-0.000 000 000 742 147 712 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 712(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
-0.000 000 000 742 147 712(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:

|-0.000 000 000 742 147 712| = 0.000 000 000 742 147 712


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: 0.000 000 000 742 147 712.

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) 0.000 000 000 742 147 712 × 2 = 0 + 0.000 000 001 484 295 424;
  • 2) 0.000 000 001 484 295 424 × 2 = 0 + 0.000 000 002 968 590 848;
  • 3) 0.000 000 002 968 590 848 × 2 = 0 + 0.000 000 005 937 181 696;
  • 4) 0.000 000 005 937 181 696 × 2 = 0 + 0.000 000 011 874 363 392;
  • 5) 0.000 000 011 874 363 392 × 2 = 0 + 0.000 000 023 748 726 784;
  • 6) 0.000 000 023 748 726 784 × 2 = 0 + 0.000 000 047 497 453 568;
  • 7) 0.000 000 047 497 453 568 × 2 = 0 + 0.000 000 094 994 907 136;
  • 8) 0.000 000 094 994 907 136 × 2 = 0 + 0.000 000 189 989 814 272;
  • 9) 0.000 000 189 989 814 272 × 2 = 0 + 0.000 000 379 979 628 544;
  • 10) 0.000 000 379 979 628 544 × 2 = 0 + 0.000 000 759 959 257 088;
  • 11) 0.000 000 759 959 257 088 × 2 = 0 + 0.000 001 519 918 514 176;
  • 12) 0.000 001 519 918 514 176 × 2 = 0 + 0.000 003 039 837 028 352;
  • 13) 0.000 003 039 837 028 352 × 2 = 0 + 0.000 006 079 674 056 704;
  • 14) 0.000 006 079 674 056 704 × 2 = 0 + 0.000 012 159 348 113 408;
  • 15) 0.000 012 159 348 113 408 × 2 = 0 + 0.000 024 318 696 226 816;
  • 16) 0.000 024 318 696 226 816 × 2 = 0 + 0.000 048 637 392 453 632;
  • 17) 0.000 048 637 392 453 632 × 2 = 0 + 0.000 097 274 784 907 264;
  • 18) 0.000 097 274 784 907 264 × 2 = 0 + 0.000 194 549 569 814 528;
  • 19) 0.000 194 549 569 814 528 × 2 = 0 + 0.000 389 099 139 629 056;
  • 20) 0.000 389 099 139 629 056 × 2 = 0 + 0.000 778 198 279 258 112;
  • 21) 0.000 778 198 279 258 112 × 2 = 0 + 0.001 556 396 558 516 224;
  • 22) 0.001 556 396 558 516 224 × 2 = 0 + 0.003 112 793 117 032 448;
  • 23) 0.003 112 793 117 032 448 × 2 = 0 + 0.006 225 586 234 064 896;
  • 24) 0.006 225 586 234 064 896 × 2 = 0 + 0.012 451 172 468 129 792;
  • 25) 0.012 451 172 468 129 792 × 2 = 0 + 0.024 902 344 936 259 584;
  • 26) 0.024 902 344 936 259 584 × 2 = 0 + 0.049 804 689 872 519 168;
  • 27) 0.049 804 689 872 519 168 × 2 = 0 + 0.099 609 379 745 038 336;
  • 28) 0.099 609 379 745 038 336 × 2 = 0 + 0.199 218 759 490 076 672;
  • 29) 0.199 218 759 490 076 672 × 2 = 0 + 0.398 437 518 980 153 344;
  • 30) 0.398 437 518 980 153 344 × 2 = 0 + 0.796 875 037 960 306 688;
  • 31) 0.796 875 037 960 306 688 × 2 = 1 + 0.593 750 075 920 613 376;
  • 32) 0.593 750 075 920 613 376 × 2 = 1 + 0.187 500 151 841 226 752;
  • 33) 0.187 500 151 841 226 752 × 2 = 0 + 0.375 000 303 682 453 504;
  • 34) 0.375 000 303 682 453 504 × 2 = 0 + 0.750 000 607 364 907 008;
  • 35) 0.750 000 607 364 907 008 × 2 = 1 + 0.500 001 214 729 814 016;
  • 36) 0.500 001 214 729 814 016 × 2 = 1 + 0.000 002 429 459 628 032;
  • 37) 0.000 002 429 459 628 032 × 2 = 0 + 0.000 004 858 919 256 064;
  • 38) 0.000 004 858 919 256 064 × 2 = 0 + 0.000 009 717 838 512 128;
  • 39) 0.000 009 717 838 512 128 × 2 = 0 + 0.000 019 435 677 024 256;
  • 40) 0.000 019 435 677 024 256 × 2 = 0 + 0.000 038 871 354 048 512;
  • 41) 0.000 038 871 354 048 512 × 2 = 0 + 0.000 077 742 708 097 024;
  • 42) 0.000 077 742 708 097 024 × 2 = 0 + 0.000 155 485 416 194 048;
  • 43) 0.000 155 485 416 194 048 × 2 = 0 + 0.000 310 970 832 388 096;
  • 44) 0.000 310 970 832 388 096 × 2 = 0 + 0.000 621 941 664 776 192;
  • 45) 0.000 621 941 664 776 192 × 2 = 0 + 0.001 243 883 329 552 384;
  • 46) 0.001 243 883 329 552 384 × 2 = 0 + 0.002 487 766 659 104 768;
  • 47) 0.002 487 766 659 104 768 × 2 = 0 + 0.004 975 533 318 209 536;
  • 48) 0.004 975 533 318 209 536 × 2 = 0 + 0.009 951 066 636 419 072;
  • 49) 0.009 951 066 636 419 072 × 2 = 0 + 0.019 902 133 272 838 144;
  • 50) 0.019 902 133 272 838 144 × 2 = 0 + 0.039 804 266 545 676 288;
  • 51) 0.039 804 266 545 676 288 × 2 = 0 + 0.079 608 533 091 352 576;
  • 52) 0.079 608 533 091 352 576 × 2 = 0 + 0.159 217 066 182 705 152;
  • 53) 0.159 217 066 182 705 152 × 2 = 0 + 0.318 434 132 365 410 304;
  • 54) 0.318 434 132 365 410 304 × 2 = 0 + 0.636 868 264 730 820 608;

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:


0.000 000 000 742 147 712(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 712(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:


0.000 000 000 742 147 712(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 -0.000 000 000 742 147 712 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 731 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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