-0.000 000 000 742 147 676 646 728 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 728 4(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
-0.000 000 000 742 147 676 646 728 4(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:

|-0.000 000 000 742 147 676 646 728 4| = 0.000 000 000 742 147 676 646 728 4


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: 0.000 000 000 742 147 676 646 728 4.

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) 0.000 000 000 742 147 676 646 728 4 × 2 = 0 + 0.000 000 001 484 295 353 293 456 8;
  • 2) 0.000 000 001 484 295 353 293 456 8 × 2 = 0 + 0.000 000 002 968 590 706 586 913 6;
  • 3) 0.000 000 002 968 590 706 586 913 6 × 2 = 0 + 0.000 000 005 937 181 413 173 827 2;
  • 4) 0.000 000 005 937 181 413 173 827 2 × 2 = 0 + 0.000 000 011 874 362 826 347 654 4;
  • 5) 0.000 000 011 874 362 826 347 654 4 × 2 = 0 + 0.000 000 023 748 725 652 695 308 8;
  • 6) 0.000 000 023 748 725 652 695 308 8 × 2 = 0 + 0.000 000 047 497 451 305 390 617 6;
  • 7) 0.000 000 047 497 451 305 390 617 6 × 2 = 0 + 0.000 000 094 994 902 610 781 235 2;
  • 8) 0.000 000 094 994 902 610 781 235 2 × 2 = 0 + 0.000 000 189 989 805 221 562 470 4;
  • 9) 0.000 000 189 989 805 221 562 470 4 × 2 = 0 + 0.000 000 379 979 610 443 124 940 8;
  • 10) 0.000 000 379 979 610 443 124 940 8 × 2 = 0 + 0.000 000 759 959 220 886 249 881 6;
  • 11) 0.000 000 759 959 220 886 249 881 6 × 2 = 0 + 0.000 001 519 918 441 772 499 763 2;
  • 12) 0.000 001 519 918 441 772 499 763 2 × 2 = 0 + 0.000 003 039 836 883 544 999 526 4;
  • 13) 0.000 003 039 836 883 544 999 526 4 × 2 = 0 + 0.000 006 079 673 767 089 999 052 8;
  • 14) 0.000 006 079 673 767 089 999 052 8 × 2 = 0 + 0.000 012 159 347 534 179 998 105 6;
  • 15) 0.000 012 159 347 534 179 998 105 6 × 2 = 0 + 0.000 024 318 695 068 359 996 211 2;
  • 16) 0.000 024 318 695 068 359 996 211 2 × 2 = 0 + 0.000 048 637 390 136 719 992 422 4;
  • 17) 0.000 048 637 390 136 719 992 422 4 × 2 = 0 + 0.000 097 274 780 273 439 984 844 8;
  • 18) 0.000 097 274 780 273 439 984 844 8 × 2 = 0 + 0.000 194 549 560 546 879 969 689 6;
  • 19) 0.000 194 549 560 546 879 969 689 6 × 2 = 0 + 0.000 389 099 121 093 759 939 379 2;
  • 20) 0.000 389 099 121 093 759 939 379 2 × 2 = 0 + 0.000 778 198 242 187 519 878 758 4;
  • 21) 0.000 778 198 242 187 519 878 758 4 × 2 = 0 + 0.001 556 396 484 375 039 757 516 8;
  • 22) 0.001 556 396 484 375 039 757 516 8 × 2 = 0 + 0.003 112 792 968 750 079 515 033 6;
  • 23) 0.003 112 792 968 750 079 515 033 6 × 2 = 0 + 0.006 225 585 937 500 159 030 067 2;
  • 24) 0.006 225 585 937 500 159 030 067 2 × 2 = 0 + 0.012 451 171 875 000 318 060 134 4;
  • 25) 0.012 451 171 875 000 318 060 134 4 × 2 = 0 + 0.024 902 343 750 000 636 120 268 8;
  • 26) 0.024 902 343 750 000 636 120 268 8 × 2 = 0 + 0.049 804 687 500 001 272 240 537 6;
  • 27) 0.049 804 687 500 001 272 240 537 6 × 2 = 0 + 0.099 609 375 000 002 544 481 075 2;
  • 28) 0.099 609 375 000 002 544 481 075 2 × 2 = 0 + 0.199 218 750 000 005 088 962 150 4;
  • 29) 0.199 218 750 000 005 088 962 150 4 × 2 = 0 + 0.398 437 500 000 010 177 924 300 8;
  • 30) 0.398 437 500 000 010 177 924 300 8 × 2 = 0 + 0.796 875 000 000 020 355 848 601 6;
  • 31) 0.796 875 000 000 020 355 848 601 6 × 2 = 1 + 0.593 750 000 000 040 711 697 203 2;
  • 32) 0.593 750 000 000 040 711 697 203 2 × 2 = 1 + 0.187 500 000 000 081 423 394 406 4;
  • 33) 0.187 500 000 000 081 423 394 406 4 × 2 = 0 + 0.375 000 000 000 162 846 788 812 8;
  • 34) 0.375 000 000 000 162 846 788 812 8 × 2 = 0 + 0.750 000 000 000 325 693 577 625 6;
  • 35) 0.750 000 000 000 325 693 577 625 6 × 2 = 1 + 0.500 000 000 000 651 387 155 251 2;
  • 36) 0.500 000 000 000 651 387 155 251 2 × 2 = 1 + 0.000 000 000 001 302 774 310 502 4;
  • 37) 0.000 000 000 001 302 774 310 502 4 × 2 = 0 + 0.000 000 000 002 605 548 621 004 8;
  • 38) 0.000 000 000 002 605 548 621 004 8 × 2 = 0 + 0.000 000 000 005 211 097 242 009 6;
  • 39) 0.000 000 000 005 211 097 242 009 6 × 2 = 0 + 0.000 000 000 010 422 194 484 019 2;
  • 40) 0.000 000 000 010 422 194 484 019 2 × 2 = 0 + 0.000 000 000 020 844 388 968 038 4;
  • 41) 0.000 000 000 020 844 388 968 038 4 × 2 = 0 + 0.000 000 000 041 688 777 936 076 8;
  • 42) 0.000 000 000 041 688 777 936 076 8 × 2 = 0 + 0.000 000 000 083 377 555 872 153 6;
  • 43) 0.000 000 000 083 377 555 872 153 6 × 2 = 0 + 0.000 000 000 166 755 111 744 307 2;
  • 44) 0.000 000 000 166 755 111 744 307 2 × 2 = 0 + 0.000 000 000 333 510 223 488 614 4;
  • 45) 0.000 000 000 333 510 223 488 614 4 × 2 = 0 + 0.000 000 000 667 020 446 977 228 8;
  • 46) 0.000 000 000 667 020 446 977 228 8 × 2 = 0 + 0.000 000 001 334 040 893 954 457 6;
  • 47) 0.000 000 001 334 040 893 954 457 6 × 2 = 0 + 0.000 000 002 668 081 787 908 915 2;
  • 48) 0.000 000 002 668 081 787 908 915 2 × 2 = 0 + 0.000 000 005 336 163 575 817 830 4;
  • 49) 0.000 000 005 336 163 575 817 830 4 × 2 = 0 + 0.000 000 010 672 327 151 635 660 8;
  • 50) 0.000 000 010 672 327 151 635 660 8 × 2 = 0 + 0.000 000 021 344 654 303 271 321 6;
  • 51) 0.000 000 021 344 654 303 271 321 6 × 2 = 0 + 0.000 000 042 689 308 606 542 643 2;
  • 52) 0.000 000 042 689 308 606 542 643 2 × 2 = 0 + 0.000 000 085 378 617 213 085 286 4;
  • 53) 0.000 000 085 378 617 213 085 286 4 × 2 = 0 + 0.000 000 170 757 234 426 170 572 8;
  • 54) 0.000 000 170 757 234 426 170 572 8 × 2 = 0 + 0.000 000 341 514 468 852 341 145 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:


0.000 000 000 742 147 676 646 728 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 728 4(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:


0.000 000 000 742 147 676 646 728 4(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 -0.000 000 000 742 147 676 646 728 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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