Convert -0.045 000 000 000 000 3 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

-0.045 000 000 000 000 3(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. Start with the positive version of the number:

|-0.045 000 000 000 000 3| = 0.045 000 000 000 000 3

2. First, convert to the binary (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 the binary (base 2) the fractional part: 0.045 000 000 000 000 3.

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) 0.045 000 000 000 000 3 × 2 = 0 + 0.090 000 000 000 000 6;
  • 2) 0.090 000 000 000 000 6 × 2 = 0 + 0.180 000 000 000 001 2;
  • 3) 0.180 000 000 000 001 2 × 2 = 0 + 0.360 000 000 000 002 4;
  • 4) 0.360 000 000 000 002 4 × 2 = 0 + 0.720 000 000 000 004 8;
  • 5) 0.720 000 000 000 004 8 × 2 = 1 + 0.440 000 000 000 009 6;
  • 6) 0.440 000 000 000 009 6 × 2 = 0 + 0.880 000 000 000 019 2;
  • 7) 0.880 000 000 000 019 2 × 2 = 1 + 0.760 000 000 000 038 4;
  • 8) 0.760 000 000 000 038 4 × 2 = 1 + 0.520 000 000 000 076 8;
  • 9) 0.520 000 000 000 076 8 × 2 = 1 + 0.040 000 000 000 153 6;
  • 10) 0.040 000 000 000 153 6 × 2 = 0 + 0.080 000 000 000 307 2;
  • 11) 0.080 000 000 000 307 2 × 2 = 0 + 0.160 000 000 000 614 4;
  • 12) 0.160 000 000 000 614 4 × 2 = 0 + 0.320 000 000 001 228 8;
  • 13) 0.320 000 000 001 228 8 × 2 = 0 + 0.640 000 000 002 457 6;
  • 14) 0.640 000 000 002 457 6 × 2 = 1 + 0.280 000 000 004 915 2;
  • 15) 0.280 000 000 004 915 2 × 2 = 0 + 0.560 000 000 009 830 4;
  • 16) 0.560 000 000 009 830 4 × 2 = 1 + 0.120 000 000 019 660 8;
  • 17) 0.120 000 000 019 660 8 × 2 = 0 + 0.240 000 000 039 321 6;
  • 18) 0.240 000 000 039 321 6 × 2 = 0 + 0.480 000 000 078 643 2;
  • 19) 0.480 000 000 078 643 2 × 2 = 0 + 0.960 000 000 157 286 4;
  • 20) 0.960 000 000 157 286 4 × 2 = 1 + 0.920 000 000 314 572 8;
  • 21) 0.920 000 000 314 572 8 × 2 = 1 + 0.840 000 000 629 145 6;
  • 22) 0.840 000 000 629 145 6 × 2 = 1 + 0.680 000 001 258 291 2;
  • 23) 0.680 000 001 258 291 2 × 2 = 1 + 0.360 000 002 516 582 4;
  • 24) 0.360 000 002 516 582 4 × 2 = 0 + 0.720 000 005 033 164 8;
  • 25) 0.720 000 005 033 164 8 × 2 = 1 + 0.440 000 010 066 329 6;
  • 26) 0.440 000 010 066 329 6 × 2 = 0 + 0.880 000 020 132 659 2;
  • 27) 0.880 000 020 132 659 2 × 2 = 1 + 0.760 000 040 265 318 4;
  • 28) 0.760 000 040 265 318 4 × 2 = 1 + 0.520 000 080 530 636 8;
  • 29) 0.520 000 080 530 636 8 × 2 = 1 + 0.040 000 161 061 273 6;
  • 30) 0.040 000 161 061 273 6 × 2 = 0 + 0.080 000 322 122 547 2;
  • 31) 0.080 000 322 122 547 2 × 2 = 0 + 0.160 000 644 245 094 4;
  • 32) 0.160 000 644 245 094 4 × 2 = 0 + 0.320 001 288 490 188 8;
  • 33) 0.320 001 288 490 188 8 × 2 = 0 + 0.640 002 576 980 377 6;
  • 34) 0.640 002 576 980 377 6 × 2 = 1 + 0.280 005 153 960 755 2;
  • 35) 0.280 005 153 960 755 2 × 2 = 0 + 0.560 010 307 921 510 4;
  • 36) 0.560 010 307 921 510 4 × 2 = 1 + 0.120 020 615 843 020 8;
  • 37) 0.120 020 615 843 020 8 × 2 = 0 + 0.240 041 231 686 041 6;
  • 38) 0.240 041 231 686 041 6 × 2 = 0 + 0.480 082 463 372 083 2;
  • 39) 0.480 082 463 372 083 2 × 2 = 0 + 0.960 164 926 744 166 4;
  • 40) 0.960 164 926 744 166 4 × 2 = 1 + 0.920 329 853 488 332 8;
  • 41) 0.920 329 853 488 332 8 × 2 = 1 + 0.840 659 706 976 665 6;
  • 42) 0.840 659 706 976 665 6 × 2 = 1 + 0.681 319 413 953 331 2;
  • 43) 0.681 319 413 953 331 2 × 2 = 1 + 0.362 638 827 906 662 4;
  • 44) 0.362 638 827 906 662 4 × 2 = 0 + 0.725 277 655 813 324 8;
  • 45) 0.725 277 655 813 324 8 × 2 = 1 + 0.450 555 311 626 649 6;
  • 46) 0.450 555 311 626 649 6 × 2 = 0 + 0.901 110 623 253 299 2;
  • 47) 0.901 110 623 253 299 2 × 2 = 1 + 0.802 221 246 506 598 4;
  • 48) 0.802 221 246 506 598 4 × 2 = 1 + 0.604 442 493 013 196 8;
  • 49) 0.604 442 493 013 196 8 × 2 = 1 + 0.208 884 986 026 393 6;
  • 50) 0.208 884 986 026 393 6 × 2 = 0 + 0.417 769 972 052 787 2;
  • 51) 0.417 769 972 052 787 2 × 2 = 0 + 0.835 539 944 105 574 4;
  • 52) 0.835 539 944 105 574 4 × 2 = 1 + 0.671 079 888 211 148 8;
  • 53) 0.671 079 888 211 148 8 × 2 = 1 + 0.342 159 776 422 297 6;
  • 54) 0.342 159 776 422 297 6 × 2 = 0 + 0.684 319 552 844 595 2;
  • 55) 0.684 319 552 844 595 2 × 2 = 1 + 0.368 639 105 689 190 4;
  • 56) 0.368 639 105 689 190 4 × 2 = 0 + 0.737 278 211 378 380 8;
  • 57) 0.737 278 211 378 380 8 × 2 = 1 + 0.474 556 422 756 761 6;

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...)


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:

0.045 000 000 000 000 3(10) =


0.0000 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 1010 1(2)


6. Positive number before normalization:

0.045 000 000 000 000 3(10) =


0.0000 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 1010 1(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right so that only one non zero digit remains to the left of it:

0.045 000 000 000 000 3(10) =


0.0000 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 1010 1(2) =


0.0000 1011 1000 0101 0001 1110 1011 1000 0101 0001 1110 1011 1001 1010 1(2) × 20 =


1.0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 0101(2) × 2-5


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): -5


Mantissa (not normalized):
1.0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-5 + 2(11-1) - 1 =


(-5 + 1 023)(10) =


1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 509 ÷ 2 = 254 + 1;
  • 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) =


1018(10) =


011 1111 1010(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. 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 0101 =


0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 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 1111 1010


Mantissa (52 bits) =
0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 0101


Number -0.045 000 000 000 000 3 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
1 - 011 1111 1010 - 0111 0000 1010 0011 1101 0111 0000 1010 0011 1101 0111 0011 0101

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 0

      54
    • 1

      53
    • 0

      52
  • Mantissa (52 bits):

    • 0

      51
    • 1

      50
    • 1

      49
    • 1

      48
    • 0

      47
    • 0

      46
    • 0

      45
    • 0

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 0

      38
    • 1

      37
    • 1

      36
    • 1

      35
    • 1

      34
    • 0

      33
    • 1

      32
    • 0

      31
    • 1

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 0

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

-0.045 000 000 000 000 4 = ? ... -0.045 000 000 000 000 2 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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