-0.016 738 891 601 562 496 530 552 869 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 869(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
-0.016 738 891 601 562 496 530 552 869(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:

|-0.016 738 891 601 562 496 530 552 869| = 0.016 738 891 601 562 496 530 552 869


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: 0.016 738 891 601 562 496 530 552 869.

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) 0.016 738 891 601 562 496 530 552 869 × 2 = 0 + 0.033 477 783 203 124 993 061 105 738;
  • 2) 0.033 477 783 203 124 993 061 105 738 × 2 = 0 + 0.066 955 566 406 249 986 122 211 476;
  • 3) 0.066 955 566 406 249 986 122 211 476 × 2 = 0 + 0.133 911 132 812 499 972 244 422 952;
  • 4) 0.133 911 132 812 499 972 244 422 952 × 2 = 0 + 0.267 822 265 624 999 944 488 845 904;
  • 5) 0.267 822 265 624 999 944 488 845 904 × 2 = 0 + 0.535 644 531 249 999 888 977 691 808;
  • 6) 0.535 644 531 249 999 888 977 691 808 × 2 = 1 + 0.071 289 062 499 999 777 955 383 616;
  • 7) 0.071 289 062 499 999 777 955 383 616 × 2 = 0 + 0.142 578 124 999 999 555 910 767 232;
  • 8) 0.142 578 124 999 999 555 910 767 232 × 2 = 0 + 0.285 156 249 999 999 111 821 534 464;
  • 9) 0.285 156 249 999 999 111 821 534 464 × 2 = 0 + 0.570 312 499 999 998 223 643 068 928;
  • 10) 0.570 312 499 999 998 223 643 068 928 × 2 = 1 + 0.140 624 999 999 996 447 286 137 856;
  • 11) 0.140 624 999 999 996 447 286 137 856 × 2 = 0 + 0.281 249 999 999 992 894 572 275 712;
  • 12) 0.281 249 999 999 992 894 572 275 712 × 2 = 0 + 0.562 499 999 999 985 789 144 551 424;
  • 13) 0.562 499 999 999 985 789 144 551 424 × 2 = 1 + 0.124 999 999 999 971 578 289 102 848;
  • 14) 0.124 999 999 999 971 578 289 102 848 × 2 = 0 + 0.249 999 999 999 943 156 578 205 696;
  • 15) 0.249 999 999 999 943 156 578 205 696 × 2 = 0 + 0.499 999 999 999 886 313 156 411 392;
  • 16) 0.499 999 999 999 886 313 156 411 392 × 2 = 0 + 0.999 999 999 999 772 626 312 822 784;
  • 17) 0.999 999 999 999 772 626 312 822 784 × 2 = 1 + 0.999 999 999 999 545 252 625 645 568;
  • 18) 0.999 999 999 999 545 252 625 645 568 × 2 = 1 + 0.999 999 999 999 090 505 251 291 136;
  • 19) 0.999 999 999 999 090 505 251 291 136 × 2 = 1 + 0.999 999 999 998 181 010 502 582 272;
  • 20) 0.999 999 999 998 181 010 502 582 272 × 2 = 1 + 0.999 999 999 996 362 021 005 164 544;
  • 21) 0.999 999 999 996 362 021 005 164 544 × 2 = 1 + 0.999 999 999 992 724 042 010 329 088;
  • 22) 0.999 999 999 992 724 042 010 329 088 × 2 = 1 + 0.999 999 999 985 448 084 020 658 176;
  • 23) 0.999 999 999 985 448 084 020 658 176 × 2 = 1 + 0.999 999 999 970 896 168 041 316 352;
  • 24) 0.999 999 999 970 896 168 041 316 352 × 2 = 1 + 0.999 999 999 941 792 336 082 632 704;
  • 25) 0.999 999 999 941 792 336 082 632 704 × 2 = 1 + 0.999 999 999 883 584 672 165 265 408;
  • 26) 0.999 999 999 883 584 672 165 265 408 × 2 = 1 + 0.999 999 999 767 169 344 330 530 816;
  • 27) 0.999 999 999 767 169 344 330 530 816 × 2 = 1 + 0.999 999 999 534 338 688 661 061 632;
  • 28) 0.999 999 999 534 338 688 661 061 632 × 2 = 1 + 0.999 999 999 068 677 377 322 123 264;
  • 29) 0.999 999 999 068 677 377 322 123 264 × 2 = 1 + 0.999 999 998 137 354 754 644 246 528;
  • 30) 0.999 999 998 137 354 754 644 246 528 × 2 = 1 + 0.999 999 996 274 709 509 288 493 056;
  • 31) 0.999 999 996 274 709 509 288 493 056 × 2 = 1 + 0.999 999 992 549 419 018 576 986 112;
  • 32) 0.999 999 992 549 419 018 576 986 112 × 2 = 1 + 0.999 999 985 098 838 037 153 972 224;
  • 33) 0.999 999 985 098 838 037 153 972 224 × 2 = 1 + 0.999 999 970 197 676 074 307 944 448;
  • 34) 0.999 999 970 197 676 074 307 944 448 × 2 = 1 + 0.999 999 940 395 352 148 615 888 896;
  • 35) 0.999 999 940 395 352 148 615 888 896 × 2 = 1 + 0.999 999 880 790 704 297 231 777 792;
  • 36) 0.999 999 880 790 704 297 231 777 792 × 2 = 1 + 0.999 999 761 581 408 594 463 555 584;
  • 37) 0.999 999 761 581 408 594 463 555 584 × 2 = 1 + 0.999 999 523 162 817 188 927 111 168;
  • 38) 0.999 999 523 162 817 188 927 111 168 × 2 = 1 + 0.999 999 046 325 634 377 854 222 336;
  • 39) 0.999 999 046 325 634 377 854 222 336 × 2 = 1 + 0.999 998 092 651 268 755 708 444 672;
  • 40) 0.999 998 092 651 268 755 708 444 672 × 2 = 1 + 0.999 996 185 302 537 511 416 889 344;
  • 41) 0.999 996 185 302 537 511 416 889 344 × 2 = 1 + 0.999 992 370 605 075 022 833 778 688;
  • 42) 0.999 992 370 605 075 022 833 778 688 × 2 = 1 + 0.999 984 741 210 150 045 667 557 376;
  • 43) 0.999 984 741 210 150 045 667 557 376 × 2 = 1 + 0.999 969 482 420 300 091 335 114 752;
  • 44) 0.999 969 482 420 300 091 335 114 752 × 2 = 1 + 0.999 938 964 840 600 182 670 229 504;
  • 45) 0.999 938 964 840 600 182 670 229 504 × 2 = 1 + 0.999 877 929 681 200 365 340 459 008;
  • 46) 0.999 877 929 681 200 365 340 459 008 × 2 = 1 + 0.999 755 859 362 400 730 680 918 016;
  • 47) 0.999 755 859 362 400 730 680 918 016 × 2 = 1 + 0.999 511 718 724 801 461 361 836 032;
  • 48) 0.999 511 718 724 801 461 361 836 032 × 2 = 1 + 0.999 023 437 449 602 922 723 672 064;
  • 49) 0.999 023 437 449 602 922 723 672 064 × 2 = 1 + 0.998 046 874 899 205 845 447 344 128;
  • 50) 0.998 046 874 899 205 845 447 344 128 × 2 = 1 + 0.996 093 749 798 411 690 894 688 256;
  • 51) 0.996 093 749 798 411 690 894 688 256 × 2 = 1 + 0.992 187 499 596 823 381 789 376 512;
  • 52) 0.992 187 499 596 823 381 789 376 512 × 2 = 1 + 0.984 374 999 193 646 763 578 753 024;
  • 53) 0.984 374 999 193 646 763 578 753 024 × 2 = 1 + 0.968 749 998 387 293 527 157 506 048;
  • 54) 0.968 749 998 387 293 527 157 506 048 × 2 = 1 + 0.937 499 996 774 587 054 315 012 096;
  • 55) 0.937 499 996 774 587 054 315 012 096 × 2 = 1 + 0.874 999 993 549 174 108 630 024 192;
  • 56) 0.874 999 993 549 174 108 630 024 192 × 2 = 1 + 0.749 999 987 098 348 217 260 048 384;
  • 57) 0.749 999 987 098 348 217 260 048 384 × 2 = 1 + 0.499 999 974 196 696 434 520 096 768;
  • 58) 0.499 999 974 196 696 434 520 096 768 × 2 = 0 + 0.999 999 948 393 392 869 040 193 536;

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:


0.016 738 891 601 562 496 530 552 869(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 869(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 530 552 869(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 496 530 552 869 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 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