-0.000 000 670 552 575 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 670 552 575(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
-0.000 000 670 552 575(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. Start with the positive version of the number:

|-0.000 000 670 552 575| = 0.000 000 670 552 575


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: 0.000 000 670 552 575.

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) 0.000 000 670 552 575 × 2 = 0 + 0.000 001 341 105 15;
  • 2) 0.000 001 341 105 15 × 2 = 0 + 0.000 002 682 210 3;
  • 3) 0.000 002 682 210 3 × 2 = 0 + 0.000 005 364 420 6;
  • 4) 0.000 005 364 420 6 × 2 = 0 + 0.000 010 728 841 2;
  • 5) 0.000 010 728 841 2 × 2 = 0 + 0.000 021 457 682 4;
  • 6) 0.000 021 457 682 4 × 2 = 0 + 0.000 042 915 364 8;
  • 7) 0.000 042 915 364 8 × 2 = 0 + 0.000 085 830 729 6;
  • 8) 0.000 085 830 729 6 × 2 = 0 + 0.000 171 661 459 2;
  • 9) 0.000 171 661 459 2 × 2 = 0 + 0.000 343 322 918 4;
  • 10) 0.000 343 322 918 4 × 2 = 0 + 0.000 686 645 836 8;
  • 11) 0.000 686 645 836 8 × 2 = 0 + 0.001 373 291 673 6;
  • 12) 0.001 373 291 673 6 × 2 = 0 + 0.002 746 583 347 2;
  • 13) 0.002 746 583 347 2 × 2 = 0 + 0.005 493 166 694 4;
  • 14) 0.005 493 166 694 4 × 2 = 0 + 0.010 986 333 388 8;
  • 15) 0.010 986 333 388 8 × 2 = 0 + 0.021 972 666 777 6;
  • 16) 0.021 972 666 777 6 × 2 = 0 + 0.043 945 333 555 2;
  • 17) 0.043 945 333 555 2 × 2 = 0 + 0.087 890 667 110 4;
  • 18) 0.087 890 667 110 4 × 2 = 0 + 0.175 781 334 220 8;
  • 19) 0.175 781 334 220 8 × 2 = 0 + 0.351 562 668 441 6;
  • 20) 0.351 562 668 441 6 × 2 = 0 + 0.703 125 336 883 2;
  • 21) 0.703 125 336 883 2 × 2 = 1 + 0.406 250 673 766 4;
  • 22) 0.406 250 673 766 4 × 2 = 0 + 0.812 501 347 532 8;
  • 23) 0.812 501 347 532 8 × 2 = 1 + 0.625 002 695 065 6;
  • 24) 0.625 002 695 065 6 × 2 = 1 + 0.250 005 390 131 2;
  • 25) 0.250 005 390 131 2 × 2 = 0 + 0.500 010 780 262 4;
  • 26) 0.500 010 780 262 4 × 2 = 1 + 0.000 021 560 524 8;
  • 27) 0.000 021 560 524 8 × 2 = 0 + 0.000 043 121 049 6;
  • 28) 0.000 043 121 049 6 × 2 = 0 + 0.000 086 242 099 2;
  • 29) 0.000 086 242 099 2 × 2 = 0 + 0.000 172 484 198 4;
  • 30) 0.000 172 484 198 4 × 2 = 0 + 0.000 344 968 396 8;
  • 31) 0.000 344 968 396 8 × 2 = 0 + 0.000 689 936 793 6;
  • 32) 0.000 689 936 793 6 × 2 = 0 + 0.001 379 873 587 2;
  • 33) 0.001 379 873 587 2 × 2 = 0 + 0.002 759 747 174 4;
  • 34) 0.002 759 747 174 4 × 2 = 0 + 0.005 519 494 348 8;
  • 35) 0.005 519 494 348 8 × 2 = 0 + 0.011 038 988 697 6;
  • 36) 0.011 038 988 697 6 × 2 = 0 + 0.022 077 977 395 2;
  • 37) 0.022 077 977 395 2 × 2 = 0 + 0.044 155 954 790 4;
  • 38) 0.044 155 954 790 4 × 2 = 0 + 0.088 311 909 580 8;
  • 39) 0.088 311 909 580 8 × 2 = 0 + 0.176 623 819 161 6;
  • 40) 0.176 623 819 161 6 × 2 = 0 + 0.353 247 638 323 2;
  • 41) 0.353 247 638 323 2 × 2 = 0 + 0.706 495 276 646 4;
  • 42) 0.706 495 276 646 4 × 2 = 1 + 0.412 990 553 292 8;
  • 43) 0.412 990 553 292 8 × 2 = 0 + 0.825 981 106 585 6;
  • 44) 0.825 981 106 585 6 × 2 = 1 + 0.651 962 213 171 2;

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:


0.000 000 670 552 575(10) =

0.0000 0000 0000 0000 0000 1011 0100 0000 0000 0000 0101(2)

6. Positive number before normalization:

0.000 000 670 552 575(10) =

0.0000 0000 0000 0000 0000 1011 0100 0000 0000 0000 0101(2)

7. 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:


0.000 000 670 552 575(10) =

0.0000 0000 0000 0000 0000 1011 0100 0000 0000 0000 0101(2) =

0.0000 0000 0000 0000 0000 1011 0100 0000 0000 0000 0101(2) × 20 =

1.0110 1000 0000 0000 0000 101(2) × 2-21


8. 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 1 (a negative number)

Exponent (unadjusted): -21

Mantissa (not normalized):
1.0110 1000 0000 0000 0000 101


9. 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)


10. 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;

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) =

106(10) =

0110 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 011 0100 0000 0000 0000 0101 =

011 0100 0000 0000 0000 0101


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

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

Exponent (8 bits) =
0110 1010

Mantissa (23 bits) =
011 0100 0000 0000 0000 0101


Decimal number -0.000 000 670 552 575 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1010 - 011 0100 0000 0000 0000 0101

Share

» Convert decimal -0.000 000 670 552 514 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. Gemini Notebook YouTube Video of 8.15 minutes

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