-0.016 738 891 601 562 496 530 453 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 453(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
-0.016 738 891 601 562 496 530 453(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:

|-0.016 738 891 601 562 496 530 453| = 0.016 738 891 601 562 496 530 453


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: 0.016 738 891 601 562 496 530 453.

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) 0.016 738 891 601 562 496 530 453 × 2 = 0 + 0.033 477 783 203 124 993 060 906;
  • 2) 0.033 477 783 203 124 993 060 906 × 2 = 0 + 0.066 955 566 406 249 986 121 812;
  • 3) 0.066 955 566 406 249 986 121 812 × 2 = 0 + 0.133 911 132 812 499 972 243 624;
  • 4) 0.133 911 132 812 499 972 243 624 × 2 = 0 + 0.267 822 265 624 999 944 487 248;
  • 5) 0.267 822 265 624 999 944 487 248 × 2 = 0 + 0.535 644 531 249 999 888 974 496;
  • 6) 0.535 644 531 249 999 888 974 496 × 2 = 1 + 0.071 289 062 499 999 777 948 992;
  • 7) 0.071 289 062 499 999 777 948 992 × 2 = 0 + 0.142 578 124 999 999 555 897 984;
  • 8) 0.142 578 124 999 999 555 897 984 × 2 = 0 + 0.285 156 249 999 999 111 795 968;
  • 9) 0.285 156 249 999 999 111 795 968 × 2 = 0 + 0.570 312 499 999 998 223 591 936;
  • 10) 0.570 312 499 999 998 223 591 936 × 2 = 1 + 0.140 624 999 999 996 447 183 872;
  • 11) 0.140 624 999 999 996 447 183 872 × 2 = 0 + 0.281 249 999 999 992 894 367 744;
  • 12) 0.281 249 999 999 992 894 367 744 × 2 = 0 + 0.562 499 999 999 985 788 735 488;
  • 13) 0.562 499 999 999 985 788 735 488 × 2 = 1 + 0.124 999 999 999 971 577 470 976;
  • 14) 0.124 999 999 999 971 577 470 976 × 2 = 0 + 0.249 999 999 999 943 154 941 952;
  • 15) 0.249 999 999 999 943 154 941 952 × 2 = 0 + 0.499 999 999 999 886 309 883 904;
  • 16) 0.499 999 999 999 886 309 883 904 × 2 = 0 + 0.999 999 999 999 772 619 767 808;
  • 17) 0.999 999 999 999 772 619 767 808 × 2 = 1 + 0.999 999 999 999 545 239 535 616;
  • 18) 0.999 999 999 999 545 239 535 616 × 2 = 1 + 0.999 999 999 999 090 479 071 232;
  • 19) 0.999 999 999 999 090 479 071 232 × 2 = 1 + 0.999 999 999 998 180 958 142 464;
  • 20) 0.999 999 999 998 180 958 142 464 × 2 = 1 + 0.999 999 999 996 361 916 284 928;
  • 21) 0.999 999 999 996 361 916 284 928 × 2 = 1 + 0.999 999 999 992 723 832 569 856;
  • 22) 0.999 999 999 992 723 832 569 856 × 2 = 1 + 0.999 999 999 985 447 665 139 712;
  • 23) 0.999 999 999 985 447 665 139 712 × 2 = 1 + 0.999 999 999 970 895 330 279 424;
  • 24) 0.999 999 999 970 895 330 279 424 × 2 = 1 + 0.999 999 999 941 790 660 558 848;
  • 25) 0.999 999 999 941 790 660 558 848 × 2 = 1 + 0.999 999 999 883 581 321 117 696;
  • 26) 0.999 999 999 883 581 321 117 696 × 2 = 1 + 0.999 999 999 767 162 642 235 392;
  • 27) 0.999 999 999 767 162 642 235 392 × 2 = 1 + 0.999 999 999 534 325 284 470 784;
  • 28) 0.999 999 999 534 325 284 470 784 × 2 = 1 + 0.999 999 999 068 650 568 941 568;
  • 29) 0.999 999 999 068 650 568 941 568 × 2 = 1 + 0.999 999 998 137 301 137 883 136;
  • 30) 0.999 999 998 137 301 137 883 136 × 2 = 1 + 0.999 999 996 274 602 275 766 272;
  • 31) 0.999 999 996 274 602 275 766 272 × 2 = 1 + 0.999 999 992 549 204 551 532 544;
  • 32) 0.999 999 992 549 204 551 532 544 × 2 = 1 + 0.999 999 985 098 409 103 065 088;
  • 33) 0.999 999 985 098 409 103 065 088 × 2 = 1 + 0.999 999 970 196 818 206 130 176;
  • 34) 0.999 999 970 196 818 206 130 176 × 2 = 1 + 0.999 999 940 393 636 412 260 352;
  • 35) 0.999 999 940 393 636 412 260 352 × 2 = 1 + 0.999 999 880 787 272 824 520 704;
  • 36) 0.999 999 880 787 272 824 520 704 × 2 = 1 + 0.999 999 761 574 545 649 041 408;
  • 37) 0.999 999 761 574 545 649 041 408 × 2 = 1 + 0.999 999 523 149 091 298 082 816;
  • 38) 0.999 999 523 149 091 298 082 816 × 2 = 1 + 0.999 999 046 298 182 596 165 632;
  • 39) 0.999 999 046 298 182 596 165 632 × 2 = 1 + 0.999 998 092 596 365 192 331 264;
  • 40) 0.999 998 092 596 365 192 331 264 × 2 = 1 + 0.999 996 185 192 730 384 662 528;
  • 41) 0.999 996 185 192 730 384 662 528 × 2 = 1 + 0.999 992 370 385 460 769 325 056;
  • 42) 0.999 992 370 385 460 769 325 056 × 2 = 1 + 0.999 984 740 770 921 538 650 112;
  • 43) 0.999 984 740 770 921 538 650 112 × 2 = 1 + 0.999 969 481 541 843 077 300 224;
  • 44) 0.999 969 481 541 843 077 300 224 × 2 = 1 + 0.999 938 963 083 686 154 600 448;
  • 45) 0.999 938 963 083 686 154 600 448 × 2 = 1 + 0.999 877 926 167 372 309 200 896;
  • 46) 0.999 877 926 167 372 309 200 896 × 2 = 1 + 0.999 755 852 334 744 618 401 792;
  • 47) 0.999 755 852 334 744 618 401 792 × 2 = 1 + 0.999 511 704 669 489 236 803 584;
  • 48) 0.999 511 704 669 489 236 803 584 × 2 = 1 + 0.999 023 409 338 978 473 607 168;
  • 49) 0.999 023 409 338 978 473 607 168 × 2 = 1 + 0.998 046 818 677 956 947 214 336;
  • 50) 0.998 046 818 677 956 947 214 336 × 2 = 1 + 0.996 093 637 355 913 894 428 672;
  • 51) 0.996 093 637 355 913 894 428 672 × 2 = 1 + 0.992 187 274 711 827 788 857 344;
  • 52) 0.992 187 274 711 827 788 857 344 × 2 = 1 + 0.984 374 549 423 655 577 714 688;
  • 53) 0.984 374 549 423 655 577 714 688 × 2 = 1 + 0.968 749 098 847 311 155 429 376;
  • 54) 0.968 749 098 847 311 155 429 376 × 2 = 1 + 0.937 498 197 694 622 310 858 752;
  • 55) 0.937 498 197 694 622 310 858 752 × 2 = 1 + 0.874 996 395 389 244 621 717 504;
  • 56) 0.874 996 395 389 244 621 717 504 × 2 = 1 + 0.749 992 790 778 489 243 435 008;
  • 57) 0.749 992 790 778 489 243 435 008 × 2 = 1 + 0.499 985 581 556 978 486 870 016;
  • 58) 0.499 985 581 556 978 486 870 016 × 2 = 0 + 0.999 971 163 113 956 973 740 032;

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:


0.016 738 891 601 562 496 530 453(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 453(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:


0.016 738 891 601 562 496 530 453(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 -0.016 738 891 601 562 496 530 453 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