-0.016 738 891 601 562 496 530 553 021 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 021(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
-0.016 738 891 601 562 496 530 553 021(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:

|-0.016 738 891 601 562 496 530 553 021| = 0.016 738 891 601 562 496 530 553 021


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: 0.016 738 891 601 562 496 530 553 021.

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) 0.016 738 891 601 562 496 530 553 021 × 2 = 0 + 0.033 477 783 203 124 993 061 106 042;
  • 2) 0.033 477 783 203 124 993 061 106 042 × 2 = 0 + 0.066 955 566 406 249 986 122 212 084;
  • 3) 0.066 955 566 406 249 986 122 212 084 × 2 = 0 + 0.133 911 132 812 499 972 244 424 168;
  • 4) 0.133 911 132 812 499 972 244 424 168 × 2 = 0 + 0.267 822 265 624 999 944 488 848 336;
  • 5) 0.267 822 265 624 999 944 488 848 336 × 2 = 0 + 0.535 644 531 249 999 888 977 696 672;
  • 6) 0.535 644 531 249 999 888 977 696 672 × 2 = 1 + 0.071 289 062 499 999 777 955 393 344;
  • 7) 0.071 289 062 499 999 777 955 393 344 × 2 = 0 + 0.142 578 124 999 999 555 910 786 688;
  • 8) 0.142 578 124 999 999 555 910 786 688 × 2 = 0 + 0.285 156 249 999 999 111 821 573 376;
  • 9) 0.285 156 249 999 999 111 821 573 376 × 2 = 0 + 0.570 312 499 999 998 223 643 146 752;
  • 10) 0.570 312 499 999 998 223 643 146 752 × 2 = 1 + 0.140 624 999 999 996 447 286 293 504;
  • 11) 0.140 624 999 999 996 447 286 293 504 × 2 = 0 + 0.281 249 999 999 992 894 572 587 008;
  • 12) 0.281 249 999 999 992 894 572 587 008 × 2 = 0 + 0.562 499 999 999 985 789 145 174 016;
  • 13) 0.562 499 999 999 985 789 145 174 016 × 2 = 1 + 0.124 999 999 999 971 578 290 348 032;
  • 14) 0.124 999 999 999 971 578 290 348 032 × 2 = 0 + 0.249 999 999 999 943 156 580 696 064;
  • 15) 0.249 999 999 999 943 156 580 696 064 × 2 = 0 + 0.499 999 999 999 886 313 161 392 128;
  • 16) 0.499 999 999 999 886 313 161 392 128 × 2 = 0 + 0.999 999 999 999 772 626 322 784 256;
  • 17) 0.999 999 999 999 772 626 322 784 256 × 2 = 1 + 0.999 999 999 999 545 252 645 568 512;
  • 18) 0.999 999 999 999 545 252 645 568 512 × 2 = 1 + 0.999 999 999 999 090 505 291 137 024;
  • 19) 0.999 999 999 999 090 505 291 137 024 × 2 = 1 + 0.999 999 999 998 181 010 582 274 048;
  • 20) 0.999 999 999 998 181 010 582 274 048 × 2 = 1 + 0.999 999 999 996 362 021 164 548 096;
  • 21) 0.999 999 999 996 362 021 164 548 096 × 2 = 1 + 0.999 999 999 992 724 042 329 096 192;
  • 22) 0.999 999 999 992 724 042 329 096 192 × 2 = 1 + 0.999 999 999 985 448 084 658 192 384;
  • 23) 0.999 999 999 985 448 084 658 192 384 × 2 = 1 + 0.999 999 999 970 896 169 316 384 768;
  • 24) 0.999 999 999 970 896 169 316 384 768 × 2 = 1 + 0.999 999 999 941 792 338 632 769 536;
  • 25) 0.999 999 999 941 792 338 632 769 536 × 2 = 1 + 0.999 999 999 883 584 677 265 539 072;
  • 26) 0.999 999 999 883 584 677 265 539 072 × 2 = 1 + 0.999 999 999 767 169 354 531 078 144;
  • 27) 0.999 999 999 767 169 354 531 078 144 × 2 = 1 + 0.999 999 999 534 338 709 062 156 288;
  • 28) 0.999 999 999 534 338 709 062 156 288 × 2 = 1 + 0.999 999 999 068 677 418 124 312 576;
  • 29) 0.999 999 999 068 677 418 124 312 576 × 2 = 1 + 0.999 999 998 137 354 836 248 625 152;
  • 30) 0.999 999 998 137 354 836 248 625 152 × 2 = 1 + 0.999 999 996 274 709 672 497 250 304;
  • 31) 0.999 999 996 274 709 672 497 250 304 × 2 = 1 + 0.999 999 992 549 419 344 994 500 608;
  • 32) 0.999 999 992 549 419 344 994 500 608 × 2 = 1 + 0.999 999 985 098 838 689 989 001 216;
  • 33) 0.999 999 985 098 838 689 989 001 216 × 2 = 1 + 0.999 999 970 197 677 379 978 002 432;
  • 34) 0.999 999 970 197 677 379 978 002 432 × 2 = 1 + 0.999 999 940 395 354 759 956 004 864;
  • 35) 0.999 999 940 395 354 759 956 004 864 × 2 = 1 + 0.999 999 880 790 709 519 912 009 728;
  • 36) 0.999 999 880 790 709 519 912 009 728 × 2 = 1 + 0.999 999 761 581 419 039 824 019 456;
  • 37) 0.999 999 761 581 419 039 824 019 456 × 2 = 1 + 0.999 999 523 162 838 079 648 038 912;
  • 38) 0.999 999 523 162 838 079 648 038 912 × 2 = 1 + 0.999 999 046 325 676 159 296 077 824;
  • 39) 0.999 999 046 325 676 159 296 077 824 × 2 = 1 + 0.999 998 092 651 352 318 592 155 648;
  • 40) 0.999 998 092 651 352 318 592 155 648 × 2 = 1 + 0.999 996 185 302 704 637 184 311 296;
  • 41) 0.999 996 185 302 704 637 184 311 296 × 2 = 1 + 0.999 992 370 605 409 274 368 622 592;
  • 42) 0.999 992 370 605 409 274 368 622 592 × 2 = 1 + 0.999 984 741 210 818 548 737 245 184;
  • 43) 0.999 984 741 210 818 548 737 245 184 × 2 = 1 + 0.999 969 482 421 637 097 474 490 368;
  • 44) 0.999 969 482 421 637 097 474 490 368 × 2 = 1 + 0.999 938 964 843 274 194 948 980 736;
  • 45) 0.999 938 964 843 274 194 948 980 736 × 2 = 1 + 0.999 877 929 686 548 389 897 961 472;
  • 46) 0.999 877 929 686 548 389 897 961 472 × 2 = 1 + 0.999 755 859 373 096 779 795 922 944;
  • 47) 0.999 755 859 373 096 779 795 922 944 × 2 = 1 + 0.999 511 718 746 193 559 591 845 888;
  • 48) 0.999 511 718 746 193 559 591 845 888 × 2 = 1 + 0.999 023 437 492 387 119 183 691 776;
  • 49) 0.999 023 437 492 387 119 183 691 776 × 2 = 1 + 0.998 046 874 984 774 238 367 383 552;
  • 50) 0.998 046 874 984 774 238 367 383 552 × 2 = 1 + 0.996 093 749 969 548 476 734 767 104;
  • 51) 0.996 093 749 969 548 476 734 767 104 × 2 = 1 + 0.992 187 499 939 096 953 469 534 208;
  • 52) 0.992 187 499 939 096 953 469 534 208 × 2 = 1 + 0.984 374 999 878 193 906 939 068 416;
  • 53) 0.984 374 999 878 193 906 939 068 416 × 2 = 1 + 0.968 749 999 756 387 813 878 136 832;
  • 54) 0.968 749 999 756 387 813 878 136 832 × 2 = 1 + 0.937 499 999 512 775 627 756 273 664;
  • 55) 0.937 499 999 512 775 627 756 273 664 × 2 = 1 + 0.874 999 999 025 551 255 512 547 328;
  • 56) 0.874 999 999 025 551 255 512 547 328 × 2 = 1 + 0.749 999 998 051 102 511 025 094 656;
  • 57) 0.749 999 998 051 102 511 025 094 656 × 2 = 1 + 0.499 999 996 102 205 022 050 189 312;
  • 58) 0.499 999 996 102 205 022 050 189 312 × 2 = 0 + 0.999 999 992 204 410 044 100 378 624;

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:


0.016 738 891 601 562 496 530 553 021(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 021(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:


0.016 738 891 601 562 496 530 553 021(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 -0.016 738 891 601 562 496 530 553 021 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