-0.000 000 000 742 147 690 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 690 1(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
-0.000 000 000 742 147 690 1(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:

|-0.000 000 000 742 147 690 1| = 0.000 000 000 742 147 690 1


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: 0.000 000 000 742 147 690 1.

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) 0.000 000 000 742 147 690 1 × 2 = 0 + 0.000 000 001 484 295 380 2;
  • 2) 0.000 000 001 484 295 380 2 × 2 = 0 + 0.000 000 002 968 590 760 4;
  • 3) 0.000 000 002 968 590 760 4 × 2 = 0 + 0.000 000 005 937 181 520 8;
  • 4) 0.000 000 005 937 181 520 8 × 2 = 0 + 0.000 000 011 874 363 041 6;
  • 5) 0.000 000 011 874 363 041 6 × 2 = 0 + 0.000 000 023 748 726 083 2;
  • 6) 0.000 000 023 748 726 083 2 × 2 = 0 + 0.000 000 047 497 452 166 4;
  • 7) 0.000 000 047 497 452 166 4 × 2 = 0 + 0.000 000 094 994 904 332 8;
  • 8) 0.000 000 094 994 904 332 8 × 2 = 0 + 0.000 000 189 989 808 665 6;
  • 9) 0.000 000 189 989 808 665 6 × 2 = 0 + 0.000 000 379 979 617 331 2;
  • 10) 0.000 000 379 979 617 331 2 × 2 = 0 + 0.000 000 759 959 234 662 4;
  • 11) 0.000 000 759 959 234 662 4 × 2 = 0 + 0.000 001 519 918 469 324 8;
  • 12) 0.000 001 519 918 469 324 8 × 2 = 0 + 0.000 003 039 836 938 649 6;
  • 13) 0.000 003 039 836 938 649 6 × 2 = 0 + 0.000 006 079 673 877 299 2;
  • 14) 0.000 006 079 673 877 299 2 × 2 = 0 + 0.000 012 159 347 754 598 4;
  • 15) 0.000 012 159 347 754 598 4 × 2 = 0 + 0.000 024 318 695 509 196 8;
  • 16) 0.000 024 318 695 509 196 8 × 2 = 0 + 0.000 048 637 391 018 393 6;
  • 17) 0.000 048 637 391 018 393 6 × 2 = 0 + 0.000 097 274 782 036 787 2;
  • 18) 0.000 097 274 782 036 787 2 × 2 = 0 + 0.000 194 549 564 073 574 4;
  • 19) 0.000 194 549 564 073 574 4 × 2 = 0 + 0.000 389 099 128 147 148 8;
  • 20) 0.000 389 099 128 147 148 8 × 2 = 0 + 0.000 778 198 256 294 297 6;
  • 21) 0.000 778 198 256 294 297 6 × 2 = 0 + 0.001 556 396 512 588 595 2;
  • 22) 0.001 556 396 512 588 595 2 × 2 = 0 + 0.003 112 793 025 177 190 4;
  • 23) 0.003 112 793 025 177 190 4 × 2 = 0 + 0.006 225 586 050 354 380 8;
  • 24) 0.006 225 586 050 354 380 8 × 2 = 0 + 0.012 451 172 100 708 761 6;
  • 25) 0.012 451 172 100 708 761 6 × 2 = 0 + 0.024 902 344 201 417 523 2;
  • 26) 0.024 902 344 201 417 523 2 × 2 = 0 + 0.049 804 688 402 835 046 4;
  • 27) 0.049 804 688 402 835 046 4 × 2 = 0 + 0.099 609 376 805 670 092 8;
  • 28) 0.099 609 376 805 670 092 8 × 2 = 0 + 0.199 218 753 611 340 185 6;
  • 29) 0.199 218 753 611 340 185 6 × 2 = 0 + 0.398 437 507 222 680 371 2;
  • 30) 0.398 437 507 222 680 371 2 × 2 = 0 + 0.796 875 014 445 360 742 4;
  • 31) 0.796 875 014 445 360 742 4 × 2 = 1 + 0.593 750 028 890 721 484 8;
  • 32) 0.593 750 028 890 721 484 8 × 2 = 1 + 0.187 500 057 781 442 969 6;
  • 33) 0.187 500 057 781 442 969 6 × 2 = 0 + 0.375 000 115 562 885 939 2;
  • 34) 0.375 000 115 562 885 939 2 × 2 = 0 + 0.750 000 231 125 771 878 4;
  • 35) 0.750 000 231 125 771 878 4 × 2 = 1 + 0.500 000 462 251 543 756 8;
  • 36) 0.500 000 462 251 543 756 8 × 2 = 1 + 0.000 000 924 503 087 513 6;
  • 37) 0.000 000 924 503 087 513 6 × 2 = 0 + 0.000 001 849 006 175 027 2;
  • 38) 0.000 001 849 006 175 027 2 × 2 = 0 + 0.000 003 698 012 350 054 4;
  • 39) 0.000 003 698 012 350 054 4 × 2 = 0 + 0.000 007 396 024 700 108 8;
  • 40) 0.000 007 396 024 700 108 8 × 2 = 0 + 0.000 014 792 049 400 217 6;
  • 41) 0.000 014 792 049 400 217 6 × 2 = 0 + 0.000 029 584 098 800 435 2;
  • 42) 0.000 029 584 098 800 435 2 × 2 = 0 + 0.000 059 168 197 600 870 4;
  • 43) 0.000 059 168 197 600 870 4 × 2 = 0 + 0.000 118 336 395 201 740 8;
  • 44) 0.000 118 336 395 201 740 8 × 2 = 0 + 0.000 236 672 790 403 481 6;
  • 45) 0.000 236 672 790 403 481 6 × 2 = 0 + 0.000 473 345 580 806 963 2;
  • 46) 0.000 473 345 580 806 963 2 × 2 = 0 + 0.000 946 691 161 613 926 4;
  • 47) 0.000 946 691 161 613 926 4 × 2 = 0 + 0.001 893 382 323 227 852 8;
  • 48) 0.001 893 382 323 227 852 8 × 2 = 0 + 0.003 786 764 646 455 705 6;
  • 49) 0.003 786 764 646 455 705 6 × 2 = 0 + 0.007 573 529 292 911 411 2;
  • 50) 0.007 573 529 292 911 411 2 × 2 = 0 + 0.015 147 058 585 822 822 4;
  • 51) 0.015 147 058 585 822 822 4 × 2 = 0 + 0.030 294 117 171 645 644 8;
  • 52) 0.030 294 117 171 645 644 8 × 2 = 0 + 0.060 588 234 343 291 289 6;
  • 53) 0.060 588 234 343 291 289 6 × 2 = 0 + 0.121 176 468 686 582 579 2;
  • 54) 0.121 176 468 686 582 579 2 × 2 = 0 + 0.242 352 937 373 165 158 4;

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:


0.000 000 000 742 147 690 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 690 1(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:


0.000 000 000 742 147 690 1(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 -0.000 000 000 742 147 690 1 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