5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173(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
5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173(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. First, convert to binary (in base 2) the integer part: 5.
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;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.

5(10) =

101(2)


3. Convert to binary (base 2) the fractional part: 0.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173.

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) 0.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173 × 2 = 0 + 0.520 674 439 859 473 196 320 192 008 139 098 816 477 056 733 110 183 318 346;
  • 2) 0.520 674 439 859 473 196 320 192 008 139 098 816 477 056 733 110 183 318 346 × 2 = 1 + 0.041 348 879 718 946 392 640 384 016 278 197 632 954 113 466 220 366 636 692;
  • 3) 0.041 348 879 718 946 392 640 384 016 278 197 632 954 113 466 220 366 636 692 × 2 = 0 + 0.082 697 759 437 892 785 280 768 032 556 395 265 908 226 932 440 733 273 384;
  • 4) 0.082 697 759 437 892 785 280 768 032 556 395 265 908 226 932 440 733 273 384 × 2 = 0 + 0.165 395 518 875 785 570 561 536 065 112 790 531 816 453 864 881 466 546 768;
  • 5) 0.165 395 518 875 785 570 561 536 065 112 790 531 816 453 864 881 466 546 768 × 2 = 0 + 0.330 791 037 751 571 141 123 072 130 225 581 063 632 907 729 762 933 093 536;
  • 6) 0.330 791 037 751 571 141 123 072 130 225 581 063 632 907 729 762 933 093 536 × 2 = 0 + 0.661 582 075 503 142 282 246 144 260 451 162 127 265 815 459 525 866 187 072;
  • 7) 0.661 582 075 503 142 282 246 144 260 451 162 127 265 815 459 525 866 187 072 × 2 = 1 + 0.323 164 151 006 284 564 492 288 520 902 324 254 531 630 919 051 732 374 144;
  • 8) 0.323 164 151 006 284 564 492 288 520 902 324 254 531 630 919 051 732 374 144 × 2 = 0 + 0.646 328 302 012 569 128 984 577 041 804 648 509 063 261 838 103 464 748 288;
  • 9) 0.646 328 302 012 569 128 984 577 041 804 648 509 063 261 838 103 464 748 288 × 2 = 1 + 0.292 656 604 025 138 257 969 154 083 609 297 018 126 523 676 206 929 496 576;
  • 10) 0.292 656 604 025 138 257 969 154 083 609 297 018 126 523 676 206 929 496 576 × 2 = 0 + 0.585 313 208 050 276 515 938 308 167 218 594 036 253 047 352 413 858 993 152;
  • 11) 0.585 313 208 050 276 515 938 308 167 218 594 036 253 047 352 413 858 993 152 × 2 = 1 + 0.170 626 416 100 553 031 876 616 334 437 188 072 506 094 704 827 717 986 304;
  • 12) 0.170 626 416 100 553 031 876 616 334 437 188 072 506 094 704 827 717 986 304 × 2 = 0 + 0.341 252 832 201 106 063 753 232 668 874 376 145 012 189 409 655 435 972 608;
  • 13) 0.341 252 832 201 106 063 753 232 668 874 376 145 012 189 409 655 435 972 608 × 2 = 0 + 0.682 505 664 402 212 127 506 465 337 748 752 290 024 378 819 310 871 945 216;
  • 14) 0.682 505 664 402 212 127 506 465 337 748 752 290 024 378 819 310 871 945 216 × 2 = 1 + 0.365 011 328 804 424 255 012 930 675 497 504 580 048 757 638 621 743 890 432;
  • 15) 0.365 011 328 804 424 255 012 930 675 497 504 580 048 757 638 621 743 890 432 × 2 = 0 + 0.730 022 657 608 848 510 025 861 350 995 009 160 097 515 277 243 487 780 864;
  • 16) 0.730 022 657 608 848 510 025 861 350 995 009 160 097 515 277 243 487 780 864 × 2 = 1 + 0.460 045 315 217 697 020 051 722 701 990 018 320 195 030 554 486 975 561 728;
  • 17) 0.460 045 315 217 697 020 051 722 701 990 018 320 195 030 554 486 975 561 728 × 2 = 0 + 0.920 090 630 435 394 040 103 445 403 980 036 640 390 061 108 973 951 123 456;
  • 18) 0.920 090 630 435 394 040 103 445 403 980 036 640 390 061 108 973 951 123 456 × 2 = 1 + 0.840 181 260 870 788 080 206 890 807 960 073 280 780 122 217 947 902 246 912;
  • 19) 0.840 181 260 870 788 080 206 890 807 960 073 280 780 122 217 947 902 246 912 × 2 = 1 + 0.680 362 521 741 576 160 413 781 615 920 146 561 560 244 435 895 804 493 824;
  • 20) 0.680 362 521 741 576 160 413 781 615 920 146 561 560 244 435 895 804 493 824 × 2 = 1 + 0.360 725 043 483 152 320 827 563 231 840 293 123 120 488 871 791 608 987 648;
  • 21) 0.360 725 043 483 152 320 827 563 231 840 293 123 120 488 871 791 608 987 648 × 2 = 0 + 0.721 450 086 966 304 641 655 126 463 680 586 246 240 977 743 583 217 975 296;
  • 22) 0.721 450 086 966 304 641 655 126 463 680 586 246 240 977 743 583 217 975 296 × 2 = 1 + 0.442 900 173 932 609 283 310 252 927 361 172 492 481 955 487 166 435 950 592;
  • 23) 0.442 900 173 932 609 283 310 252 927 361 172 492 481 955 487 166 435 950 592 × 2 = 0 + 0.885 800 347 865 218 566 620 505 854 722 344 984 963 910 974 332 871 901 184;
  • 24) 0.885 800 347 865 218 566 620 505 854 722 344 984 963 910 974 332 871 901 184 × 2 = 1 + 0.771 600 695 730 437 133 241 011 709 444 689 969 927 821 948 665 743 802 368;

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).


4. 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:


0.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173(10) =

0.0100 0010 1010 0101 0111 0101(2)

5. Positive number before normalization:

5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173(10) =

101.0100 0010 1010 0101 0111 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173(10) =

101.0100 0010 1010 0101 0111 0101(2) =

101.0100 0010 1010 0101 0111 0101(2) × 20 =

1.0101 0000 1010 1001 0101 1101 01(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.0101 0000 1010 1001 0101 1101 01


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

2 + 2(8-1) - 1 =

(2 + 127)(10) =

129(10)


9. 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;
  • 129 ÷ 2 = 64 + 1;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

129(10) =

1000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 010 1000 0101 0100 1010 1110 101 =

010 1000 0101 0100 1010 1110


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
1000 0001

Mantissa (23 bits) =
010 1000 0101 0100 1010 1110


Decimal number 5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 173 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0001 - 010 1000 0101 0100 1010 1110

Share

» Convert decimal 5.260 337 219 929 736 598 160 096 004 069 549 408 238 528 366 555 091 659 101 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 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. Gemini Notebook YouTube Video of 7.50 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