918 582 738 944 × 2 = 1 + 0.837 165 477 888;
  • 62) 0.837 165 477 888 × 2 = 1 + 0.674 330 955 776;
  • 63) 0.674 330 955 776 × 2 = 1 + 0.348 661 911 552;
  • 64) 0.348 661 911 552 × 2 = 0 + 0.697 323 823 104;

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:


0.000 282 006 419(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110(2)

6. Positive number before normalization:

0.000 282 006 419(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 006 419(10) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110(2) × 20 =

1.0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110 =

0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110


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 0011

Mantissa (52 bits) =
0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110


Decimal number -0.000 282 006 419 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 1000 0101 1000 1100 1100 0100 0111 1100 1110

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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