-0.000 000 000 747 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 747 3(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
-0.000 000 000 747 3(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:

|-0.000 000 000 747 3| = 0.000 000 000 747 3


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: 0.000 000 000 747 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) 0.000 000 000 747 3 × 2 = 0 + 0.000 000 001 494 6;
  • 2) 0.000 000 001 494 6 × 2 = 0 + 0.000 000 002 989 2;
  • 3) 0.000 000 002 989 2 × 2 = 0 + 0.000 000 005 978 4;
  • 4) 0.000 000 005 978 4 × 2 = 0 + 0.000 000 011 956 8;
  • 5) 0.000 000 011 956 8 × 2 = 0 + 0.000 000 023 913 6;
  • 6) 0.000 000 023 913 6 × 2 = 0 + 0.000 000 047 827 2;
  • 7) 0.000 000 047 827 2 × 2 = 0 + 0.000 000 095 654 4;
  • 8) 0.000 000 095 654 4 × 2 = 0 + 0.000 000 191 308 8;
  • 9) 0.000 000 191 308 8 × 2 = 0 + 0.000 000 382 617 6;
  • 10) 0.000 000 382 617 6 × 2 = 0 + 0.000 000 765 235 2;
  • 11) 0.000 000 765 235 2 × 2 = 0 + 0.000 001 530 470 4;
  • 12) 0.000 001 530 470 4 × 2 = 0 + 0.000 003 060 940 8;
  • 13) 0.000 003 060 940 8 × 2 = 0 + 0.000 006 121 881 6;
  • 14) 0.000 006 121 881 6 × 2 = 0 + 0.000 012 243 763 2;
  • 15) 0.000 012 243 763 2 × 2 = 0 + 0.000 024 487 526 4;
  • 16) 0.000 024 487 526 4 × 2 = 0 + 0.000 048 975 052 8;
  • 17) 0.000 048 975 052 8 × 2 = 0 + 0.000 097 950 105 6;
  • 18) 0.000 097 950 105 6 × 2 = 0 + 0.000 195 900 211 2;
  • 19) 0.000 195 900 211 2 × 2 = 0 + 0.000 391 800 422 4;
  • 20) 0.000 391 800 422 4 × 2 = 0 + 0.000 783 600 844 8;
  • 21) 0.000 783 600 844 8 × 2 = 0 + 0.001 567 201 689 6;
  • 22) 0.001 567 201 689 6 × 2 = 0 + 0.003 134 403 379 2;
  • 23) 0.003 134 403 379 2 × 2 = 0 + 0.006 268 806 758 4;
  • 24) 0.006 268 806 758 4 × 2 = 0 + 0.012 537 613 516 8;
  • 25) 0.012 537 613 516 8 × 2 = 0 + 0.025 075 227 033 6;
  • 26) 0.025 075 227 033 6 × 2 = 0 + 0.050 150 454 067 2;
  • 27) 0.050 150 454 067 2 × 2 = 0 + 0.100 300 908 134 4;
  • 28) 0.100 300 908 134 4 × 2 = 0 + 0.200 601 816 268 8;
  • 29) 0.200 601 816 268 8 × 2 = 0 + 0.401 203 632 537 6;
  • 30) 0.401 203 632 537 6 × 2 = 0 + 0.802 407 265 075 2;
  • 31) 0.802 407 265 075 2 × 2 = 1 + 0.604 814 530 150 4;
  • 32) 0.604 814 530 150 4 × 2 = 1 + 0.209 629 060 300 8;
  • 33) 0.209 629 060 300 8 × 2 = 0 + 0.419 258 120 601 6;
  • 34) 0.419 258 120 601 6 × 2 = 0 + 0.838 516 241 203 2;
  • 35) 0.838 516 241 203 2 × 2 = 1 + 0.677 032 482 406 4;
  • 36) 0.677 032 482 406 4 × 2 = 1 + 0.354 064 964 812 8;
  • 37) 0.354 064 964 812 8 × 2 = 0 + 0.708 129 929 625 6;
  • 38) 0.708 129 929 625 6 × 2 = 1 + 0.416 259 859 251 2;
  • 39) 0.416 259 859 251 2 × 2 = 0 + 0.832 519 718 502 4;
  • 40) 0.832 519 718 502 4 × 2 = 1 + 0.665 039 437 004 8;
  • 41) 0.665 039 437 004 8 × 2 = 1 + 0.330 078 874 009 6;
  • 42) 0.330 078 874 009 6 × 2 = 0 + 0.660 157 748 019 2;
  • 43) 0.660 157 748 019 2 × 2 = 1 + 0.320 315 496 038 4;
  • 44) 0.320 315 496 038 4 × 2 = 0 + 0.640 630 992 076 8;
  • 45) 0.640 630 992 076 8 × 2 = 1 + 0.281 261 984 153 6;
  • 46) 0.281 261 984 153 6 × 2 = 0 + 0.562 523 968 307 2;
  • 47) 0.562 523 968 307 2 × 2 = 1 + 0.125 047 936 614 4;
  • 48) 0.125 047 936 614 4 × 2 = 0 + 0.250 095 873 228 8;
  • 49) 0.250 095 873 228 8 × 2 = 0 + 0.500 191 746 457 6;
  • 50) 0.500 191 746 457 6 × 2 = 1 + 0.000 383 492 915 2;
  • 51) 0.000 383 492 915 2 × 2 = 0 + 0.000 766 985 830 4;
  • 52) 0.000 766 985 830 4 × 2 = 0 + 0.001 533 971 660 8;
  • 53) 0.001 533 971 660 8 × 2 = 0 + 0.003 067 943 321 6;
  • 54) 0.003 067 943 321 6 × 2 = 0 + 0.006 135 886 643 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:


0.000 000 000 747 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1010 1010 0100 00(2)

6. Positive number before normalization:

0.000 000 000 747 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1010 1010 0100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 747 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1010 1010 0100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0101 1010 1010 0100 00(2) × 20 =

1.1001 1010 1101 0101 0010 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1010 1101 0101 0010 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1101 0110 1010 1001 0000 =

100 1101 0110 1010 1001 0000


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 0000

Mantissa (23 bits) =
100 1101 0110 1010 1001 0000


Decimal number -0.000 000 000 747 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1101 0110 1010 1001 0000

Share

» Convert decimal -0.000 000 000 737 7 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 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