-0.000 000 000 742 147 676 647 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 72(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
-0.000 000 000 742 147 676 647 72(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:

|-0.000 000 000 742 147 676 647 72| = 0.000 000 000 742 147 676 647 72


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: 0.000 000 000 742 147 676 647 72.

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) 0.000 000 000 742 147 676 647 72 × 2 = 0 + 0.000 000 001 484 295 353 295 44;
  • 2) 0.000 000 001 484 295 353 295 44 × 2 = 0 + 0.000 000 002 968 590 706 590 88;
  • 3) 0.000 000 002 968 590 706 590 88 × 2 = 0 + 0.000 000 005 937 181 413 181 76;
  • 4) 0.000 000 005 937 181 413 181 76 × 2 = 0 + 0.000 000 011 874 362 826 363 52;
  • 5) 0.000 000 011 874 362 826 363 52 × 2 = 0 + 0.000 000 023 748 725 652 727 04;
  • 6) 0.000 000 023 748 725 652 727 04 × 2 = 0 + 0.000 000 047 497 451 305 454 08;
  • 7) 0.000 000 047 497 451 305 454 08 × 2 = 0 + 0.000 000 094 994 902 610 908 16;
  • 8) 0.000 000 094 994 902 610 908 16 × 2 = 0 + 0.000 000 189 989 805 221 816 32;
  • 9) 0.000 000 189 989 805 221 816 32 × 2 = 0 + 0.000 000 379 979 610 443 632 64;
  • 10) 0.000 000 379 979 610 443 632 64 × 2 = 0 + 0.000 000 759 959 220 887 265 28;
  • 11) 0.000 000 759 959 220 887 265 28 × 2 = 0 + 0.000 001 519 918 441 774 530 56;
  • 12) 0.000 001 519 918 441 774 530 56 × 2 = 0 + 0.000 003 039 836 883 549 061 12;
  • 13) 0.000 003 039 836 883 549 061 12 × 2 = 0 + 0.000 006 079 673 767 098 122 24;
  • 14) 0.000 006 079 673 767 098 122 24 × 2 = 0 + 0.000 012 159 347 534 196 244 48;
  • 15) 0.000 012 159 347 534 196 244 48 × 2 = 0 + 0.000 024 318 695 068 392 488 96;
  • 16) 0.000 024 318 695 068 392 488 96 × 2 = 0 + 0.000 048 637 390 136 784 977 92;
  • 17) 0.000 048 637 390 136 784 977 92 × 2 = 0 + 0.000 097 274 780 273 569 955 84;
  • 18) 0.000 097 274 780 273 569 955 84 × 2 = 0 + 0.000 194 549 560 547 139 911 68;
  • 19) 0.000 194 549 560 547 139 911 68 × 2 = 0 + 0.000 389 099 121 094 279 823 36;
  • 20) 0.000 389 099 121 094 279 823 36 × 2 = 0 + 0.000 778 198 242 188 559 646 72;
  • 21) 0.000 778 198 242 188 559 646 72 × 2 = 0 + 0.001 556 396 484 377 119 293 44;
  • 22) 0.001 556 396 484 377 119 293 44 × 2 = 0 + 0.003 112 792 968 754 238 586 88;
  • 23) 0.003 112 792 968 754 238 586 88 × 2 = 0 + 0.006 225 585 937 508 477 173 76;
  • 24) 0.006 225 585 937 508 477 173 76 × 2 = 0 + 0.012 451 171 875 016 954 347 52;
  • 25) 0.012 451 171 875 016 954 347 52 × 2 = 0 + 0.024 902 343 750 033 908 695 04;
  • 26) 0.024 902 343 750 033 908 695 04 × 2 = 0 + 0.049 804 687 500 067 817 390 08;
  • 27) 0.049 804 687 500 067 817 390 08 × 2 = 0 + 0.099 609 375 000 135 634 780 16;
  • 28) 0.099 609 375 000 135 634 780 16 × 2 = 0 + 0.199 218 750 000 271 269 560 32;
  • 29) 0.199 218 750 000 271 269 560 32 × 2 = 0 + 0.398 437 500 000 542 539 120 64;
  • 30) 0.398 437 500 000 542 539 120 64 × 2 = 0 + 0.796 875 000 001 085 078 241 28;
  • 31) 0.796 875 000 001 085 078 241 28 × 2 = 1 + 0.593 750 000 002 170 156 482 56;
  • 32) 0.593 750 000 002 170 156 482 56 × 2 = 1 + 0.187 500 000 004 340 312 965 12;
  • 33) 0.187 500 000 004 340 312 965 12 × 2 = 0 + 0.375 000 000 008 680 625 930 24;
  • 34) 0.375 000 000 008 680 625 930 24 × 2 = 0 + 0.750 000 000 017 361 251 860 48;
  • 35) 0.750 000 000 017 361 251 860 48 × 2 = 1 + 0.500 000 000 034 722 503 720 96;
  • 36) 0.500 000 000 034 722 503 720 96 × 2 = 1 + 0.000 000 000 069 445 007 441 92;
  • 37) 0.000 000 000 069 445 007 441 92 × 2 = 0 + 0.000 000 000 138 890 014 883 84;
  • 38) 0.000 000 000 138 890 014 883 84 × 2 = 0 + 0.000 000 000 277 780 029 767 68;
  • 39) 0.000 000 000 277 780 029 767 68 × 2 = 0 + 0.000 000 000 555 560 059 535 36;
  • 40) 0.000 000 000 555 560 059 535 36 × 2 = 0 + 0.000 000 001 111 120 119 070 72;
  • 41) 0.000 000 001 111 120 119 070 72 × 2 = 0 + 0.000 000 002 222 240 238 141 44;
  • 42) 0.000 000 002 222 240 238 141 44 × 2 = 0 + 0.000 000 004 444 480 476 282 88;
  • 43) 0.000 000 004 444 480 476 282 88 × 2 = 0 + 0.000 000 008 888 960 952 565 76;
  • 44) 0.000 000 008 888 960 952 565 76 × 2 = 0 + 0.000 000 017 777 921 905 131 52;
  • 45) 0.000 000 017 777 921 905 131 52 × 2 = 0 + 0.000 000 035 555 843 810 263 04;
  • 46) 0.000 000 035 555 843 810 263 04 × 2 = 0 + 0.000 000 071 111 687 620 526 08;
  • 47) 0.000 000 071 111 687 620 526 08 × 2 = 0 + 0.000 000 142 223 375 241 052 16;
  • 48) 0.000 000 142 223 375 241 052 16 × 2 = 0 + 0.000 000 284 446 750 482 104 32;
  • 49) 0.000 000 284 446 750 482 104 32 × 2 = 0 + 0.000 000 568 893 500 964 208 64;
  • 50) 0.000 000 568 893 500 964 208 64 × 2 = 0 + 0.000 001 137 787 001 928 417 28;
  • 51) 0.000 001 137 787 001 928 417 28 × 2 = 0 + 0.000 002 275 574 003 856 834 56;
  • 52) 0.000 002 275 574 003 856 834 56 × 2 = 0 + 0.000 004 551 148 007 713 669 12;
  • 53) 0.000 004 551 148 007 713 669 12 × 2 = 0 + 0.000 009 102 296 015 427 338 24;
  • 54) 0.000 009 102 296 015 427 338 24 × 2 = 0 + 0.000 018 204 592 030 854 676 48;

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:


0.000 000 000 742 147 676 647 72(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 647 72(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:


0.000 000 000 742 147 676 647 72(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 -0.000 000 000 742 147 676 647 72 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