-0.000 000 000 001 05 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 001 05(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
-0.000 000 000 001 05(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:

|-0.000 000 000 001 05| = 0.000 000 000 001 05


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: 0.000 000 000 001 05.

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) 0.000 000 000 001 05 × 2 = 0 + 0.000 000 000 002 1;
  • 2) 0.000 000 000 002 1 × 2 = 0 + 0.000 000 000 004 2;
  • 3) 0.000 000 000 004 2 × 2 = 0 + 0.000 000 000 008 4;
  • 4) 0.000 000 000 008 4 × 2 = 0 + 0.000 000 000 016 8;
  • 5) 0.000 000 000 016 8 × 2 = 0 + 0.000 000 000 033 6;
  • 6) 0.000 000 000 033 6 × 2 = 0 + 0.000 000 000 067 2;
  • 7) 0.000 000 000 067 2 × 2 = 0 + 0.000 000 000 134 4;
  • 8) 0.000 000 000 134 4 × 2 = 0 + 0.000 000 000 268 8;
  • 9) 0.000 000 000 268 8 × 2 = 0 + 0.000 000 000 537 6;
  • 10) 0.000 000 000 537 6 × 2 = 0 + 0.000 000 001 075 2;
  • 11) 0.000 000 001 075 2 × 2 = 0 + 0.000 000 002 150 4;
  • 12) 0.000 000 002 150 4 × 2 = 0 + 0.000 000 004 300 8;
  • 13) 0.000 000 004 300 8 × 2 = 0 + 0.000 000 008 601 6;
  • 14) 0.000 000 008 601 6 × 2 = 0 + 0.000 000 017 203 2;
  • 15) 0.000 000 017 203 2 × 2 = 0 + 0.000 000 034 406 4;
  • 16) 0.000 000 034 406 4 × 2 = 0 + 0.000 000 068 812 8;
  • 17) 0.000 000 068 812 8 × 2 = 0 + 0.000 000 137 625 6;
  • 18) 0.000 000 137 625 6 × 2 = 0 + 0.000 000 275 251 2;
  • 19) 0.000 000 275 251 2 × 2 = 0 + 0.000 000 550 502 4;
  • 20) 0.000 000 550 502 4 × 2 = 0 + 0.000 001 101 004 8;
  • 21) 0.000 001 101 004 8 × 2 = 0 + 0.000 002 202 009 6;
  • 22) 0.000 002 202 009 6 × 2 = 0 + 0.000 004 404 019 2;
  • 23) 0.000 004 404 019 2 × 2 = 0 + 0.000 008 808 038 4;
  • 24) 0.000 008 808 038 4 × 2 = 0 + 0.000 017 616 076 8;
  • 25) 0.000 017 616 076 8 × 2 = 0 + 0.000 035 232 153 6;
  • 26) 0.000 035 232 153 6 × 2 = 0 + 0.000 070 464 307 2;
  • 27) 0.000 070 464 307 2 × 2 = 0 + 0.000 140 928 614 4;
  • 28) 0.000 140 928 614 4 × 2 = 0 + 0.000 281 857 228 8;
  • 29) 0.000 281 857 228 8 × 2 = 0 + 0.000 563 714 457 6;
  • 30) 0.000 563 714 457 6 × 2 = 0 + 0.001 127 428 915 2;
  • 31) 0.001 127 428 915 2 × 2 = 0 + 0.002 254 857 830 4;
  • 32) 0.002 254 857 830 4 × 2 = 0 + 0.004 509 715 660 8;
  • 33) 0.004 509 715 660 8 × 2 = 0 + 0.009 019 431 321 6;
  • 34) 0.009 019 431 321 6 × 2 = 0 + 0.018 038 862 643 2;
  • 35) 0.018 038 862 643 2 × 2 = 0 + 0.036 077 725 286 4;
  • 36) 0.036 077 725 286 4 × 2 = 0 + 0.072 155 450 572 8;
  • 37) 0.072 155 450 572 8 × 2 = 0 + 0.144 310 901 145 6;
  • 38) 0.144 310 901 145 6 × 2 = 0 + 0.288 621 802 291 2;
  • 39) 0.288 621 802 291 2 × 2 = 0 + 0.577 243 604 582 4;
  • 40) 0.577 243 604 582 4 × 2 = 1 + 0.154 487 209 164 8;
  • 41) 0.154 487 209 164 8 × 2 = 0 + 0.308 974 418 329 6;
  • 42) 0.308 974 418 329 6 × 2 = 0 + 0.617 948 836 659 2;
  • 43) 0.617 948 836 659 2 × 2 = 1 + 0.235 897 673 318 4;
  • 44) 0.235 897 673 318 4 × 2 = 0 + 0.471 795 346 636 8;
  • 45) 0.471 795 346 636 8 × 2 = 0 + 0.943 590 693 273 6;
  • 46) 0.943 590 693 273 6 × 2 = 1 + 0.887 181 386 547 2;
  • 47) 0.887 181 386 547 2 × 2 = 1 + 0.774 362 773 094 4;
  • 48) 0.774 362 773 094 4 × 2 = 1 + 0.548 725 546 188 8;
  • 49) 0.548 725 546 188 8 × 2 = 1 + 0.097 451 092 377 6;
  • 50) 0.097 451 092 377 6 × 2 = 0 + 0.194 902 184 755 2;
  • 51) 0.194 902 184 755 2 × 2 = 0 + 0.389 804 369 510 4;
  • 52) 0.389 804 369 510 4 × 2 = 0 + 0.779 608 739 020 8;
  • 53) 0.779 608 739 020 8 × 2 = 1 + 0.559 217 478 041 6;
  • 54) 0.559 217 478 041 6 × 2 = 1 + 0.118 434 956 083 2;
  • 55) 0.118 434 956 083 2 × 2 = 0 + 0.236 869 912 166 4;
  • 56) 0.236 869 912 166 4 × 2 = 0 + 0.473 739 824 332 8;
  • 57) 0.473 739 824 332 8 × 2 = 0 + 0.947 479 648 665 6;
  • 58) 0.947 479 648 665 6 × 2 = 1 + 0.894 959 297 331 2;
  • 59) 0.894 959 297 331 2 × 2 = 1 + 0.789 918 594 662 4;
  • 60) 0.789 918 594 662 4 × 2 = 1 + 0.579 837 189 324 8;
  • 61) 0.579 837 189 324 8 × 2 = 1 + 0.159 674 378 649 6;
  • 62) 0.159 674 378 649 6 × 2 = 0 + 0.319 348 757 299 2;
  • 63) 0.319 348 757 299 2 × 2 = 0 + 0.638 697 514 598 4;
  • 64) 0.638 697 514 598 4 × 2 = 1 + 0.277 395 029 196 8;
  • 65) 0.277 395 029 196 8 × 2 = 0 + 0.554 790 058 393 6;
  • 66) 0.554 790 058 393 6 × 2 = 1 + 0.109 580 116 787 2;
  • 67) 0.109 580 116 787 2 × 2 = 0 + 0.219 160 233 574 4;
  • 68) 0.219 160 233 574 4 × 2 = 0 + 0.438 320 467 148 8;
  • 69) 0.438 320 467 148 8 × 2 = 0 + 0.876 640 934 297 6;
  • 70) 0.876 640 934 297 6 × 2 = 1 + 0.753 281 868 595 2;
  • 71) 0.753 281 868 595 2 × 2 = 1 + 0.506 563 737 190 4;
  • 72) 0.506 563 737 190 4 × 2 = 1 + 0.013 127 474 380 8;
  • 73) 0.013 127 474 380 8 × 2 = 0 + 0.026 254 948 761 6;
  • 74) 0.026 254 948 761 6 × 2 = 0 + 0.052 509 897 523 2;
  • 75) 0.052 509 897 523 2 × 2 = 0 + 0.105 019 795 046 4;
  • 76) 0.105 019 795 046 4 × 2 = 0 + 0.210 039 590 092 8;
  • 77) 0.210 039 590 092 8 × 2 = 0 + 0.420 079 180 185 6;
  • 78) 0.420 079 180 185 6 × 2 = 0 + 0.840 158 360 371 2;
  • 79) 0.840 158 360 371 2 × 2 = 1 + 0.680 316 720 742 4;
  • 80) 0.680 316 720 742 4 × 2 = 1 + 0.360 633 441 484 8;
  • 81) 0.360 633 441 484 8 × 2 = 0 + 0.721 266 882 969 6;
  • 82) 0.721 266 882 969 6 × 2 = 1 + 0.442 533 765 939 2;
  • 83) 0.442 533 765 939 2 × 2 = 0 + 0.885 067 531 878 4;
  • 84) 0.885 067 531 878 4 × 2 = 1 + 0.770 135 063 756 8;
  • 85) 0.770 135 063 756 8 × 2 = 1 + 0.540 270 127 513 6;
  • 86) 0.540 270 127 513 6 × 2 = 1 + 0.080 540 255 027 2;
  • 87) 0.080 540 255 027 2 × 2 = 0 + 0.161 080 510 054 4;
  • 88) 0.161 080 510 054 4 × 2 = 0 + 0.322 161 020 108 8;
  • 89) 0.322 161 020 108 8 × 2 = 0 + 0.644 322 040 217 6;
  • 90) 0.644 322 040 217 6 × 2 = 1 + 0.288 644 080 435 2;
  • 91) 0.288 644 080 435 2 × 2 = 0 + 0.577 288 160 870 4;
  • 92) 0.577 288 160 870 4 × 2 = 1 + 0.154 576 321 740 8;

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:


0.000 000 000 001 05(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101(2)

6. Positive number before normalization:

0.000 000 000 001 05(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 40 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 001 05(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101(2) × 20 =

1.0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101(2) × 2-40


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): -40

Mantissa (not normalized):
1.0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-40 + 2(11-1) - 1 =

(-40 + 1 023)(10) =

983(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;
  • 983 ÷ 2 = 491 + 1;
  • 491 ÷ 2 = 245 + 1;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

983(10) =

011 1101 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101 =

0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101


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 1101 0111

Mantissa (52 bits) =
0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101


Decimal number -0.000 000 000 001 05 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0111 - 0010 0111 1000 1100 0111 1001 0100 0111 0000 0011 0101 1100 0101

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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