0.000 000 619 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 619(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
0.000 000 619(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: 0.000 000 619.

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) 0.000 000 619 × 2 = 0 + 0.000 001 238;
  • 2) 0.000 001 238 × 2 = 0 + 0.000 002 476;
  • 3) 0.000 002 476 × 2 = 0 + 0.000 004 952;
  • 4) 0.000 004 952 × 2 = 0 + 0.000 009 904;
  • 5) 0.000 009 904 × 2 = 0 + 0.000 019 808;
  • 6) 0.000 019 808 × 2 = 0 + 0.000 039 616;
  • 7) 0.000 039 616 × 2 = 0 + 0.000 079 232;
  • 8) 0.000 079 232 × 2 = 0 + 0.000 158 464;
  • 9) 0.000 158 464 × 2 = 0 + 0.000 316 928;
  • 10) 0.000 316 928 × 2 = 0 + 0.000 633 856;
  • 11) 0.000 633 856 × 2 = 0 + 0.001 267 712;
  • 12) 0.001 267 712 × 2 = 0 + 0.002 535 424;
  • 13) 0.002 535 424 × 2 = 0 + 0.005 070 848;
  • 14) 0.005 070 848 × 2 = 0 + 0.010 141 696;
  • 15) 0.010 141 696 × 2 = 0 + 0.020 283 392;
  • 16) 0.020 283 392 × 2 = 0 + 0.040 566 784;
  • 17) 0.040 566 784 × 2 = 0 + 0.081 133 568;
  • 18) 0.081 133 568 × 2 = 0 + 0.162 267 136;
  • 19) 0.162 267 136 × 2 = 0 + 0.324 534 272;
  • 20) 0.324 534 272 × 2 = 0 + 0.649 068 544;
  • 21) 0.649 068 544 × 2 = 1 + 0.298 137 088;
  • 22) 0.298 137 088 × 2 = 0 + 0.596 274 176;
  • 23) 0.596 274 176 × 2 = 1 + 0.192 548 352;
  • 24) 0.192 548 352 × 2 = 0 + 0.385 096 704;
  • 25) 0.385 096 704 × 2 = 0 + 0.770 193 408;
  • 26) 0.770 193 408 × 2 = 1 + 0.540 386 816;
  • 27) 0.540 386 816 × 2 = 1 + 0.080 773 632;
  • 28) 0.080 773 632 × 2 = 0 + 0.161 547 264;
  • 29) 0.161 547 264 × 2 = 0 + 0.323 094 528;
  • 30) 0.323 094 528 × 2 = 0 + 0.646 189 056;
  • 31) 0.646 189 056 × 2 = 1 + 0.292 378 112;
  • 32) 0.292 378 112 × 2 = 0 + 0.584 756 224;
  • 33) 0.584 756 224 × 2 = 1 + 0.169 512 448;
  • 34) 0.169 512 448 × 2 = 0 + 0.339 024 896;
  • 35) 0.339 024 896 × 2 = 0 + 0.678 049 792;
  • 36) 0.678 049 792 × 2 = 1 + 0.356 099 584;
  • 37) 0.356 099 584 × 2 = 0 + 0.712 199 168;
  • 38) 0.712 199 168 × 2 = 1 + 0.424 398 336;
  • 39) 0.424 398 336 × 2 = 0 + 0.848 796 672;
  • 40) 0.848 796 672 × 2 = 1 + 0.697 593 344;
  • 41) 0.697 593 344 × 2 = 1 + 0.395 186 688;
  • 42) 0.395 186 688 × 2 = 0 + 0.790 373 376;
  • 43) 0.790 373 376 × 2 = 1 + 0.580 746 752;
  • 44) 0.580 746 752 × 2 = 1 + 0.161 493 504;

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:


0.000 000 619(10) =

0.0000 0000 0000 0000 0000 1010 0110 0010 1001 0101 1011(2)

5. Positive number before normalization:

0.000 000 619(10) =

0.0000 0000 0000 0000 0000 1010 0110 0010 1001 0101 1011(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:


0.000 000 619(10) =

0.0000 0000 0000 0000 0000 1010 0110 0010 1001 0101 1011(2) =

0.0000 0000 0000 0000 0000 1010 0110 0010 1001 0101 1011(2) × 20 =

1.0100 1100 0101 0010 1011 011(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):
1.0100 1100 0101 0010 1011 011


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. 010 0110 0010 1001 0101 1011 =

010 0110 0010 1001 0101 1011


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) =
010 0110 0010 1001 0101 1011


Decimal number 0.000 000 619 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 010 0110 0010 1001 0101 1011

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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