-0.000 000 000 742 147 823 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 823(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
-0.000 000 000 742 147 823(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:

|-0.000 000 000 742 147 823| = 0.000 000 000 742 147 823


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: 0.000 000 000 742 147 823.

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) 0.000 000 000 742 147 823 × 2 = 0 + 0.000 000 001 484 295 646;
  • 2) 0.000 000 001 484 295 646 × 2 = 0 + 0.000 000 002 968 591 292;
  • 3) 0.000 000 002 968 591 292 × 2 = 0 + 0.000 000 005 937 182 584;
  • 4) 0.000 000 005 937 182 584 × 2 = 0 + 0.000 000 011 874 365 168;
  • 5) 0.000 000 011 874 365 168 × 2 = 0 + 0.000 000 023 748 730 336;
  • 6) 0.000 000 023 748 730 336 × 2 = 0 + 0.000 000 047 497 460 672;
  • 7) 0.000 000 047 497 460 672 × 2 = 0 + 0.000 000 094 994 921 344;
  • 8) 0.000 000 094 994 921 344 × 2 = 0 + 0.000 000 189 989 842 688;
  • 9) 0.000 000 189 989 842 688 × 2 = 0 + 0.000 000 379 979 685 376;
  • 10) 0.000 000 379 979 685 376 × 2 = 0 + 0.000 000 759 959 370 752;
  • 11) 0.000 000 759 959 370 752 × 2 = 0 + 0.000 001 519 918 741 504;
  • 12) 0.000 001 519 918 741 504 × 2 = 0 + 0.000 003 039 837 483 008;
  • 13) 0.000 003 039 837 483 008 × 2 = 0 + 0.000 006 079 674 966 016;
  • 14) 0.000 006 079 674 966 016 × 2 = 0 + 0.000 012 159 349 932 032;
  • 15) 0.000 012 159 349 932 032 × 2 = 0 + 0.000 024 318 699 864 064;
  • 16) 0.000 024 318 699 864 064 × 2 = 0 + 0.000 048 637 399 728 128;
  • 17) 0.000 048 637 399 728 128 × 2 = 0 + 0.000 097 274 799 456 256;
  • 18) 0.000 097 274 799 456 256 × 2 = 0 + 0.000 194 549 598 912 512;
  • 19) 0.000 194 549 598 912 512 × 2 = 0 + 0.000 389 099 197 825 024;
  • 20) 0.000 389 099 197 825 024 × 2 = 0 + 0.000 778 198 395 650 048;
  • 21) 0.000 778 198 395 650 048 × 2 = 0 + 0.001 556 396 791 300 096;
  • 22) 0.001 556 396 791 300 096 × 2 = 0 + 0.003 112 793 582 600 192;
  • 23) 0.003 112 793 582 600 192 × 2 = 0 + 0.006 225 587 165 200 384;
  • 24) 0.006 225 587 165 200 384 × 2 = 0 + 0.012 451 174 330 400 768;
  • 25) 0.012 451 174 330 400 768 × 2 = 0 + 0.024 902 348 660 801 536;
  • 26) 0.024 902 348 660 801 536 × 2 = 0 + 0.049 804 697 321 603 072;
  • 27) 0.049 804 697 321 603 072 × 2 = 0 + 0.099 609 394 643 206 144;
  • 28) 0.099 609 394 643 206 144 × 2 = 0 + 0.199 218 789 286 412 288;
  • 29) 0.199 218 789 286 412 288 × 2 = 0 + 0.398 437 578 572 824 576;
  • 30) 0.398 437 578 572 824 576 × 2 = 0 + 0.796 875 157 145 649 152;
  • 31) 0.796 875 157 145 649 152 × 2 = 1 + 0.593 750 314 291 298 304;
  • 32) 0.593 750 314 291 298 304 × 2 = 1 + 0.187 500 628 582 596 608;
  • 33) 0.187 500 628 582 596 608 × 2 = 0 + 0.375 001 257 165 193 216;
  • 34) 0.375 001 257 165 193 216 × 2 = 0 + 0.750 002 514 330 386 432;
  • 35) 0.750 002 514 330 386 432 × 2 = 1 + 0.500 005 028 660 772 864;
  • 36) 0.500 005 028 660 772 864 × 2 = 1 + 0.000 010 057 321 545 728;
  • 37) 0.000 010 057 321 545 728 × 2 = 0 + 0.000 020 114 643 091 456;
  • 38) 0.000 020 114 643 091 456 × 2 = 0 + 0.000 040 229 286 182 912;
  • 39) 0.000 040 229 286 182 912 × 2 = 0 + 0.000 080 458 572 365 824;
  • 40) 0.000 080 458 572 365 824 × 2 = 0 + 0.000 160 917 144 731 648;
  • 41) 0.000 160 917 144 731 648 × 2 = 0 + 0.000 321 834 289 463 296;
  • 42) 0.000 321 834 289 463 296 × 2 = 0 + 0.000 643 668 578 926 592;
  • 43) 0.000 643 668 578 926 592 × 2 = 0 + 0.001 287 337 157 853 184;
  • 44) 0.001 287 337 157 853 184 × 2 = 0 + 0.002 574 674 315 706 368;
  • 45) 0.002 574 674 315 706 368 × 2 = 0 + 0.005 149 348 631 412 736;
  • 46) 0.005 149 348 631 412 736 × 2 = 0 + 0.010 298 697 262 825 472;
  • 47) 0.010 298 697 262 825 472 × 2 = 0 + 0.020 597 394 525 650 944;
  • 48) 0.020 597 394 525 650 944 × 2 = 0 + 0.041 194 789 051 301 888;
  • 49) 0.041 194 789 051 301 888 × 2 = 0 + 0.082 389 578 102 603 776;
  • 50) 0.082 389 578 102 603 776 × 2 = 0 + 0.164 779 156 205 207 552;
  • 51) 0.164 779 156 205 207 552 × 2 = 0 + 0.329 558 312 410 415 104;
  • 52) 0.329 558 312 410 415 104 × 2 = 0 + 0.659 116 624 820 830 208;
  • 53) 0.659 116 624 820 830 208 × 2 = 1 + 0.318 233 249 641 660 416;
  • 54) 0.318 233 249 641 660 416 × 2 = 0 + 0.636 466 499 283 320 832;

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:


0.000 000 000 742 147 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 823(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 823 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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