-284.011 100 000 001 110 001 110 081 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 081(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
-284.011 100 000 001 110 001 110 081(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:

|-284.011 100 000 001 110 001 110 081| = 284.011 100 000 001 110 001 110 081


2. First, convert to binary (in base 2) the integer part: 284.
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;
  • 284 ÷ 2 = 142 + 0;
  • 142 ÷ 2 = 71 + 0;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.

284(10) =

1 0001 1100(2)


4. Convert to binary (base 2) the fractional part: 0.011 100 000 001 110 001 110 081.

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) 0.011 100 000 001 110 001 110 081 × 2 = 0 + 0.022 200 000 002 220 002 220 162;
  • 2) 0.022 200 000 002 220 002 220 162 × 2 = 0 + 0.044 400 000 004 440 004 440 324;
  • 3) 0.044 400 000 004 440 004 440 324 × 2 = 0 + 0.088 800 000 008 880 008 880 648;
  • 4) 0.088 800 000 008 880 008 880 648 × 2 = 0 + 0.177 600 000 017 760 017 761 296;
  • 5) 0.177 600 000 017 760 017 761 296 × 2 = 0 + 0.355 200 000 035 520 035 522 592;
  • 6) 0.355 200 000 035 520 035 522 592 × 2 = 0 + 0.710 400 000 071 040 071 045 184;
  • 7) 0.710 400 000 071 040 071 045 184 × 2 = 1 + 0.420 800 000 142 080 142 090 368;
  • 8) 0.420 800 000 142 080 142 090 368 × 2 = 0 + 0.841 600 000 284 160 284 180 736;
  • 9) 0.841 600 000 284 160 284 180 736 × 2 = 1 + 0.683 200 000 568 320 568 361 472;
  • 10) 0.683 200 000 568 320 568 361 472 × 2 = 1 + 0.366 400 001 136 641 136 722 944;
  • 11) 0.366 400 001 136 641 136 722 944 × 2 = 0 + 0.732 800 002 273 282 273 445 888;
  • 12) 0.732 800 002 273 282 273 445 888 × 2 = 1 + 0.465 600 004 546 564 546 891 776;
  • 13) 0.465 600 004 546 564 546 891 776 × 2 = 0 + 0.931 200 009 093 129 093 783 552;
  • 14) 0.931 200 009 093 129 093 783 552 × 2 = 1 + 0.862 400 018 186 258 187 567 104;
  • 15) 0.862 400 018 186 258 187 567 104 × 2 = 1 + 0.724 800 036 372 516 375 134 208;
  • 16) 0.724 800 036 372 516 375 134 208 × 2 = 1 + 0.449 600 072 745 032 750 268 416;
  • 17) 0.449 600 072 745 032 750 268 416 × 2 = 0 + 0.899 200 145 490 065 500 536 832;
  • 18) 0.899 200 145 490 065 500 536 832 × 2 = 1 + 0.798 400 290 980 131 001 073 664;
  • 19) 0.798 400 290 980 131 001 073 664 × 2 = 1 + 0.596 800 581 960 262 002 147 328;
  • 20) 0.596 800 581 960 262 002 147 328 × 2 = 1 + 0.193 601 163 920 524 004 294 656;
  • 21) 0.193 601 163 920 524 004 294 656 × 2 = 0 + 0.387 202 327 841 048 008 589 312;
  • 22) 0.387 202 327 841 048 008 589 312 × 2 = 0 + 0.774 404 655 682 096 017 178 624;
  • 23) 0.774 404 655 682 096 017 178 624 × 2 = 1 + 0.548 809 311 364 192 034 357 248;
  • 24) 0.548 809 311 364 192 034 357 248 × 2 = 1 + 0.097 618 622 728 384 068 714 496;
  • 25) 0.097 618 622 728 384 068 714 496 × 2 = 0 + 0.195 237 245 456 768 137 428 992;
  • 26) 0.195 237 245 456 768 137 428 992 × 2 = 0 + 0.390 474 490 913 536 274 857 984;
  • 27) 0.390 474 490 913 536 274 857 984 × 2 = 0 + 0.780 948 981 827 072 549 715 968;
  • 28) 0.780 948 981 827 072 549 715 968 × 2 = 1 + 0.561 897 963 654 145 099 431 936;
  • 29) 0.561 897 963 654 145 099 431 936 × 2 = 1 + 0.123 795 927 308 290 198 863 872;
  • 30) 0.123 795 927 308 290 198 863 872 × 2 = 0 + 0.247 591 854 616 580 397 727 744;
  • 31) 0.247 591 854 616 580 397 727 744 × 2 = 0 + 0.495 183 709 233 160 795 455 488;
  • 32) 0.495 183 709 233 160 795 455 488 × 2 = 0 + 0.990 367 418 466 321 590 910 976;
  • 33) 0.990 367 418 466 321 590 910 976 × 2 = 1 + 0.980 734 836 932 643 181 821 952;
  • 34) 0.980 734 836 932 643 181 821 952 × 2 = 1 + 0.961 469 673 865 286 363 643 904;
  • 35) 0.961 469 673 865 286 363 643 904 × 2 = 1 + 0.922 939 347 730 572 727 287 808;
  • 36) 0.922 939 347 730 572 727 287 808 × 2 = 1 + 0.845 878 695 461 145 454 575 616;
  • 37) 0.845 878 695 461 145 454 575 616 × 2 = 1 + 0.691 757 390 922 290 909 151 232;
  • 38) 0.691 757 390 922 290 909 151 232 × 2 = 1 + 0.383 514 781 844 581 818 302 464;
  • 39) 0.383 514 781 844 581 818 302 464 × 2 = 0 + 0.767 029 563 689 163 636 604 928;
  • 40) 0.767 029 563 689 163 636 604 928 × 2 = 1 + 0.534 059 127 378 327 273 209 856;
  • 41) 0.534 059 127 378 327 273 209 856 × 2 = 1 + 0.068 118 254 756 654 546 419 712;
  • 42) 0.068 118 254 756 654 546 419 712 × 2 = 0 + 0.136 236 509 513 309 092 839 424;
  • 43) 0.136 236 509 513 309 092 839 424 × 2 = 0 + 0.272 473 019 026 618 185 678 848;
  • 44) 0.272 473 019 026 618 185 678 848 × 2 = 0 + 0.544 946 038 053 236 371 357 696;
  • 45) 0.544 946 038 053 236 371 357 696 × 2 = 1 + 0.089 892 076 106 472 742 715 392;
  • 46) 0.089 892 076 106 472 742 715 392 × 2 = 0 + 0.179 784 152 212 945 485 430 784;
  • 47) 0.179 784 152 212 945 485 430 784 × 2 = 0 + 0.359 568 304 425 890 970 861 568;
  • 48) 0.359 568 304 425 890 970 861 568 × 2 = 0 + 0.719 136 608 851 781 941 723 136;
  • 49) 0.719 136 608 851 781 941 723 136 × 2 = 1 + 0.438 273 217 703 563 883 446 272;
  • 50) 0.438 273 217 703 563 883 446 272 × 2 = 0 + 0.876 546 435 407 127 766 892 544;
  • 51) 0.876 546 435 407 127 766 892 544 × 2 = 1 + 0.753 092 870 814 255 533 785 088;
  • 52) 0.753 092 870 814 255 533 785 088 × 2 = 1 + 0.506 185 741 628 511 067 570 176;
  • 53) 0.506 185 741 628 511 067 570 176 × 2 = 1 + 0.012 371 483 257 022 135 140 352;

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:


0.011 100 000 001 110 001 110 081(10) =

0.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2)

6. Positive number before normalization:

284.011 100 000 001 110 001 110 081(10) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left, so that only one non zero digit remains to the left of it:


284.011 100 000 001 110 001 110 081(10) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) × 20 =

1.0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) × 28


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): 8

Mantissa (not normalized):
1.0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

8 + 2(11-1) - 1 =

(8 + 1 023)(10) =

1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1031(10) =

100 0000 0111(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. 0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1 0001 0111 =

0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000


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 0111

Mantissa (52 bits) =
0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000


Decimal number -284.011 100 000 001 110 001 110 081 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0111 - 0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000

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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