-0.000 000 000 742 147 676 647 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 7(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 7(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 7| = 0.000 000 000 742 147 676 647 7


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 7.

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 7 × 2 = 0 + 0.000 000 001 484 295 353 295 4;
  • 2) 0.000 000 001 484 295 353 295 4 × 2 = 0 + 0.000 000 002 968 590 706 590 8;
  • 3) 0.000 000 002 968 590 706 590 8 × 2 = 0 + 0.000 000 005 937 181 413 181 6;
  • 4) 0.000 000 005 937 181 413 181 6 × 2 = 0 + 0.000 000 011 874 362 826 363 2;
  • 5) 0.000 000 011 874 362 826 363 2 × 2 = 0 + 0.000 000 023 748 725 652 726 4;
  • 6) 0.000 000 023 748 725 652 726 4 × 2 = 0 + 0.000 000 047 497 451 305 452 8;
  • 7) 0.000 000 047 497 451 305 452 8 × 2 = 0 + 0.000 000 094 994 902 610 905 6;
  • 8) 0.000 000 094 994 902 610 905 6 × 2 = 0 + 0.000 000 189 989 805 221 811 2;
  • 9) 0.000 000 189 989 805 221 811 2 × 2 = 0 + 0.000 000 379 979 610 443 622 4;
  • 10) 0.000 000 379 979 610 443 622 4 × 2 = 0 + 0.000 000 759 959 220 887 244 8;
  • 11) 0.000 000 759 959 220 887 244 8 × 2 = 0 + 0.000 001 519 918 441 774 489 6;
  • 12) 0.000 001 519 918 441 774 489 6 × 2 = 0 + 0.000 003 039 836 883 548 979 2;
  • 13) 0.000 003 039 836 883 548 979 2 × 2 = 0 + 0.000 006 079 673 767 097 958 4;
  • 14) 0.000 006 079 673 767 097 958 4 × 2 = 0 + 0.000 012 159 347 534 195 916 8;
  • 15) 0.000 012 159 347 534 195 916 8 × 2 = 0 + 0.000 024 318 695 068 391 833 6;
  • 16) 0.000 024 318 695 068 391 833 6 × 2 = 0 + 0.000 048 637 390 136 783 667 2;
  • 17) 0.000 048 637 390 136 783 667 2 × 2 = 0 + 0.000 097 274 780 273 567 334 4;
  • 18) 0.000 097 274 780 273 567 334 4 × 2 = 0 + 0.000 194 549 560 547 134 668 8;
  • 19) 0.000 194 549 560 547 134 668 8 × 2 = 0 + 0.000 389 099 121 094 269 337 6;
  • 20) 0.000 389 099 121 094 269 337 6 × 2 = 0 + 0.000 778 198 242 188 538 675 2;
  • 21) 0.000 778 198 242 188 538 675 2 × 2 = 0 + 0.001 556 396 484 377 077 350 4;
  • 22) 0.001 556 396 484 377 077 350 4 × 2 = 0 + 0.003 112 792 968 754 154 700 8;
  • 23) 0.003 112 792 968 754 154 700 8 × 2 = 0 + 0.006 225 585 937 508 309 401 6;
  • 24) 0.006 225 585 937 508 309 401 6 × 2 = 0 + 0.012 451 171 875 016 618 803 2;
  • 25) 0.012 451 171 875 016 618 803 2 × 2 = 0 + 0.024 902 343 750 033 237 606 4;
  • 26) 0.024 902 343 750 033 237 606 4 × 2 = 0 + 0.049 804 687 500 066 475 212 8;
  • 27) 0.049 804 687 500 066 475 212 8 × 2 = 0 + 0.099 609 375 000 132 950 425 6;
  • 28) 0.099 609 375 000 132 950 425 6 × 2 = 0 + 0.199 218 750 000 265 900 851 2;
  • 29) 0.199 218 750 000 265 900 851 2 × 2 = 0 + 0.398 437 500 000 531 801 702 4;
  • 30) 0.398 437 500 000 531 801 702 4 × 2 = 0 + 0.796 875 000 001 063 603 404 8;
  • 31) 0.796 875 000 001 063 603 404 8 × 2 = 1 + 0.593 750 000 002 127 206 809 6;
  • 32) 0.593 750 000 002 127 206 809 6 × 2 = 1 + 0.187 500 000 004 254 413 619 2;
  • 33) 0.187 500 000 004 254 413 619 2 × 2 = 0 + 0.375 000 000 008 508 827 238 4;
  • 34) 0.375 000 000 008 508 827 238 4 × 2 = 0 + 0.750 000 000 017 017 654 476 8;
  • 35) 0.750 000 000 017 017 654 476 8 × 2 = 1 + 0.500 000 000 034 035 308 953 6;
  • 36) 0.500 000 000 034 035 308 953 6 × 2 = 1 + 0.000 000 000 068 070 617 907 2;
  • 37) 0.000 000 000 068 070 617 907 2 × 2 = 0 + 0.000 000 000 136 141 235 814 4;
  • 38) 0.000 000 000 136 141 235 814 4 × 2 = 0 + 0.000 000 000 272 282 471 628 8;
  • 39) 0.000 000 000 272 282 471 628 8 × 2 = 0 + 0.000 000 000 544 564 943 257 6;
  • 40) 0.000 000 000 544 564 943 257 6 × 2 = 0 + 0.000 000 001 089 129 886 515 2;
  • 41) 0.000 000 001 089 129 886 515 2 × 2 = 0 + 0.000 000 002 178 259 773 030 4;
  • 42) 0.000 000 002 178 259 773 030 4 × 2 = 0 + 0.000 000 004 356 519 546 060 8;
  • 43) 0.000 000 004 356 519 546 060 8 × 2 = 0 + 0.000 000 008 713 039 092 121 6;
  • 44) 0.000 000 008 713 039 092 121 6 × 2 = 0 + 0.000 000 017 426 078 184 243 2;
  • 45) 0.000 000 017 426 078 184 243 2 × 2 = 0 + 0.000 000 034 852 156 368 486 4;
  • 46) 0.000 000 034 852 156 368 486 4 × 2 = 0 + 0.000 000 069 704 312 736 972 8;
  • 47) 0.000 000 069 704 312 736 972 8 × 2 = 0 + 0.000 000 139 408 625 473 945 6;
  • 48) 0.000 000 139 408 625 473 945 6 × 2 = 0 + 0.000 000 278 817 250 947 891 2;
  • 49) 0.000 000 278 817 250 947 891 2 × 2 = 0 + 0.000 000 557 634 501 895 782 4;
  • 50) 0.000 000 557 634 501 895 782 4 × 2 = 0 + 0.000 001 115 269 003 791 564 8;
  • 51) 0.000 001 115 269 003 791 564 8 × 2 = 0 + 0.000 002 230 538 007 583 129 6;
  • 52) 0.000 002 230 538 007 583 129 6 × 2 = 0 + 0.000 004 461 076 015 166 259 2;
  • 53) 0.000 004 461 076 015 166 259 2 × 2 = 0 + 0.000 008 922 152 030 332 518 4;
  • 54) 0.000 008 922 152 030 332 518 4 × 2 = 0 + 0.000 017 844 304 060 665 036 8;

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 7(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 7(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 7(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 7 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