-0.016 738 891 601 562 496 530 381 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 381(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
-0.016 738 891 601 562 496 530 381(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:

|-0.016 738 891 601 562 496 530 381| = 0.016 738 891 601 562 496 530 381


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: 0.016 738 891 601 562 496 530 381.

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) 0.016 738 891 601 562 496 530 381 × 2 = 0 + 0.033 477 783 203 124 993 060 762;
  • 2) 0.033 477 783 203 124 993 060 762 × 2 = 0 + 0.066 955 566 406 249 986 121 524;
  • 3) 0.066 955 566 406 249 986 121 524 × 2 = 0 + 0.133 911 132 812 499 972 243 048;
  • 4) 0.133 911 132 812 499 972 243 048 × 2 = 0 + 0.267 822 265 624 999 944 486 096;
  • 5) 0.267 822 265 624 999 944 486 096 × 2 = 0 + 0.535 644 531 249 999 888 972 192;
  • 6) 0.535 644 531 249 999 888 972 192 × 2 = 1 + 0.071 289 062 499 999 777 944 384;
  • 7) 0.071 289 062 499 999 777 944 384 × 2 = 0 + 0.142 578 124 999 999 555 888 768;
  • 8) 0.142 578 124 999 999 555 888 768 × 2 = 0 + 0.285 156 249 999 999 111 777 536;
  • 9) 0.285 156 249 999 999 111 777 536 × 2 = 0 + 0.570 312 499 999 998 223 555 072;
  • 10) 0.570 312 499 999 998 223 555 072 × 2 = 1 + 0.140 624 999 999 996 447 110 144;
  • 11) 0.140 624 999 999 996 447 110 144 × 2 = 0 + 0.281 249 999 999 992 894 220 288;
  • 12) 0.281 249 999 999 992 894 220 288 × 2 = 0 + 0.562 499 999 999 985 788 440 576;
  • 13) 0.562 499 999 999 985 788 440 576 × 2 = 1 + 0.124 999 999 999 971 576 881 152;
  • 14) 0.124 999 999 999 971 576 881 152 × 2 = 0 + 0.249 999 999 999 943 153 762 304;
  • 15) 0.249 999 999 999 943 153 762 304 × 2 = 0 + 0.499 999 999 999 886 307 524 608;
  • 16) 0.499 999 999 999 886 307 524 608 × 2 = 0 + 0.999 999 999 999 772 615 049 216;
  • 17) 0.999 999 999 999 772 615 049 216 × 2 = 1 + 0.999 999 999 999 545 230 098 432;
  • 18) 0.999 999 999 999 545 230 098 432 × 2 = 1 + 0.999 999 999 999 090 460 196 864;
  • 19) 0.999 999 999 999 090 460 196 864 × 2 = 1 + 0.999 999 999 998 180 920 393 728;
  • 20) 0.999 999 999 998 180 920 393 728 × 2 = 1 + 0.999 999 999 996 361 840 787 456;
  • 21) 0.999 999 999 996 361 840 787 456 × 2 = 1 + 0.999 999 999 992 723 681 574 912;
  • 22) 0.999 999 999 992 723 681 574 912 × 2 = 1 + 0.999 999 999 985 447 363 149 824;
  • 23) 0.999 999 999 985 447 363 149 824 × 2 = 1 + 0.999 999 999 970 894 726 299 648;
  • 24) 0.999 999 999 970 894 726 299 648 × 2 = 1 + 0.999 999 999 941 789 452 599 296;
  • 25) 0.999 999 999 941 789 452 599 296 × 2 = 1 + 0.999 999 999 883 578 905 198 592;
  • 26) 0.999 999 999 883 578 905 198 592 × 2 = 1 + 0.999 999 999 767 157 810 397 184;
  • 27) 0.999 999 999 767 157 810 397 184 × 2 = 1 + 0.999 999 999 534 315 620 794 368;
  • 28) 0.999 999 999 534 315 620 794 368 × 2 = 1 + 0.999 999 999 068 631 241 588 736;
  • 29) 0.999 999 999 068 631 241 588 736 × 2 = 1 + 0.999 999 998 137 262 483 177 472;
  • 30) 0.999 999 998 137 262 483 177 472 × 2 = 1 + 0.999 999 996 274 524 966 354 944;
  • 31) 0.999 999 996 274 524 966 354 944 × 2 = 1 + 0.999 999 992 549 049 932 709 888;
  • 32) 0.999 999 992 549 049 932 709 888 × 2 = 1 + 0.999 999 985 098 099 865 419 776;
  • 33) 0.999 999 985 098 099 865 419 776 × 2 = 1 + 0.999 999 970 196 199 730 839 552;
  • 34) 0.999 999 970 196 199 730 839 552 × 2 = 1 + 0.999 999 940 392 399 461 679 104;
  • 35) 0.999 999 940 392 399 461 679 104 × 2 = 1 + 0.999 999 880 784 798 923 358 208;
  • 36) 0.999 999 880 784 798 923 358 208 × 2 = 1 + 0.999 999 761 569 597 846 716 416;
  • 37) 0.999 999 761 569 597 846 716 416 × 2 = 1 + 0.999 999 523 139 195 693 432 832;
  • 38) 0.999 999 523 139 195 693 432 832 × 2 = 1 + 0.999 999 046 278 391 386 865 664;
  • 39) 0.999 999 046 278 391 386 865 664 × 2 = 1 + 0.999 998 092 556 782 773 731 328;
  • 40) 0.999 998 092 556 782 773 731 328 × 2 = 1 + 0.999 996 185 113 565 547 462 656;
  • 41) 0.999 996 185 113 565 547 462 656 × 2 = 1 + 0.999 992 370 227 131 094 925 312;
  • 42) 0.999 992 370 227 131 094 925 312 × 2 = 1 + 0.999 984 740 454 262 189 850 624;
  • 43) 0.999 984 740 454 262 189 850 624 × 2 = 1 + 0.999 969 480 908 524 379 701 248;
  • 44) 0.999 969 480 908 524 379 701 248 × 2 = 1 + 0.999 938 961 817 048 759 402 496;
  • 45) 0.999 938 961 817 048 759 402 496 × 2 = 1 + 0.999 877 923 634 097 518 804 992;
  • 46) 0.999 877 923 634 097 518 804 992 × 2 = 1 + 0.999 755 847 268 195 037 609 984;
  • 47) 0.999 755 847 268 195 037 609 984 × 2 = 1 + 0.999 511 694 536 390 075 219 968;
  • 48) 0.999 511 694 536 390 075 219 968 × 2 = 1 + 0.999 023 389 072 780 150 439 936;
  • 49) 0.999 023 389 072 780 150 439 936 × 2 = 1 + 0.998 046 778 145 560 300 879 872;
  • 50) 0.998 046 778 145 560 300 879 872 × 2 = 1 + 0.996 093 556 291 120 601 759 744;
  • 51) 0.996 093 556 291 120 601 759 744 × 2 = 1 + 0.992 187 112 582 241 203 519 488;
  • 52) 0.992 187 112 582 241 203 519 488 × 2 = 1 + 0.984 374 225 164 482 407 038 976;
  • 53) 0.984 374 225 164 482 407 038 976 × 2 = 1 + 0.968 748 450 328 964 814 077 952;
  • 54) 0.968 748 450 328 964 814 077 952 × 2 = 1 + 0.937 496 900 657 929 628 155 904;
  • 55) 0.937 496 900 657 929 628 155 904 × 2 = 1 + 0.874 993 801 315 859 256 311 808;
  • 56) 0.874 993 801 315 859 256 311 808 × 2 = 1 + 0.749 987 602 631 718 512 623 616;
  • 57) 0.749 987 602 631 718 512 623 616 × 2 = 1 + 0.499 975 205 263 437 025 247 232;
  • 58) 0.499 975 205 263 437 025 247 232 × 2 = 0 + 0.999 950 410 526 874 050 494 464;

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:


0.016 738 891 601 562 496 530 381(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 381(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:


0.016 738 891 601 562 496 530 381(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 -0.016 738 891 601 562 496 530 381 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