-284.011 100 000 001 110 001 110 099 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 099(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
-284.011 100 000 001 110 001 110 099(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:

|-284.011 100 000 001 110 001 110 099| = 284.011 100 000 001 110 001 110 099


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: 0.011 100 000 001 110 001 110 099.

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) 0.011 100 000 001 110 001 110 099 × 2 = 0 + 0.022 200 000 002 220 002 220 198;
  • 2) 0.022 200 000 002 220 002 220 198 × 2 = 0 + 0.044 400 000 004 440 004 440 396;
  • 3) 0.044 400 000 004 440 004 440 396 × 2 = 0 + 0.088 800 000 008 880 008 880 792;
  • 4) 0.088 800 000 008 880 008 880 792 × 2 = 0 + 0.177 600 000 017 760 017 761 584;
  • 5) 0.177 600 000 017 760 017 761 584 × 2 = 0 + 0.355 200 000 035 520 035 523 168;
  • 6) 0.355 200 000 035 520 035 523 168 × 2 = 0 + 0.710 400 000 071 040 071 046 336;
  • 7) 0.710 400 000 071 040 071 046 336 × 2 = 1 + 0.420 800 000 142 080 142 092 672;
  • 8) 0.420 800 000 142 080 142 092 672 × 2 = 0 + 0.841 600 000 284 160 284 185 344;
  • 9) 0.841 600 000 284 160 284 185 344 × 2 = 1 + 0.683 200 000 568 320 568 370 688;
  • 10) 0.683 200 000 568 320 568 370 688 × 2 = 1 + 0.366 400 001 136 641 136 741 376;
  • 11) 0.366 400 001 136 641 136 741 376 × 2 = 0 + 0.732 800 002 273 282 273 482 752;
  • 12) 0.732 800 002 273 282 273 482 752 × 2 = 1 + 0.465 600 004 546 564 546 965 504;
  • 13) 0.465 600 004 546 564 546 965 504 × 2 = 0 + 0.931 200 009 093 129 093 931 008;
  • 14) 0.931 200 009 093 129 093 931 008 × 2 = 1 + 0.862 400 018 186 258 187 862 016;
  • 15) 0.862 400 018 186 258 187 862 016 × 2 = 1 + 0.724 800 036 372 516 375 724 032;
  • 16) 0.724 800 036 372 516 375 724 032 × 2 = 1 + 0.449 600 072 745 032 751 448 064;
  • 17) 0.449 600 072 745 032 751 448 064 × 2 = 0 + 0.899 200 145 490 065 502 896 128;
  • 18) 0.899 200 145 490 065 502 896 128 × 2 = 1 + 0.798 400 290 980 131 005 792 256;
  • 19) 0.798 400 290 980 131 005 792 256 × 2 = 1 + 0.596 800 581 960 262 011 584 512;
  • 20) 0.596 800 581 960 262 011 584 512 × 2 = 1 + 0.193 601 163 920 524 023 169 024;
  • 21) 0.193 601 163 920 524 023 169 024 × 2 = 0 + 0.387 202 327 841 048 046 338 048;
  • 22) 0.387 202 327 841 048 046 338 048 × 2 = 0 + 0.774 404 655 682 096 092 676 096;
  • 23) 0.774 404 655 682 096 092 676 096 × 2 = 1 + 0.548 809 311 364 192 185 352 192;
  • 24) 0.548 809 311 364 192 185 352 192 × 2 = 1 + 0.097 618 622 728 384 370 704 384;
  • 25) 0.097 618 622 728 384 370 704 384 × 2 = 0 + 0.195 237 245 456 768 741 408 768;
  • 26) 0.195 237 245 456 768 741 408 768 × 2 = 0 + 0.390 474 490 913 537 482 817 536;
  • 27) 0.390 474 490 913 537 482 817 536 × 2 = 0 + 0.780 948 981 827 074 965 635 072;
  • 28) 0.780 948 981 827 074 965 635 072 × 2 = 1 + 0.561 897 963 654 149 931 270 144;
  • 29) 0.561 897 963 654 149 931 270 144 × 2 = 1 + 0.123 795 927 308 299 862 540 288;
  • 30) 0.123 795 927 308 299 862 540 288 × 2 = 0 + 0.247 591 854 616 599 725 080 576;
  • 31) 0.247 591 854 616 599 725 080 576 × 2 = 0 + 0.495 183 709 233 199 450 161 152;
  • 32) 0.495 183 709 233 199 450 161 152 × 2 = 0 + 0.990 367 418 466 398 900 322 304;
  • 33) 0.990 367 418 466 398 900 322 304 × 2 = 1 + 0.980 734 836 932 797 800 644 608;
  • 34) 0.980 734 836 932 797 800 644 608 × 2 = 1 + 0.961 469 673 865 595 601 289 216;
  • 35) 0.961 469 673 865 595 601 289 216 × 2 = 1 + 0.922 939 347 731 191 202 578 432;
  • 36) 0.922 939 347 731 191 202 578 432 × 2 = 1 + 0.845 878 695 462 382 405 156 864;
  • 37) 0.845 878 695 462 382 405 156 864 × 2 = 1 + 0.691 757 390 924 764 810 313 728;
  • 38) 0.691 757 390 924 764 810 313 728 × 2 = 1 + 0.383 514 781 849 529 620 627 456;
  • 39) 0.383 514 781 849 529 620 627 456 × 2 = 0 + 0.767 029 563 699 059 241 254 912;
  • 40) 0.767 029 563 699 059 241 254 912 × 2 = 1 + 0.534 059 127 398 118 482 509 824;
  • 41) 0.534 059 127 398 118 482 509 824 × 2 = 1 + 0.068 118 254 796 236 965 019 648;
  • 42) 0.068 118 254 796 236 965 019 648 × 2 = 0 + 0.136 236 509 592 473 930 039 296;
  • 43) 0.136 236 509 592 473 930 039 296 × 2 = 0 + 0.272 473 019 184 947 860 078 592;
  • 44) 0.272 473 019 184 947 860 078 592 × 2 = 0 + 0.544 946 038 369 895 720 157 184;
  • 45) 0.544 946 038 369 895 720 157 184 × 2 = 1 + 0.089 892 076 739 791 440 314 368;
  • 46) 0.089 892 076 739 791 440 314 368 × 2 = 0 + 0.179 784 153 479 582 880 628 736;
  • 47) 0.179 784 153 479 582 880 628 736 × 2 = 0 + 0.359 568 306 959 165 761 257 472;
  • 48) 0.359 568 306 959 165 761 257 472 × 2 = 0 + 0.719 136 613 918 331 522 514 944;
  • 49) 0.719 136 613 918 331 522 514 944 × 2 = 1 + 0.438 273 227 836 663 045 029 888;
  • 50) 0.438 273 227 836 663 045 029 888 × 2 = 0 + 0.876 546 455 673 326 090 059 776;
  • 51) 0.876 546 455 673 326 090 059 776 × 2 = 1 + 0.753 092 911 346 652 180 119 552;
  • 52) 0.753 092 911 346 652 180 119 552 × 2 = 1 + 0.506 185 822 693 304 360 239 104;
  • 53) 0.506 185 822 693 304 360 239 104 × 2 = 1 + 0.012 371 645 386 608 720 478 208;

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:


0.011 100 000 001 110 001 110 099(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 099(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:


284.011 100 000 001 110 001 110 099(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 -284.011 100 000 001 110 001 110 099 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

» Convert decimal -284.011 100 000 001 110 001 110 034 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