-0.000 282 006 007 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 007(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
-0.000 282 006 007(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:

|-0.000 282 006 007| = 0.000 282 006 007


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: 0.000 282 006 007.

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) 0.000 282 006 007 × 2 = 0 + 0.000 564 012 014;
  • 2) 0.000 564 012 014 × 2 = 0 + 0.001 128 024 028;
  • 3) 0.001 128 024 028 × 2 = 0 + 0.002 256 048 056;
  • 4) 0.002 256 048 056 × 2 = 0 + 0.004 512 096 112;
  • 5) 0.004 512 096 112 × 2 = 0 + 0.009 024 192 224;
  • 6) 0.009 024 192 224 × 2 = 0 + 0.018 048 384 448;
  • 7) 0.018 048 384 448 × 2 = 0 + 0.036 096 768 896;
  • 8) 0.036 096 768 896 × 2 = 0 + 0.072 193 537 792;
  • 9) 0.072 193 537 792 × 2 = 0 + 0.144 387 075 584;
  • 10) 0.144 387 075 584 × 2 = 0 + 0.288 774 151 168;
  • 11) 0.288 774 151 168 × 2 = 0 + 0.577 548 302 336;
  • 12) 0.577 548 302 336 × 2 = 1 + 0.155 096 604 672;
  • 13) 0.155 096 604 672 × 2 = 0 + 0.310 193 209 344;
  • 14) 0.310 193 209 344 × 2 = 0 + 0.620 386 418 688;
  • 15) 0.620 386 418 688 × 2 = 1 + 0.240 772 837 376;
  • 16) 0.240 772 837 376 × 2 = 0 + 0.481 545 674 752;
  • 17) 0.481 545 674 752 × 2 = 0 + 0.963 091 349 504;
  • 18) 0.963 091 349 504 × 2 = 1 + 0.926 182 699 008;
  • 19) 0.926 182 699 008 × 2 = 1 + 0.852 365 398 016;
  • 20) 0.852 365 398 016 × 2 = 1 + 0.704 730 796 032;
  • 21) 0.704 730 796 032 × 2 = 1 + 0.409 461 592 064;
  • 22) 0.409 461 592 064 × 2 = 0 + 0.818 923 184 128;
  • 23) 0.818 923 184 128 × 2 = 1 + 0.637 846 368 256;
  • 24) 0.637 846 368 256 × 2 = 1 + 0.275 692 736 512;
  • 25) 0.275 692 736 512 × 2 = 0 + 0.551 385 473 024;
  • 26) 0.551 385 473 024 × 2 = 1 + 0.102 770 946 048;
  • 27) 0.102 770 946 048 × 2 = 0 + 0.205 541 892 096;
  • 28) 0.205 541 892 096 × 2 = 0 + 0.411 083 784 192;
  • 29) 0.411 083 784 192 × 2 = 0 + 0.822 167 568 384;
  • 30) 0.822 167 568 384 × 2 = 1 + 0.644 335 136 768;
  • 31) 0.644 335 136 768 × 2 = 1 + 0.288 670 273 536;
  • 32) 0.288 670 273 536 × 2 = 0 + 0.577 340 547 072;
  • 33) 0.577 340 547 072 × 2 = 1 + 0.154 681 094 144;
  • 34) 0.154 681 094 144 × 2 = 0 + 0.309 362 188 288;
  • 35) 0.309 362 188 288 × 2 = 0 + 0.618 724 376 576;
  • 36) 0.618 724 376 576 × 2 = 1 + 0.237 448 753 152;
  • 37) 0.237 448 753 152 × 2 = 0 + 0.474 897 506 304;
  • 38) 0.474 897 506 304 × 2 = 0 + 0.949 795 012 608;
  • 39) 0.949 795 012 608 × 2 = 1 + 0.899 590 025 216;
  • 40) 0.899 590 025 216 × 2 = 1 + 0.799 180 050 432;
  • 41) 0.799 180 050 432 × 2 = 1 + 0.598 360 100 864;
  • 42) 0.598 360 100 864 × 2 = 1 + 0.196 720 201 728;
  • 43) 0.196 720 201 728 × 2 = 0 + 0.393 440 403 456;
  • 44) 0.393 440 403 456 × 2 = 0 + 0.786 880 806 912;
  • 45) 0.786 880 806 912 × 2 = 1 + 0.573 761 613 824;
  • 46) 0.573 761 613 824 × 2 = 1 + 0.147 523 227 648;
  • 47) 0.147 523 227 648 × 2 = 0 + 0.295 046 455 296;
  • 48) 0.295 046 455 296 × 2 = 0 + 0.590 092 910 592;
  • 49) 0.590 092 910 592 × 2 = 1 + 0.180 185 821 184;
  • 50) 0.180 185 821 184 × 2 = 0 + 0.360 371 642 368;
  • 51) 0.360 371 642 368 × 2 = 0 + 0.720 743 284 736;
  • 52) 0.720 743 284 736 × 2 = 1 + 0.441 486 569 472;
  • 53) 0.441 486 569 472 × 2 = 0 + 0.882 973 138 944;
  • 54) 0.882 973 138 944 × 2 = 1 + 0.765 946 277 888;
  • 55) 0.765 946 277 888 × 2 = 1 + 0.531 892 555 776;
  • 56) 0.531 892 555 776 × 2 = 1 + 0.063 785 111 552;
  • 57) 0.063 785 111 552 × 2 = 0 + 0.127 570 223 104;
  • 58) 0.127 570 223 104 × 2 = 0 + 0.255 140 446 208;
  • 59) 0.255 140 446 208 × 2 = 0 + 0.510 280 892 416;
  • 60) 0.510 280 892 416 × 2 = 1 + 0.020 561 784 832;
  • 61) 0.020 561 784 832 × 2 = 0 + 0.041 123 569 664;
  • 62) 0.041 123 569 664 × 2 = 0 + 0.082 247 139 328;
  • 63) 0.082 247 139 328 × 2 = 0 + 0.164 494 278 656;
  • 64) 0.164 494 278 656 × 2 = 0 + 0.328 988 557 312;

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:


0.000 282 006 007(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000(2)

6. Positive number before normalization:

0.000 282 006 007(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000(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:


0.000 282 006 007(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000(2) × 20 =

1.0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000(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 0110 1001 0011 1100 1100 1001 0111 0001 0000


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 0110 1001 0011 1100 1100 1001 0111 0001 0000 =

0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 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) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 0000


Decimal number -0.000 282 006 007 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1001 0011 1100 1100 1001 0111 0001 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