-0.000 000 000 742 147 676 646 721 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 721 9(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
-0.000 000 000 742 147 676 646 721 9(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:

|-0.000 000 000 742 147 676 646 721 9| = 0.000 000 000 742 147 676 646 721 9


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: 0.000 000 000 742 147 676 646 721 9.

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) 0.000 000 000 742 147 676 646 721 9 × 2 = 0 + 0.000 000 001 484 295 353 293 443 8;
  • 2) 0.000 000 001 484 295 353 293 443 8 × 2 = 0 + 0.000 000 002 968 590 706 586 887 6;
  • 3) 0.000 000 002 968 590 706 586 887 6 × 2 = 0 + 0.000 000 005 937 181 413 173 775 2;
  • 4) 0.000 000 005 937 181 413 173 775 2 × 2 = 0 + 0.000 000 011 874 362 826 347 550 4;
  • 5) 0.000 000 011 874 362 826 347 550 4 × 2 = 0 + 0.000 000 023 748 725 652 695 100 8;
  • 6) 0.000 000 023 748 725 652 695 100 8 × 2 = 0 + 0.000 000 047 497 451 305 390 201 6;
  • 7) 0.000 000 047 497 451 305 390 201 6 × 2 = 0 + 0.000 000 094 994 902 610 780 403 2;
  • 8) 0.000 000 094 994 902 610 780 403 2 × 2 = 0 + 0.000 000 189 989 805 221 560 806 4;
  • 9) 0.000 000 189 989 805 221 560 806 4 × 2 = 0 + 0.000 000 379 979 610 443 121 612 8;
  • 10) 0.000 000 379 979 610 443 121 612 8 × 2 = 0 + 0.000 000 759 959 220 886 243 225 6;
  • 11) 0.000 000 759 959 220 886 243 225 6 × 2 = 0 + 0.000 001 519 918 441 772 486 451 2;
  • 12) 0.000 001 519 918 441 772 486 451 2 × 2 = 0 + 0.000 003 039 836 883 544 972 902 4;
  • 13) 0.000 003 039 836 883 544 972 902 4 × 2 = 0 + 0.000 006 079 673 767 089 945 804 8;
  • 14) 0.000 006 079 673 767 089 945 804 8 × 2 = 0 + 0.000 012 159 347 534 179 891 609 6;
  • 15) 0.000 012 159 347 534 179 891 609 6 × 2 = 0 + 0.000 024 318 695 068 359 783 219 2;
  • 16) 0.000 024 318 695 068 359 783 219 2 × 2 = 0 + 0.000 048 637 390 136 719 566 438 4;
  • 17) 0.000 048 637 390 136 719 566 438 4 × 2 = 0 + 0.000 097 274 780 273 439 132 876 8;
  • 18) 0.000 097 274 780 273 439 132 876 8 × 2 = 0 + 0.000 194 549 560 546 878 265 753 6;
  • 19) 0.000 194 549 560 546 878 265 753 6 × 2 = 0 + 0.000 389 099 121 093 756 531 507 2;
  • 20) 0.000 389 099 121 093 756 531 507 2 × 2 = 0 + 0.000 778 198 242 187 513 063 014 4;
  • 21) 0.000 778 198 242 187 513 063 014 4 × 2 = 0 + 0.001 556 396 484 375 026 126 028 8;
  • 22) 0.001 556 396 484 375 026 126 028 8 × 2 = 0 + 0.003 112 792 968 750 052 252 057 6;
  • 23) 0.003 112 792 968 750 052 252 057 6 × 2 = 0 + 0.006 225 585 937 500 104 504 115 2;
  • 24) 0.006 225 585 937 500 104 504 115 2 × 2 = 0 + 0.012 451 171 875 000 209 008 230 4;
  • 25) 0.012 451 171 875 000 209 008 230 4 × 2 = 0 + 0.024 902 343 750 000 418 016 460 8;
  • 26) 0.024 902 343 750 000 418 016 460 8 × 2 = 0 + 0.049 804 687 500 000 836 032 921 6;
  • 27) 0.049 804 687 500 000 836 032 921 6 × 2 = 0 + 0.099 609 375 000 001 672 065 843 2;
  • 28) 0.099 609 375 000 001 672 065 843 2 × 2 = 0 + 0.199 218 750 000 003 344 131 686 4;
  • 29) 0.199 218 750 000 003 344 131 686 4 × 2 = 0 + 0.398 437 500 000 006 688 263 372 8;
  • 30) 0.398 437 500 000 006 688 263 372 8 × 2 = 0 + 0.796 875 000 000 013 376 526 745 6;
  • 31) 0.796 875 000 000 013 376 526 745 6 × 2 = 1 + 0.593 750 000 000 026 753 053 491 2;
  • 32) 0.593 750 000 000 026 753 053 491 2 × 2 = 1 + 0.187 500 000 000 053 506 106 982 4;
  • 33) 0.187 500 000 000 053 506 106 982 4 × 2 = 0 + 0.375 000 000 000 107 012 213 964 8;
  • 34) 0.375 000 000 000 107 012 213 964 8 × 2 = 0 + 0.750 000 000 000 214 024 427 929 6;
  • 35) 0.750 000 000 000 214 024 427 929 6 × 2 = 1 + 0.500 000 000 000 428 048 855 859 2;
  • 36) 0.500 000 000 000 428 048 855 859 2 × 2 = 1 + 0.000 000 000 000 856 097 711 718 4;
  • 37) 0.000 000 000 000 856 097 711 718 4 × 2 = 0 + 0.000 000 000 001 712 195 423 436 8;
  • 38) 0.000 000 000 001 712 195 423 436 8 × 2 = 0 + 0.000 000 000 003 424 390 846 873 6;
  • 39) 0.000 000 000 003 424 390 846 873 6 × 2 = 0 + 0.000 000 000 006 848 781 693 747 2;
  • 40) 0.000 000 000 006 848 781 693 747 2 × 2 = 0 + 0.000 000 000 013 697 563 387 494 4;
  • 41) 0.000 000 000 013 697 563 387 494 4 × 2 = 0 + 0.000 000 000 027 395 126 774 988 8;
  • 42) 0.000 000 000 027 395 126 774 988 8 × 2 = 0 + 0.000 000 000 054 790 253 549 977 6;
  • 43) 0.000 000 000 054 790 253 549 977 6 × 2 = 0 + 0.000 000 000 109 580 507 099 955 2;
  • 44) 0.000 000 000 109 580 507 099 955 2 × 2 = 0 + 0.000 000 000 219 161 014 199 910 4;
  • 45) 0.000 000 000 219 161 014 199 910 4 × 2 = 0 + 0.000 000 000 438 322 028 399 820 8;
  • 46) 0.000 000 000 438 322 028 399 820 8 × 2 = 0 + 0.000 000 000 876 644 056 799 641 6;
  • 47) 0.000 000 000 876 644 056 799 641 6 × 2 = 0 + 0.000 000 001 753 288 113 599 283 2;
  • 48) 0.000 000 001 753 288 113 599 283 2 × 2 = 0 + 0.000 000 003 506 576 227 198 566 4;
  • 49) 0.000 000 003 506 576 227 198 566 4 × 2 = 0 + 0.000 000 007 013 152 454 397 132 8;
  • 50) 0.000 000 007 013 152 454 397 132 8 × 2 = 0 + 0.000 000 014 026 304 908 794 265 6;
  • 51) 0.000 000 014 026 304 908 794 265 6 × 2 = 0 + 0.000 000 028 052 609 817 588 531 2;
  • 52) 0.000 000 028 052 609 817 588 531 2 × 2 = 0 + 0.000 000 056 105 219 635 177 062 4;
  • 53) 0.000 000 056 105 219 635 177 062 4 × 2 = 0 + 0.000 000 112 210 439 270 354 124 8;
  • 54) 0.000 000 112 210 439 270 354 124 8 × 2 = 0 + 0.000 000 224 420 878 540 708 249 6;

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:


0.000 000 000 742 147 676 646 721 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 721 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 676 646 721 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 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 1000 0000 0000 0000 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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 721 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 676 646 718 3 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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