-0.000 000 000 742 147 75 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 75(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
-0.000 000 000 742 147 75(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:

|-0.000 000 000 742 147 75| = 0.000 000 000 742 147 75


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: 0.000 000 000 742 147 75.

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) 0.000 000 000 742 147 75 × 2 = 0 + 0.000 000 001 484 295 5;
  • 2) 0.000 000 001 484 295 5 × 2 = 0 + 0.000 000 002 968 591;
  • 3) 0.000 000 002 968 591 × 2 = 0 + 0.000 000 005 937 182;
  • 4) 0.000 000 005 937 182 × 2 = 0 + 0.000 000 011 874 364;
  • 5) 0.000 000 011 874 364 × 2 = 0 + 0.000 000 023 748 728;
  • 6) 0.000 000 023 748 728 × 2 = 0 + 0.000 000 047 497 456;
  • 7) 0.000 000 047 497 456 × 2 = 0 + 0.000 000 094 994 912;
  • 8) 0.000 000 094 994 912 × 2 = 0 + 0.000 000 189 989 824;
  • 9) 0.000 000 189 989 824 × 2 = 0 + 0.000 000 379 979 648;
  • 10) 0.000 000 379 979 648 × 2 = 0 + 0.000 000 759 959 296;
  • 11) 0.000 000 759 959 296 × 2 = 0 + 0.000 001 519 918 592;
  • 12) 0.000 001 519 918 592 × 2 = 0 + 0.000 003 039 837 184;
  • 13) 0.000 003 039 837 184 × 2 = 0 + 0.000 006 079 674 368;
  • 14) 0.000 006 079 674 368 × 2 = 0 + 0.000 012 159 348 736;
  • 15) 0.000 012 159 348 736 × 2 = 0 + 0.000 024 318 697 472;
  • 16) 0.000 024 318 697 472 × 2 = 0 + 0.000 048 637 394 944;
  • 17) 0.000 048 637 394 944 × 2 = 0 + 0.000 097 274 789 888;
  • 18) 0.000 097 274 789 888 × 2 = 0 + 0.000 194 549 579 776;
  • 19) 0.000 194 549 579 776 × 2 = 0 + 0.000 389 099 159 552;
  • 20) 0.000 389 099 159 552 × 2 = 0 + 0.000 778 198 319 104;
  • 21) 0.000 778 198 319 104 × 2 = 0 + 0.001 556 396 638 208;
  • 22) 0.001 556 396 638 208 × 2 = 0 + 0.003 112 793 276 416;
  • 23) 0.003 112 793 276 416 × 2 = 0 + 0.006 225 586 552 832;
  • 24) 0.006 225 586 552 832 × 2 = 0 + 0.012 451 173 105 664;
  • 25) 0.012 451 173 105 664 × 2 = 0 + 0.024 902 346 211 328;
  • 26) 0.024 902 346 211 328 × 2 = 0 + 0.049 804 692 422 656;
  • 27) 0.049 804 692 422 656 × 2 = 0 + 0.099 609 384 845 312;
  • 28) 0.099 609 384 845 312 × 2 = 0 + 0.199 218 769 690 624;
  • 29) 0.199 218 769 690 624 × 2 = 0 + 0.398 437 539 381 248;
  • 30) 0.398 437 539 381 248 × 2 = 0 + 0.796 875 078 762 496;
  • 31) 0.796 875 078 762 496 × 2 = 1 + 0.593 750 157 524 992;
  • 32) 0.593 750 157 524 992 × 2 = 1 + 0.187 500 315 049 984;
  • 33) 0.187 500 315 049 984 × 2 = 0 + 0.375 000 630 099 968;
  • 34) 0.375 000 630 099 968 × 2 = 0 + 0.750 001 260 199 936;
  • 35) 0.750 001 260 199 936 × 2 = 1 + 0.500 002 520 399 872;
  • 36) 0.500 002 520 399 872 × 2 = 1 + 0.000 005 040 799 744;
  • 37) 0.000 005 040 799 744 × 2 = 0 + 0.000 010 081 599 488;
  • 38) 0.000 010 081 599 488 × 2 = 0 + 0.000 020 163 198 976;
  • 39) 0.000 020 163 198 976 × 2 = 0 + 0.000 040 326 397 952;
  • 40) 0.000 040 326 397 952 × 2 = 0 + 0.000 080 652 795 904;
  • 41) 0.000 080 652 795 904 × 2 = 0 + 0.000 161 305 591 808;
  • 42) 0.000 161 305 591 808 × 2 = 0 + 0.000 322 611 183 616;
  • 43) 0.000 322 611 183 616 × 2 = 0 + 0.000 645 222 367 232;
  • 44) 0.000 645 222 367 232 × 2 = 0 + 0.001 290 444 734 464;
  • 45) 0.001 290 444 734 464 × 2 = 0 + 0.002 580 889 468 928;
  • 46) 0.002 580 889 468 928 × 2 = 0 + 0.005 161 778 937 856;
  • 47) 0.005 161 778 937 856 × 2 = 0 + 0.010 323 557 875 712;
  • 48) 0.010 323 557 875 712 × 2 = 0 + 0.020 647 115 751 424;
  • 49) 0.020 647 115 751 424 × 2 = 0 + 0.041 294 231 502 848;
  • 50) 0.041 294 231 502 848 × 2 = 0 + 0.082 588 463 005 696;
  • 51) 0.082 588 463 005 696 × 2 = 0 + 0.165 176 926 011 392;
  • 52) 0.165 176 926 011 392 × 2 = 0 + 0.330 353 852 022 784;
  • 53) 0.330 353 852 022 784 × 2 = 0 + 0.660 707 704 045 568;
  • 54) 0.660 707 704 045 568 × 2 = 1 + 0.321 415 408 091 136;

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:


0.000 000 000 742 147 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(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:


0.000 000 000 742 147 75(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 01(2) × 20 =

1.1001 1000 0000 0000 0000 001(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 1000 0000 0000 0000 001


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 1100 0000 0000 0000 0001 =

100 1100 0000 0000 0000 0001


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 1100 0000 0000 0000 0001


Decimal number -0.000 000 000 742 147 75 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

Share

» Convert decimal -0.000 000 000 742 146 79 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 Hidden Requirement in Every Single Job Description. NotebookLM 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