0.000 000 594 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 594(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 594(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. 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;

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 binary (base 2) the fractional part: 0.000 000 594.

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) 0.000 000 594 × 2 = 0 + 0.000 001 188;
  • 2) 0.000 001 188 × 2 = 0 + 0.000 002 376;
  • 3) 0.000 002 376 × 2 = 0 + 0.000 004 752;
  • 4) 0.000 004 752 × 2 = 0 + 0.000 009 504;
  • 5) 0.000 009 504 × 2 = 0 + 0.000 019 008;
  • 6) 0.000 019 008 × 2 = 0 + 0.000 038 016;
  • 7) 0.000 038 016 × 2 = 0 + 0.000 076 032;
  • 8) 0.000 076 032 × 2 = 0 + 0.000 152 064;
  • 9) 0.000 152 064 × 2 = 0 + 0.000 304 128;
  • 10) 0.000 304 128 × 2 = 0 + 0.000 608 256;
  • 11) 0.000 608 256 × 2 = 0 + 0.001 216 512;
  • 12) 0.001 216 512 × 2 = 0 + 0.002 433 024;
  • 13) 0.002 433 024 × 2 = 0 + 0.004 866 048;
  • 14) 0.004 866 048 × 2 = 0 + 0.009 732 096;
  • 15) 0.009 732 096 × 2 = 0 + 0.019 464 192;
  • 16) 0.019 464 192 × 2 = 0 + 0.038 928 384;
  • 17) 0.038 928 384 × 2 = 0 + 0.077 856 768;
  • 18) 0.077 856 768 × 2 = 0 + 0.155 713 536;
  • 19) 0.155 713 536 × 2 = 0 + 0.311 427 072;
  • 20) 0.311 427 072 × 2 = 0 + 0.622 854 144;
  • 21) 0.622 854 144 × 2 = 1 + 0.245 708 288;
  • 22) 0.245 708 288 × 2 = 0 + 0.491 416 576;
  • 23) 0.491 416 576 × 2 = 0 + 0.982 833 152;
  • 24) 0.982 833 152 × 2 = 1 + 0.965 666 304;
  • 25) 0.965 666 304 × 2 = 1 + 0.931 332 608;
  • 26) 0.931 332 608 × 2 = 1 + 0.862 665 216;
  • 27) 0.862 665 216 × 2 = 1 + 0.725 330 432;
  • 28) 0.725 330 432 × 2 = 1 + 0.450 660 864;
  • 29) 0.450 660 864 × 2 = 0 + 0.901 321 728;
  • 30) 0.901 321 728 × 2 = 1 + 0.802 643 456;
  • 31) 0.802 643 456 × 2 = 1 + 0.605 286 912;
  • 32) 0.605 286 912 × 2 = 1 + 0.210 573 824;
  • 33) 0.210 573 824 × 2 = 0 + 0.421 147 648;
  • 34) 0.421 147 648 × 2 = 0 + 0.842 295 296;
  • 35) 0.842 295 296 × 2 = 1 + 0.684 590 592;
  • 36) 0.684 590 592 × 2 = 1 + 0.369 181 184;
  • 37) 0.369 181 184 × 2 = 0 + 0.738 362 368;
  • 38) 0.738 362 368 × 2 = 1 + 0.476 724 736;
  • 39) 0.476 724 736 × 2 = 0 + 0.953 449 472;
  • 40) 0.953 449 472 × 2 = 1 + 0.906 898 944;
  • 41) 0.906 898 944 × 2 = 1 + 0.813 797 888;
  • 42) 0.813 797 888 × 2 = 1 + 0.627 595 776;
  • 43) 0.627 595 776 × 2 = 1 + 0.255 191 552;
  • 44) 0.255 191 552 × 2 = 0 + 0.510 383 104;

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).


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:


0.000 000 594(10) =

0.0000 0000 0000 0000 0000 1001 1111 0111 0011 0101 1110(2)

5. Positive number before normalization:

0.000 000 594(10) =

0.0000 0000 0000 0000 0000 1001 1111 0111 0011 0101 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 594(10) =

0.0000 0000 0000 0000 0000 1001 1111 0111 0011 0101 1110(2) =

0.0000 0000 0000 0000 0000 1001 1111 0111 0011 0101 1110(2) × 20 =

1.0011 1110 1110 0110 1011 110(2) × 2-21


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): -21

Mantissa (not normalized):
1.0011 1110 1110 0110 1011 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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. 001 1111 0111 0011 0101 1110 =

001 1111 0111 0011 0101 1110


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 1010

Mantissa (23 bits) =
001 1111 0111 0011 0101 1110


Decimal number 0.000 000 594 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 001 1111 0111 0011 0101 1110

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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