-0.000 000 000 742 147 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 99(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 99(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 99| = 0.000 000 000 742 147 99


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 99.

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 99 × 2 = 0 + 0.000 000 001 484 295 98;
  • 2) 0.000 000 001 484 295 98 × 2 = 0 + 0.000 000 002 968 591 96;
  • 3) 0.000 000 002 968 591 96 × 2 = 0 + 0.000 000 005 937 183 92;
  • 4) 0.000 000 005 937 183 92 × 2 = 0 + 0.000 000 011 874 367 84;
  • 5) 0.000 000 011 874 367 84 × 2 = 0 + 0.000 000 023 748 735 68;
  • 6) 0.000 000 023 748 735 68 × 2 = 0 + 0.000 000 047 497 471 36;
  • 7) 0.000 000 047 497 471 36 × 2 = 0 + 0.000 000 094 994 942 72;
  • 8) 0.000 000 094 994 942 72 × 2 = 0 + 0.000 000 189 989 885 44;
  • 9) 0.000 000 189 989 885 44 × 2 = 0 + 0.000 000 379 979 770 88;
  • 10) 0.000 000 379 979 770 88 × 2 = 0 + 0.000 000 759 959 541 76;
  • 11) 0.000 000 759 959 541 76 × 2 = 0 + 0.000 001 519 919 083 52;
  • 12) 0.000 001 519 919 083 52 × 2 = 0 + 0.000 003 039 838 167 04;
  • 13) 0.000 003 039 838 167 04 × 2 = 0 + 0.000 006 079 676 334 08;
  • 14) 0.000 006 079 676 334 08 × 2 = 0 + 0.000 012 159 352 668 16;
  • 15) 0.000 012 159 352 668 16 × 2 = 0 + 0.000 024 318 705 336 32;
  • 16) 0.000 024 318 705 336 32 × 2 = 0 + 0.000 048 637 410 672 64;
  • 17) 0.000 048 637 410 672 64 × 2 = 0 + 0.000 097 274 821 345 28;
  • 18) 0.000 097 274 821 345 28 × 2 = 0 + 0.000 194 549 642 690 56;
  • 19) 0.000 194 549 642 690 56 × 2 = 0 + 0.000 389 099 285 381 12;
  • 20) 0.000 389 099 285 381 12 × 2 = 0 + 0.000 778 198 570 762 24;
  • 21) 0.000 778 198 570 762 24 × 2 = 0 + 0.001 556 397 141 524 48;
  • 22) 0.001 556 397 141 524 48 × 2 = 0 + 0.003 112 794 283 048 96;
  • 23) 0.003 112 794 283 048 96 × 2 = 0 + 0.006 225 588 566 097 92;
  • 24) 0.006 225 588 566 097 92 × 2 = 0 + 0.012 451 177 132 195 84;
  • 25) 0.012 451 177 132 195 84 × 2 = 0 + 0.024 902 354 264 391 68;
  • 26) 0.024 902 354 264 391 68 × 2 = 0 + 0.049 804 708 528 783 36;
  • 27) 0.049 804 708 528 783 36 × 2 = 0 + 0.099 609 417 057 566 72;
  • 28) 0.099 609 417 057 566 72 × 2 = 0 + 0.199 218 834 115 133 44;
  • 29) 0.199 218 834 115 133 44 × 2 = 0 + 0.398 437 668 230 266 88;
  • 30) 0.398 437 668 230 266 88 × 2 = 0 + 0.796 875 336 460 533 76;
  • 31) 0.796 875 336 460 533 76 × 2 = 1 + 0.593 750 672 921 067 52;
  • 32) 0.593 750 672 921 067 52 × 2 = 1 + 0.187 501 345 842 135 04;
  • 33) 0.187 501 345 842 135 04 × 2 = 0 + 0.375 002 691 684 270 08;
  • 34) 0.375 002 691 684 270 08 × 2 = 0 + 0.750 005 383 368 540 16;
  • 35) 0.750 005 383 368 540 16 × 2 = 1 + 0.500 010 766 737 080 32;
  • 36) 0.500 010 766 737 080 32 × 2 = 1 + 0.000 021 533 474 160 64;
  • 37) 0.000 021 533 474 160 64 × 2 = 0 + 0.000 043 066 948 321 28;
  • 38) 0.000 043 066 948 321 28 × 2 = 0 + 0.000 086 133 896 642 56;
  • 39) 0.000 086 133 896 642 56 × 2 = 0 + 0.000 172 267 793 285 12;
  • 40) 0.000 172 267 793 285 12 × 2 = 0 + 0.000 344 535 586 570 24;
  • 41) 0.000 344 535 586 570 24 × 2 = 0 + 0.000 689 071 173 140 48;
  • 42) 0.000 689 071 173 140 48 × 2 = 0 + 0.001 378 142 346 280 96;
  • 43) 0.001 378 142 346 280 96 × 2 = 0 + 0.002 756 284 692 561 92;
  • 44) 0.002 756 284 692 561 92 × 2 = 0 + 0.005 512 569 385 123 84;
  • 45) 0.005 512 569 385 123 84 × 2 = 0 + 0.011 025 138 770 247 68;
  • 46) 0.011 025 138 770 247 68 × 2 = 0 + 0.022 050 277 540 495 36;
  • 47) 0.022 050 277 540 495 36 × 2 = 0 + 0.044 100 555 080 990 72;
  • 48) 0.044 100 555 080 990 72 × 2 = 0 + 0.088 201 110 161 981 44;
  • 49) 0.088 201 110 161 981 44 × 2 = 0 + 0.176 402 220 323 962 88;
  • 50) 0.176 402 220 323 962 88 × 2 = 0 + 0.352 804 440 647 925 76;
  • 51) 0.352 804 440 647 925 76 × 2 = 0 + 0.705 608 881 295 851 52;
  • 52) 0.705 608 881 295 851 52 × 2 = 1 + 0.411 217 762 591 703 04;
  • 53) 0.411 217 762 591 703 04 × 2 = 0 + 0.822 435 525 183 406 08;
  • 54) 0.822 435 525 183 406 08 × 2 = 1 + 0.644 871 050 366 812 16;

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 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 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 99(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0001 01(2) × 20 =

1.1001 1000 0000 0000 0000 101(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 101


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 0101 =

100 1100 0000 0000 0000 0101


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 0101


Decimal number -0.000 000 000 742 147 99 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0101

Share

» Convert decimal -0.000 000 000 742 147 41 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 minutes

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