-0.000 000 000 742 147 878 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 878(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
-0.000 000 000 742 147 878(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:

|-0.000 000 000 742 147 878| = 0.000 000 000 742 147 878


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: 0.000 000 000 742 147 878.

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) 0.000 000 000 742 147 878 × 2 = 0 + 0.000 000 001 484 295 756;
  • 2) 0.000 000 001 484 295 756 × 2 = 0 + 0.000 000 002 968 591 512;
  • 3) 0.000 000 002 968 591 512 × 2 = 0 + 0.000 000 005 937 183 024;
  • 4) 0.000 000 005 937 183 024 × 2 = 0 + 0.000 000 011 874 366 048;
  • 5) 0.000 000 011 874 366 048 × 2 = 0 + 0.000 000 023 748 732 096;
  • 6) 0.000 000 023 748 732 096 × 2 = 0 + 0.000 000 047 497 464 192;
  • 7) 0.000 000 047 497 464 192 × 2 = 0 + 0.000 000 094 994 928 384;
  • 8) 0.000 000 094 994 928 384 × 2 = 0 + 0.000 000 189 989 856 768;
  • 9) 0.000 000 189 989 856 768 × 2 = 0 + 0.000 000 379 979 713 536;
  • 10) 0.000 000 379 979 713 536 × 2 = 0 + 0.000 000 759 959 427 072;
  • 11) 0.000 000 759 959 427 072 × 2 = 0 + 0.000 001 519 918 854 144;
  • 12) 0.000 001 519 918 854 144 × 2 = 0 + 0.000 003 039 837 708 288;
  • 13) 0.000 003 039 837 708 288 × 2 = 0 + 0.000 006 079 675 416 576;
  • 14) 0.000 006 079 675 416 576 × 2 = 0 + 0.000 012 159 350 833 152;
  • 15) 0.000 012 159 350 833 152 × 2 = 0 + 0.000 024 318 701 666 304;
  • 16) 0.000 024 318 701 666 304 × 2 = 0 + 0.000 048 637 403 332 608;
  • 17) 0.000 048 637 403 332 608 × 2 = 0 + 0.000 097 274 806 665 216;
  • 18) 0.000 097 274 806 665 216 × 2 = 0 + 0.000 194 549 613 330 432;
  • 19) 0.000 194 549 613 330 432 × 2 = 0 + 0.000 389 099 226 660 864;
  • 20) 0.000 389 099 226 660 864 × 2 = 0 + 0.000 778 198 453 321 728;
  • 21) 0.000 778 198 453 321 728 × 2 = 0 + 0.001 556 396 906 643 456;
  • 22) 0.001 556 396 906 643 456 × 2 = 0 + 0.003 112 793 813 286 912;
  • 23) 0.003 112 793 813 286 912 × 2 = 0 + 0.006 225 587 626 573 824;
  • 24) 0.006 225 587 626 573 824 × 2 = 0 + 0.012 451 175 253 147 648;
  • 25) 0.012 451 175 253 147 648 × 2 = 0 + 0.024 902 350 506 295 296;
  • 26) 0.024 902 350 506 295 296 × 2 = 0 + 0.049 804 701 012 590 592;
  • 27) 0.049 804 701 012 590 592 × 2 = 0 + 0.099 609 402 025 181 184;
  • 28) 0.099 609 402 025 181 184 × 2 = 0 + 0.199 218 804 050 362 368;
  • 29) 0.199 218 804 050 362 368 × 2 = 0 + 0.398 437 608 100 724 736;
  • 30) 0.398 437 608 100 724 736 × 2 = 0 + 0.796 875 216 201 449 472;
  • 31) 0.796 875 216 201 449 472 × 2 = 1 + 0.593 750 432 402 898 944;
  • 32) 0.593 750 432 402 898 944 × 2 = 1 + 0.187 500 864 805 797 888;
  • 33) 0.187 500 864 805 797 888 × 2 = 0 + 0.375 001 729 611 595 776;
  • 34) 0.375 001 729 611 595 776 × 2 = 0 + 0.750 003 459 223 191 552;
  • 35) 0.750 003 459 223 191 552 × 2 = 1 + 0.500 006 918 446 383 104;
  • 36) 0.500 006 918 446 383 104 × 2 = 1 + 0.000 013 836 892 766 208;
  • 37) 0.000 013 836 892 766 208 × 2 = 0 + 0.000 027 673 785 532 416;
  • 38) 0.000 027 673 785 532 416 × 2 = 0 + 0.000 055 347 571 064 832;
  • 39) 0.000 055 347 571 064 832 × 2 = 0 + 0.000 110 695 142 129 664;
  • 40) 0.000 110 695 142 129 664 × 2 = 0 + 0.000 221 390 284 259 328;
  • 41) 0.000 221 390 284 259 328 × 2 = 0 + 0.000 442 780 568 518 656;
  • 42) 0.000 442 780 568 518 656 × 2 = 0 + 0.000 885 561 137 037 312;
  • 43) 0.000 885 561 137 037 312 × 2 = 0 + 0.001 771 122 274 074 624;
  • 44) 0.001 771 122 274 074 624 × 2 = 0 + 0.003 542 244 548 149 248;
  • 45) 0.003 542 244 548 149 248 × 2 = 0 + 0.007 084 489 096 298 496;
  • 46) 0.007 084 489 096 298 496 × 2 = 0 + 0.014 168 978 192 596 992;
  • 47) 0.014 168 978 192 596 992 × 2 = 0 + 0.028 337 956 385 193 984;
  • 48) 0.028 337 956 385 193 984 × 2 = 0 + 0.056 675 912 770 387 968;
  • 49) 0.056 675 912 770 387 968 × 2 = 0 + 0.113 351 825 540 775 936;
  • 50) 0.113 351 825 540 775 936 × 2 = 0 + 0.226 703 651 081 551 872;
  • 51) 0.226 703 651 081 551 872 × 2 = 0 + 0.453 407 302 163 103 744;
  • 52) 0.453 407 302 163 103 744 × 2 = 0 + 0.906 814 604 326 207 488;
  • 53) 0.906 814 604 326 207 488 × 2 = 1 + 0.813 629 208 652 414 976;
  • 54) 0.813 629 208 652 414 976 × 2 = 1 + 0.627 258 417 304 829 952;

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:


0.000 000 000 742 147 878(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 878(10) =

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


0.000 000 000 742 147 878(10) =

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

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

1.1001 1000 0000 0000 0000 011(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 011


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 0011 =

100 1100 0000 0000 0000 0011


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 0011


Decimal number -0.000 000 000 742 147 878 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0011

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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