-0.000 000 000 742 147 676 646 707 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 707(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
-0.000 000 000 742 147 676 646 707(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:

|-0.000 000 000 742 147 676 646 707| = 0.000 000 000 742 147 676 646 707


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: 0.000 000 000 742 147 676 646 707.

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) 0.000 000 000 742 147 676 646 707 × 2 = 0 + 0.000 000 001 484 295 353 293 414;
  • 2) 0.000 000 001 484 295 353 293 414 × 2 = 0 + 0.000 000 002 968 590 706 586 828;
  • 3) 0.000 000 002 968 590 706 586 828 × 2 = 0 + 0.000 000 005 937 181 413 173 656;
  • 4) 0.000 000 005 937 181 413 173 656 × 2 = 0 + 0.000 000 011 874 362 826 347 312;
  • 5) 0.000 000 011 874 362 826 347 312 × 2 = 0 + 0.000 000 023 748 725 652 694 624;
  • 6) 0.000 000 023 748 725 652 694 624 × 2 = 0 + 0.000 000 047 497 451 305 389 248;
  • 7) 0.000 000 047 497 451 305 389 248 × 2 = 0 + 0.000 000 094 994 902 610 778 496;
  • 8) 0.000 000 094 994 902 610 778 496 × 2 = 0 + 0.000 000 189 989 805 221 556 992;
  • 9) 0.000 000 189 989 805 221 556 992 × 2 = 0 + 0.000 000 379 979 610 443 113 984;
  • 10) 0.000 000 379 979 610 443 113 984 × 2 = 0 + 0.000 000 759 959 220 886 227 968;
  • 11) 0.000 000 759 959 220 886 227 968 × 2 = 0 + 0.000 001 519 918 441 772 455 936;
  • 12) 0.000 001 519 918 441 772 455 936 × 2 = 0 + 0.000 003 039 836 883 544 911 872;
  • 13) 0.000 003 039 836 883 544 911 872 × 2 = 0 + 0.000 006 079 673 767 089 823 744;
  • 14) 0.000 006 079 673 767 089 823 744 × 2 = 0 + 0.000 012 159 347 534 179 647 488;
  • 15) 0.000 012 159 347 534 179 647 488 × 2 = 0 + 0.000 024 318 695 068 359 294 976;
  • 16) 0.000 024 318 695 068 359 294 976 × 2 = 0 + 0.000 048 637 390 136 718 589 952;
  • 17) 0.000 048 637 390 136 718 589 952 × 2 = 0 + 0.000 097 274 780 273 437 179 904;
  • 18) 0.000 097 274 780 273 437 179 904 × 2 = 0 + 0.000 194 549 560 546 874 359 808;
  • 19) 0.000 194 549 560 546 874 359 808 × 2 = 0 + 0.000 389 099 121 093 748 719 616;
  • 20) 0.000 389 099 121 093 748 719 616 × 2 = 0 + 0.000 778 198 242 187 497 439 232;
  • 21) 0.000 778 198 242 187 497 439 232 × 2 = 0 + 0.001 556 396 484 374 994 878 464;
  • 22) 0.001 556 396 484 374 994 878 464 × 2 = 0 + 0.003 112 792 968 749 989 756 928;
  • 23) 0.003 112 792 968 749 989 756 928 × 2 = 0 + 0.006 225 585 937 499 979 513 856;
  • 24) 0.006 225 585 937 499 979 513 856 × 2 = 0 + 0.012 451 171 874 999 959 027 712;
  • 25) 0.012 451 171 874 999 959 027 712 × 2 = 0 + 0.024 902 343 749 999 918 055 424;
  • 26) 0.024 902 343 749 999 918 055 424 × 2 = 0 + 0.049 804 687 499 999 836 110 848;
  • 27) 0.049 804 687 499 999 836 110 848 × 2 = 0 + 0.099 609 374 999 999 672 221 696;
  • 28) 0.099 609 374 999 999 672 221 696 × 2 = 0 + 0.199 218 749 999 999 344 443 392;
  • 29) 0.199 218 749 999 999 344 443 392 × 2 = 0 + 0.398 437 499 999 998 688 886 784;
  • 30) 0.398 437 499 999 998 688 886 784 × 2 = 0 + 0.796 874 999 999 997 377 773 568;
  • 31) 0.796 874 999 999 997 377 773 568 × 2 = 1 + 0.593 749 999 999 994 755 547 136;
  • 32) 0.593 749 999 999 994 755 547 136 × 2 = 1 + 0.187 499 999 999 989 511 094 272;
  • 33) 0.187 499 999 999 989 511 094 272 × 2 = 0 + 0.374 999 999 999 979 022 188 544;
  • 34) 0.374 999 999 999 979 022 188 544 × 2 = 0 + 0.749 999 999 999 958 044 377 088;
  • 35) 0.749 999 999 999 958 044 377 088 × 2 = 1 + 0.499 999 999 999 916 088 754 176;
  • 36) 0.499 999 999 999 916 088 754 176 × 2 = 0 + 0.999 999 999 999 832 177 508 352;
  • 37) 0.999 999 999 999 832 177 508 352 × 2 = 1 + 0.999 999 999 999 664 355 016 704;
  • 38) 0.999 999 999 999 664 355 016 704 × 2 = 1 + 0.999 999 999 999 328 710 033 408;
  • 39) 0.999 999 999 999 328 710 033 408 × 2 = 1 + 0.999 999 999 998 657 420 066 816;
  • 40) 0.999 999 999 998 657 420 066 816 × 2 = 1 + 0.999 999 999 997 314 840 133 632;
  • 41) 0.999 999 999 997 314 840 133 632 × 2 = 1 + 0.999 999 999 994 629 680 267 264;
  • 42) 0.999 999 999 994 629 680 267 264 × 2 = 1 + 0.999 999 999 989 259 360 534 528;
  • 43) 0.999 999 999 989 259 360 534 528 × 2 = 1 + 0.999 999 999 978 518 721 069 056;
  • 44) 0.999 999 999 978 518 721 069 056 × 2 = 1 + 0.999 999 999 957 037 442 138 112;
  • 45) 0.999 999 999 957 037 442 138 112 × 2 = 1 + 0.999 999 999 914 074 884 276 224;
  • 46) 0.999 999 999 914 074 884 276 224 × 2 = 1 + 0.999 999 999 828 149 768 552 448;
  • 47) 0.999 999 999 828 149 768 552 448 × 2 = 1 + 0.999 999 999 656 299 537 104 896;
  • 48) 0.999 999 999 656 299 537 104 896 × 2 = 1 + 0.999 999 999 312 599 074 209 792;
  • 49) 0.999 999 999 312 599 074 209 792 × 2 = 1 + 0.999 999 998 625 198 148 419 584;
  • 50) 0.999 999 998 625 198 148 419 584 × 2 = 1 + 0.999 999 997 250 396 296 839 168;
  • 51) 0.999 999 997 250 396 296 839 168 × 2 = 1 + 0.999 999 994 500 792 593 678 336;
  • 52) 0.999 999 994 500 792 593 678 336 × 2 = 1 + 0.999 999 989 001 585 187 356 672;
  • 53) 0.999 999 989 001 585 187 356 672 × 2 = 1 + 0.999 999 978 003 170 374 713 344;
  • 54) 0.999 999 978 003 170 374 713 344 × 2 = 1 + 0.999 999 956 006 340 749 426 688;

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:


0.000 000 000 742 147 676 646 707(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 707(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 646 707(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 707 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 646 666 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