-0.000 000 000 742 214 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 214(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
-0.000 000 000 742 214(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:

|-0.000 000 000 742 214| = 0.000 000 000 742 214


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: 0.000 000 000 742 214.

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) 0.000 000 000 742 214 × 2 = 0 + 0.000 000 001 484 428;
  • 2) 0.000 000 001 484 428 × 2 = 0 + 0.000 000 002 968 856;
  • 3) 0.000 000 002 968 856 × 2 = 0 + 0.000 000 005 937 712;
  • 4) 0.000 000 005 937 712 × 2 = 0 + 0.000 000 011 875 424;
  • 5) 0.000 000 011 875 424 × 2 = 0 + 0.000 000 023 750 848;
  • 6) 0.000 000 023 750 848 × 2 = 0 + 0.000 000 047 501 696;
  • 7) 0.000 000 047 501 696 × 2 = 0 + 0.000 000 095 003 392;
  • 8) 0.000 000 095 003 392 × 2 = 0 + 0.000 000 190 006 784;
  • 9) 0.000 000 190 006 784 × 2 = 0 + 0.000 000 380 013 568;
  • 10) 0.000 000 380 013 568 × 2 = 0 + 0.000 000 760 027 136;
  • 11) 0.000 000 760 027 136 × 2 = 0 + 0.000 001 520 054 272;
  • 12) 0.000 001 520 054 272 × 2 = 0 + 0.000 003 040 108 544;
  • 13) 0.000 003 040 108 544 × 2 = 0 + 0.000 006 080 217 088;
  • 14) 0.000 006 080 217 088 × 2 = 0 + 0.000 012 160 434 176;
  • 15) 0.000 012 160 434 176 × 2 = 0 + 0.000 024 320 868 352;
  • 16) 0.000 024 320 868 352 × 2 = 0 + 0.000 048 641 736 704;
  • 17) 0.000 048 641 736 704 × 2 = 0 + 0.000 097 283 473 408;
  • 18) 0.000 097 283 473 408 × 2 = 0 + 0.000 194 566 946 816;
  • 19) 0.000 194 566 946 816 × 2 = 0 + 0.000 389 133 893 632;
  • 20) 0.000 389 133 893 632 × 2 = 0 + 0.000 778 267 787 264;
  • 21) 0.000 778 267 787 264 × 2 = 0 + 0.001 556 535 574 528;
  • 22) 0.001 556 535 574 528 × 2 = 0 + 0.003 113 071 149 056;
  • 23) 0.003 113 071 149 056 × 2 = 0 + 0.006 226 142 298 112;
  • 24) 0.006 226 142 298 112 × 2 = 0 + 0.012 452 284 596 224;
  • 25) 0.012 452 284 596 224 × 2 = 0 + 0.024 904 569 192 448;
  • 26) 0.024 904 569 192 448 × 2 = 0 + 0.049 809 138 384 896;
  • 27) 0.049 809 138 384 896 × 2 = 0 + 0.099 618 276 769 792;
  • 28) 0.099 618 276 769 792 × 2 = 0 + 0.199 236 553 539 584;
  • 29) 0.199 236 553 539 584 × 2 = 0 + 0.398 473 107 079 168;
  • 30) 0.398 473 107 079 168 × 2 = 0 + 0.796 946 214 158 336;
  • 31) 0.796 946 214 158 336 × 2 = 1 + 0.593 892 428 316 672;
  • 32) 0.593 892 428 316 672 × 2 = 1 + 0.187 784 856 633 344;
  • 33) 0.187 784 856 633 344 × 2 = 0 + 0.375 569 713 266 688;
  • 34) 0.375 569 713 266 688 × 2 = 0 + 0.751 139 426 533 376;
  • 35) 0.751 139 426 533 376 × 2 = 1 + 0.502 278 853 066 752;
  • 36) 0.502 278 853 066 752 × 2 = 1 + 0.004 557 706 133 504;
  • 37) 0.004 557 706 133 504 × 2 = 0 + 0.009 115 412 267 008;
  • 38) 0.009 115 412 267 008 × 2 = 0 + 0.018 230 824 534 016;
  • 39) 0.018 230 824 534 016 × 2 = 0 + 0.036 461 649 068 032;
  • 40) 0.036 461 649 068 032 × 2 = 0 + 0.072 923 298 136 064;
  • 41) 0.072 923 298 136 064 × 2 = 0 + 0.145 846 596 272 128;
  • 42) 0.145 846 596 272 128 × 2 = 0 + 0.291 693 192 544 256;
  • 43) 0.291 693 192 544 256 × 2 = 0 + 0.583 386 385 088 512;
  • 44) 0.583 386 385 088 512 × 2 = 1 + 0.166 772 770 177 024;
  • 45) 0.166 772 770 177 024 × 2 = 0 + 0.333 545 540 354 048;
  • 46) 0.333 545 540 354 048 × 2 = 0 + 0.667 091 080 708 096;
  • 47) 0.667 091 080 708 096 × 2 = 1 + 0.334 182 161 416 192;
  • 48) 0.334 182 161 416 192 × 2 = 0 + 0.668 364 322 832 384;
  • 49) 0.668 364 322 832 384 × 2 = 1 + 0.336 728 645 664 768;
  • 50) 0.336 728 645 664 768 × 2 = 0 + 0.673 457 291 329 536;
  • 51) 0.673 457 291 329 536 × 2 = 1 + 0.346 914 582 659 072;
  • 52) 0.346 914 582 659 072 × 2 = 0 + 0.693 829 165 318 144;
  • 53) 0.693 829 165 318 144 × 2 = 1 + 0.387 658 330 636 288;
  • 54) 0.387 658 330 636 288 × 2 = 0 + 0.775 316 661 272 576;

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:


0.000 000 000 742 214(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0010 1010 10(2)

6. Positive number before normalization:

0.000 000 000 742 214(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0010 1010 10(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:


0.000 000 000 742 214(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0010 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0010 1010 10(2) × 20 =

1.1001 1000 0000 1001 0101 010(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 1001 0101 010


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 0100 1010 1010 =

100 1100 0000 0100 1010 1010


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 0100 1010 1010


Decimal number -0.000 000 000 742 214 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0100 1010 1010

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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