Convert 0.000 000 009 703 326 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

0.000 000 009 703 326(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. 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;

2. 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)


3. Convert to the binary (base 2) the fractional part: 0.000 000 009 703 326.

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) 0.000 000 009 703 326 × 2 = 0 + 0.000 000 019 406 652;
  • 2) 0.000 000 019 406 652 × 2 = 0 + 0.000 000 038 813 304;
  • 3) 0.000 000 038 813 304 × 2 = 0 + 0.000 000 077 626 608;
  • 4) 0.000 000 077 626 608 × 2 = 0 + 0.000 000 155 253 216;
  • 5) 0.000 000 155 253 216 × 2 = 0 + 0.000 000 310 506 432;
  • 6) 0.000 000 310 506 432 × 2 = 0 + 0.000 000 621 012 864;
  • 7) 0.000 000 621 012 864 × 2 = 0 + 0.000 001 242 025 728;
  • 8) 0.000 001 242 025 728 × 2 = 0 + 0.000 002 484 051 456;
  • 9) 0.000 002 484 051 456 × 2 = 0 + 0.000 004 968 102 912;
  • 10) 0.000 004 968 102 912 × 2 = 0 + 0.000 009 936 205 824;
  • 11) 0.000 009 936 205 824 × 2 = 0 + 0.000 019 872 411 648;
  • 12) 0.000 019 872 411 648 × 2 = 0 + 0.000 039 744 823 296;
  • 13) 0.000 039 744 823 296 × 2 = 0 + 0.000 079 489 646 592;
  • 14) 0.000 079 489 646 592 × 2 = 0 + 0.000 158 979 293 184;
  • 15) 0.000 158 979 293 184 × 2 = 0 + 0.000 317 958 586 368;
  • 16) 0.000 317 958 586 368 × 2 = 0 + 0.000 635 917 172 736;
  • 17) 0.000 635 917 172 736 × 2 = 0 + 0.001 271 834 345 472;
  • 18) 0.001 271 834 345 472 × 2 = 0 + 0.002 543 668 690 944;
  • 19) 0.002 543 668 690 944 × 2 = 0 + 0.005 087 337 381 888;
  • 20) 0.005 087 337 381 888 × 2 = 0 + 0.010 174 674 763 776;
  • 21) 0.010 174 674 763 776 × 2 = 0 + 0.020 349 349 527 552;
  • 22) 0.020 349 349 527 552 × 2 = 0 + 0.040 698 699 055 104;
  • 23) 0.040 698 699 055 104 × 2 = 0 + 0.081 397 398 110 208;
  • 24) 0.081 397 398 110 208 × 2 = 0 + 0.162 794 796 220 416;
  • 25) 0.162 794 796 220 416 × 2 = 0 + 0.325 589 592 440 832;
  • 26) 0.325 589 592 440 832 × 2 = 0 + 0.651 179 184 881 664;
  • 27) 0.651 179 184 881 664 × 2 = 1 + 0.302 358 369 763 328;
  • 28) 0.302 358 369 763 328 × 2 = 0 + 0.604 716 739 526 656;
  • 29) 0.604 716 739 526 656 × 2 = 1 + 0.209 433 479 053 312;
  • 30) 0.209 433 479 053 312 × 2 = 0 + 0.418 866 958 106 624;
  • 31) 0.418 866 958 106 624 × 2 = 0 + 0.837 733 916 213 248;
  • 32) 0.837 733 916 213 248 × 2 = 1 + 0.675 467 832 426 496;
  • 33) 0.675 467 832 426 496 × 2 = 1 + 0.350 935 664 852 992;
  • 34) 0.350 935 664 852 992 × 2 = 0 + 0.701 871 329 705 984;
  • 35) 0.701 871 329 705 984 × 2 = 1 + 0.403 742 659 411 968;
  • 36) 0.403 742 659 411 968 × 2 = 0 + 0.807 485 318 823 936;
  • 37) 0.807 485 318 823 936 × 2 = 1 + 0.614 970 637 647 872;
  • 38) 0.614 970 637 647 872 × 2 = 1 + 0.229 941 275 295 744;
  • 39) 0.229 941 275 295 744 × 2 = 0 + 0.459 882 550 591 488;
  • 40) 0.459 882 550 591 488 × 2 = 0 + 0.919 765 101 182 976;
  • 41) 0.919 765 101 182 976 × 2 = 1 + 0.839 530 202 365 952;
  • 42) 0.839 530 202 365 952 × 2 = 1 + 0.679 060 404 731 904;
  • 43) 0.679 060 404 731 904 × 2 = 1 + 0.358 120 809 463 808;
  • 44) 0.358 120 809 463 808 × 2 = 0 + 0.716 241 618 927 616;
  • 45) 0.716 241 618 927 616 × 2 = 1 + 0.432 483 237 855 232;
  • 46) 0.432 483 237 855 232 × 2 = 0 + 0.864 966 475 710 464;
  • 47) 0.864 966 475 710 464 × 2 = 1 + 0.729 932 951 420 928;
  • 48) 0.729 932 951 420 928 × 2 = 1 + 0.459 865 902 841 856;
  • 49) 0.459 865 902 841 856 × 2 = 0 + 0.919 731 805 683 712;
  • 50) 0.919 731 805 683 712 × 2 = 1 + 0.839 463 611 367 424;

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...)


4. 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:

0.000 000 009 703 326(10) =


0.0000 0000 0000 0000 0000 0000 0010 1001 1010 1100 1110 1011 01(2)


5. Positive number before normalization:

0.000 000 009 703 326(10) =


0.0000 0000 0000 0000 0000 0000 0010 1001 1010 1100 1110 1011 01(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the right so that only one non zero digit remains to the left of it:

0.000 000 009 703 326(10) =


0.0000 0000 0000 0000 0000 0000 0010 1001 1010 1100 1110 1011 01(2) =


0.0000 0000 0000 0000 0000 0000 0010 1001 1010 1100 1110 1011 01(2) × 20 =


1.0100 1101 0110 0111 0101 101(2) × 2-27


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -27


Mantissa (not normalized):
1.0100 1101 0110 0111 0101 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-27 + 2(8-1) - 1 =


(-27 + 127)(10) =


100(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:

  • division = quotient + remainder;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:

Exponent (adjusted) =


100(10) =


0110 0100(2)


11. 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 23 bits, only if necessary (not the case here).

Mantissa (normalized) =


1. 010 0110 1011 0011 1010 1101 =


010 0110 1011 0011 1010 1101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
0110 0100


Mantissa (23 bits) =
010 0110 1011 0011 1010 1101


Number 0.000 000 009 703 326 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 0110 0100 - 010 0110 1011 0011 1010 1101

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 0

      27
    • 0

      26
    • 1

      25
    • 0

      24
    • 0

      23
  • Mantissa (23 bits):

    • 0

      22
    • 1

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 0

      16
    • 1

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 0

      11
    • 0

      10
    • 1

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 1

      0

More operations of this kind:

0.000 000 009 703 325 = ? ... 0.000 000 009 703 327 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

0.000 000 009 703 326 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:20 UTC (GMT)
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1 000 110 010 001 099 999 999 999 999 985 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:20 UTC (GMT)
1.326 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:20 UTC (GMT)
0.001 100 010 7 to 32 bit single precision IEEE 754 binary floating point = ? Oct 28 10:19 UTC (GMT)
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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111