-0.000 000 000 742 147 744 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 744(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
-0.000 000 000 742 147 744(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:

|-0.000 000 000 742 147 744| = 0.000 000 000 742 147 744


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: 0.000 000 000 742 147 744.

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) 0.000 000 000 742 147 744 × 2 = 0 + 0.000 000 001 484 295 488;
  • 2) 0.000 000 001 484 295 488 × 2 = 0 + 0.000 000 002 968 590 976;
  • 3) 0.000 000 002 968 590 976 × 2 = 0 + 0.000 000 005 937 181 952;
  • 4) 0.000 000 005 937 181 952 × 2 = 0 + 0.000 000 011 874 363 904;
  • 5) 0.000 000 011 874 363 904 × 2 = 0 + 0.000 000 023 748 727 808;
  • 6) 0.000 000 023 748 727 808 × 2 = 0 + 0.000 000 047 497 455 616;
  • 7) 0.000 000 047 497 455 616 × 2 = 0 + 0.000 000 094 994 911 232;
  • 8) 0.000 000 094 994 911 232 × 2 = 0 + 0.000 000 189 989 822 464;
  • 9) 0.000 000 189 989 822 464 × 2 = 0 + 0.000 000 379 979 644 928;
  • 10) 0.000 000 379 979 644 928 × 2 = 0 + 0.000 000 759 959 289 856;
  • 11) 0.000 000 759 959 289 856 × 2 = 0 + 0.000 001 519 918 579 712;
  • 12) 0.000 001 519 918 579 712 × 2 = 0 + 0.000 003 039 837 159 424;
  • 13) 0.000 003 039 837 159 424 × 2 = 0 + 0.000 006 079 674 318 848;
  • 14) 0.000 006 079 674 318 848 × 2 = 0 + 0.000 012 159 348 637 696;
  • 15) 0.000 012 159 348 637 696 × 2 = 0 + 0.000 024 318 697 275 392;
  • 16) 0.000 024 318 697 275 392 × 2 = 0 + 0.000 048 637 394 550 784;
  • 17) 0.000 048 637 394 550 784 × 2 = 0 + 0.000 097 274 789 101 568;
  • 18) 0.000 097 274 789 101 568 × 2 = 0 + 0.000 194 549 578 203 136;
  • 19) 0.000 194 549 578 203 136 × 2 = 0 + 0.000 389 099 156 406 272;
  • 20) 0.000 389 099 156 406 272 × 2 = 0 + 0.000 778 198 312 812 544;
  • 21) 0.000 778 198 312 812 544 × 2 = 0 + 0.001 556 396 625 625 088;
  • 22) 0.001 556 396 625 625 088 × 2 = 0 + 0.003 112 793 251 250 176;
  • 23) 0.003 112 793 251 250 176 × 2 = 0 + 0.006 225 586 502 500 352;
  • 24) 0.006 225 586 502 500 352 × 2 = 0 + 0.012 451 173 005 000 704;
  • 25) 0.012 451 173 005 000 704 × 2 = 0 + 0.024 902 346 010 001 408;
  • 26) 0.024 902 346 010 001 408 × 2 = 0 + 0.049 804 692 020 002 816;
  • 27) 0.049 804 692 020 002 816 × 2 = 0 + 0.099 609 384 040 005 632;
  • 28) 0.099 609 384 040 005 632 × 2 = 0 + 0.199 218 768 080 011 264;
  • 29) 0.199 218 768 080 011 264 × 2 = 0 + 0.398 437 536 160 022 528;
  • 30) 0.398 437 536 160 022 528 × 2 = 0 + 0.796 875 072 320 045 056;
  • 31) 0.796 875 072 320 045 056 × 2 = 1 + 0.593 750 144 640 090 112;
  • 32) 0.593 750 144 640 090 112 × 2 = 1 + 0.187 500 289 280 180 224;
  • 33) 0.187 500 289 280 180 224 × 2 = 0 + 0.375 000 578 560 360 448;
  • 34) 0.375 000 578 560 360 448 × 2 = 0 + 0.750 001 157 120 720 896;
  • 35) 0.750 001 157 120 720 896 × 2 = 1 + 0.500 002 314 241 441 792;
  • 36) 0.500 002 314 241 441 792 × 2 = 1 + 0.000 004 628 482 883 584;
  • 37) 0.000 004 628 482 883 584 × 2 = 0 + 0.000 009 256 965 767 168;
  • 38) 0.000 009 256 965 767 168 × 2 = 0 + 0.000 018 513 931 534 336;
  • 39) 0.000 018 513 931 534 336 × 2 = 0 + 0.000 037 027 863 068 672;
  • 40) 0.000 037 027 863 068 672 × 2 = 0 + 0.000 074 055 726 137 344;
  • 41) 0.000 074 055 726 137 344 × 2 = 0 + 0.000 148 111 452 274 688;
  • 42) 0.000 148 111 452 274 688 × 2 = 0 + 0.000 296 222 904 549 376;
  • 43) 0.000 296 222 904 549 376 × 2 = 0 + 0.000 592 445 809 098 752;
  • 44) 0.000 592 445 809 098 752 × 2 = 0 + 0.001 184 891 618 197 504;
  • 45) 0.001 184 891 618 197 504 × 2 = 0 + 0.002 369 783 236 395 008;
  • 46) 0.002 369 783 236 395 008 × 2 = 0 + 0.004 739 566 472 790 016;
  • 47) 0.004 739 566 472 790 016 × 2 = 0 + 0.009 479 132 945 580 032;
  • 48) 0.009 479 132 945 580 032 × 2 = 0 + 0.018 958 265 891 160 064;
  • 49) 0.018 958 265 891 160 064 × 2 = 0 + 0.037 916 531 782 320 128;
  • 50) 0.037 916 531 782 320 128 × 2 = 0 + 0.075 833 063 564 640 256;
  • 51) 0.075 833 063 564 640 256 × 2 = 0 + 0.151 666 127 129 280 512;
  • 52) 0.151 666 127 129 280 512 × 2 = 0 + 0.303 332 254 258 561 024;
  • 53) 0.303 332 254 258 561 024 × 2 = 0 + 0.606 664 508 517 122 048;
  • 54) 0.606 664 508 517 122 048 × 2 = 1 + 0.213 329 017 034 244 096;

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:


0.000 000 000 742 147 744(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 744(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 744(10) =

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

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

1.1001 1000 0000 0000 0000 001(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 001


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 0001 =

100 1100 0000 0000 0000 0001


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 0001


Decimal number -0.000 000 000 742 147 744 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0001

Share

» Convert decimal -0.000 000 000 742 147 796 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