-0.000 000 000 742 147 773 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 773(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
-0.000 000 000 742 147 773(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:

|-0.000 000 000 742 147 773| = 0.000 000 000 742 147 773


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: 0.000 000 000 742 147 773.

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) 0.000 000 000 742 147 773 × 2 = 0 + 0.000 000 001 484 295 546;
  • 2) 0.000 000 001 484 295 546 × 2 = 0 + 0.000 000 002 968 591 092;
  • 3) 0.000 000 002 968 591 092 × 2 = 0 + 0.000 000 005 937 182 184;
  • 4) 0.000 000 005 937 182 184 × 2 = 0 + 0.000 000 011 874 364 368;
  • 5) 0.000 000 011 874 364 368 × 2 = 0 + 0.000 000 023 748 728 736;
  • 6) 0.000 000 023 748 728 736 × 2 = 0 + 0.000 000 047 497 457 472;
  • 7) 0.000 000 047 497 457 472 × 2 = 0 + 0.000 000 094 994 914 944;
  • 8) 0.000 000 094 994 914 944 × 2 = 0 + 0.000 000 189 989 829 888;
  • 9) 0.000 000 189 989 829 888 × 2 = 0 + 0.000 000 379 979 659 776;
  • 10) 0.000 000 379 979 659 776 × 2 = 0 + 0.000 000 759 959 319 552;
  • 11) 0.000 000 759 959 319 552 × 2 = 0 + 0.000 001 519 918 639 104;
  • 12) 0.000 001 519 918 639 104 × 2 = 0 + 0.000 003 039 837 278 208;
  • 13) 0.000 003 039 837 278 208 × 2 = 0 + 0.000 006 079 674 556 416;
  • 14) 0.000 006 079 674 556 416 × 2 = 0 + 0.000 012 159 349 112 832;
  • 15) 0.000 012 159 349 112 832 × 2 = 0 + 0.000 024 318 698 225 664;
  • 16) 0.000 024 318 698 225 664 × 2 = 0 + 0.000 048 637 396 451 328;
  • 17) 0.000 048 637 396 451 328 × 2 = 0 + 0.000 097 274 792 902 656;
  • 18) 0.000 097 274 792 902 656 × 2 = 0 + 0.000 194 549 585 805 312;
  • 19) 0.000 194 549 585 805 312 × 2 = 0 + 0.000 389 099 171 610 624;
  • 20) 0.000 389 099 171 610 624 × 2 = 0 + 0.000 778 198 343 221 248;
  • 21) 0.000 778 198 343 221 248 × 2 = 0 + 0.001 556 396 686 442 496;
  • 22) 0.001 556 396 686 442 496 × 2 = 0 + 0.003 112 793 372 884 992;
  • 23) 0.003 112 793 372 884 992 × 2 = 0 + 0.006 225 586 745 769 984;
  • 24) 0.006 225 586 745 769 984 × 2 = 0 + 0.012 451 173 491 539 968;
  • 25) 0.012 451 173 491 539 968 × 2 = 0 + 0.024 902 346 983 079 936;
  • 26) 0.024 902 346 983 079 936 × 2 = 0 + 0.049 804 693 966 159 872;
  • 27) 0.049 804 693 966 159 872 × 2 = 0 + 0.099 609 387 932 319 744;
  • 28) 0.099 609 387 932 319 744 × 2 = 0 + 0.199 218 775 864 639 488;
  • 29) 0.199 218 775 864 639 488 × 2 = 0 + 0.398 437 551 729 278 976;
  • 30) 0.398 437 551 729 278 976 × 2 = 0 + 0.796 875 103 458 557 952;
  • 31) 0.796 875 103 458 557 952 × 2 = 1 + 0.593 750 206 917 115 904;
  • 32) 0.593 750 206 917 115 904 × 2 = 1 + 0.187 500 413 834 231 808;
  • 33) 0.187 500 413 834 231 808 × 2 = 0 + 0.375 000 827 668 463 616;
  • 34) 0.375 000 827 668 463 616 × 2 = 0 + 0.750 001 655 336 927 232;
  • 35) 0.750 001 655 336 927 232 × 2 = 1 + 0.500 003 310 673 854 464;
  • 36) 0.500 003 310 673 854 464 × 2 = 1 + 0.000 006 621 347 708 928;
  • 37) 0.000 006 621 347 708 928 × 2 = 0 + 0.000 013 242 695 417 856;
  • 38) 0.000 013 242 695 417 856 × 2 = 0 + 0.000 026 485 390 835 712;
  • 39) 0.000 026 485 390 835 712 × 2 = 0 + 0.000 052 970 781 671 424;
  • 40) 0.000 052 970 781 671 424 × 2 = 0 + 0.000 105 941 563 342 848;
  • 41) 0.000 105 941 563 342 848 × 2 = 0 + 0.000 211 883 126 685 696;
  • 42) 0.000 211 883 126 685 696 × 2 = 0 + 0.000 423 766 253 371 392;
  • 43) 0.000 423 766 253 371 392 × 2 = 0 + 0.000 847 532 506 742 784;
  • 44) 0.000 847 532 506 742 784 × 2 = 0 + 0.001 695 065 013 485 568;
  • 45) 0.001 695 065 013 485 568 × 2 = 0 + 0.003 390 130 026 971 136;
  • 46) 0.003 390 130 026 971 136 × 2 = 0 + 0.006 780 260 053 942 272;
  • 47) 0.006 780 260 053 942 272 × 2 = 0 + 0.013 560 520 107 884 544;
  • 48) 0.013 560 520 107 884 544 × 2 = 0 + 0.027 121 040 215 769 088;
  • 49) 0.027 121 040 215 769 088 × 2 = 0 + 0.054 242 080 431 538 176;
  • 50) 0.054 242 080 431 538 176 × 2 = 0 + 0.108 484 160 863 076 352;
  • 51) 0.108 484 160 863 076 352 × 2 = 0 + 0.216 968 321 726 152 704;
  • 52) 0.216 968 321 726 152 704 × 2 = 0 + 0.433 936 643 452 305 408;
  • 53) 0.433 936 643 452 305 408 × 2 = 0 + 0.867 873 286 904 610 816;
  • 54) 0.867 873 286 904 610 816 × 2 = 1 + 0.735 746 573 809 221 632;

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:


0.000 000 000 742 147 773(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 773(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:


0.000 000 000 742 147 773(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 -0.000 000 000 742 147 773 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

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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